How-To: Various Fields of Math

(I was told that putting how-to tutorials for math here, instead of in a blog, was acceptable - though I may have to fix one or two other small things if I missed anything.)

How-To: Specific Areas of Algebra, Calculus, or Statistics
(and maybe more in the future, who can say?)

Greetings. My name is Eclipse, and I'd like to write math tutorials for you all. However, I'm not sure what areas that people here struggle with the most in math, I will be writing these tutorials upon request.

There's a list of specific areas of math further down in this post (in the spoiler tag) that I can write tutorials on. If you have a request, you can post in this topic, as a comment in this blog post, or via private message. (If you have specific questions on an area rather than broader ones, send me a private message instead.)

I plan to post these tutorials on a fairly regular basis (planned for either every week, or every 2 weeks, dependent upon my spare time), detailing a certain area of a subject, how it is done, how it is used, and so on - I will include as much relevant information as I can.

If I don't get any new requests, I'll just pick one and write on that, but I will prioritise all requests first. Also, if you feel I haven't explained something fully, or missed a few details, let me know and I will include what you think is missing.

Do note, the following is not an exhaustive list. If you have questions about anything in those subjects that I haven't covered, that is also fair game.

The subjects I can offer help on include, but are not limited to, the following (they advance in difficulty the more you head down each list):

Spoiler

Algebra
Functions of a variable (expressed graphically)
Linear functions, slopes of functions, and rates of change
Piecewise functions
Intersection points for functions (e.g. x- and y-intercepts)
Exponents (positive, negative, zero, and fractional)
Unit conversions (e.g. miles per hour to feet per second)
Logarithms and orders of magnitude
Linear and exponential growth and decay, including compound interest (Post #7, requested by @Azure)
Rules of logarithms and exponential equations (I plan to do this one next, maybe.)
Polynomials, and how to factor them (Post #4, requested by @Azure)
Quadratic functions, expressed graphically (I plan to do this one next, maybe.)
Solving quadratic equations with factoring, and the quadratic formula (Post #5, requested by @Azure)
Piecewise (aka composite) functions (Post #6, requested by @Azure)
Inverse functions
(Remember, this is not an exhaustive list. Let me know if you have a request, including one not found here.)

Calculus
Definition of derivatives
Rules of derivation, plus a few common derivatives (may be included with the above)
How to use derivatives to sketch a function (by hand)
Integration (anti-derivatives) and basic techniques, plus a few common integrals (Post #8, requested by @Megarai111)
U-substitution (for integration)
Integration by parts
Trigonometric integrals
L'Hopital's Rule
(Remember, this is not an exhaustive list. Let me know if you have a request, including one not found here.)

Statistics
Probability
Types of data (nominal, ordinal, interval)
Measures of center (mean, median, and mode), and measures of variability (range, variance, standard deviation)
Population parameters and sample statistics
Percentiles and quartiles
Independent vs dependent events
Joint, marginal, and conditional probabilities
Discrete vs continuous random variables
Probability distributions and cumulative distributions (the ones done with small tables)
Binomial distributions
Poisson distributions
Continuous distributions and probability density functions (e.g. uniform)
Normal distributions, including the standard normal distribution
Sampling distributions
Hypothesis testing (introduction)
Z values vs T values
Hypothesis testing of a mean
Hypothesis testing of a proportion
Hypothesis testing of a variance
Goodness-of-fit tests & contingency table tests
Linear regression
Coefficients of correlation and determination (may include with the above)
(Remember, this is not an exhaustive list. Let me know if you have a request, including one not found here.)

:::: :::: ::::

If you have a specific request for any 'how-to' on these areas, leave your comment, and I shall get right on it! I do hope that my math services will be of use to you.

 

It shall be done. As you appear to have a preference for the factoring one, I'll write that one first, then work on the other two.

 

Polynomials, and how to factor them
Requested by @[member="AzureEdge"]

Introduction

Polynomials and quadratic equations! One of the many banes of algebra class, and one that comes a lot, both in it and future classes as well. They're used to find solutions for graphs where they touch the x-axis, and also serve as an aid in helping graphs be drawn as well. And they're a real pain to learn, but thankfully how they work isn't too esoteric. So let me get to explaining.

A quadratic equation is an equation in the format "ax2 + bx + c", where a doesn't equal 0. (If a did equal 0, this would just be a straight line.) If graphed out, equations like these look like parabolas. Usually, a, b, and c will all be integers (that is, numbers with no decimals), though sometimes they will be rational numbers (fractions where the numerator and denominator are both integers). You will rarely (if ever) deal with a, b, or c being an irrational number, but sometimes the solutions might be irrational. (I'll deal with that in another section.)

Though, before I get into factoring, let's start with the opposite - multiplying.


Multiplying Binomials

Do you remember how to multiply 2-digit or 3-digit numbers together? Like 24 * 76? You start with multiplying 24 by 6, then you multiply 24 by 70 (24 times 7, then moved 1 place to the left), and then add the results up:

Code:

    24
  * 76
______
   144
  168
______
  1824

The same kind of principle applies when you multiply things like "x - 4" and "x + 6" together. (Those kind of expressions are called 'binomials', because they're the sum of 2 terms.) To illustrate, I'll multiply those two terms together; first I'm going to multiply "x - 4" by x, and then again by positive 6, then add those results together:

Code:

x - 4
x + 6
_____
x^2 - 4x
6x - 24
_____
x^2 + 2x - 24

And since it's multiplying two numbers, I can do this in either order, and get the same result:

Code:

x + 6
x - 4
_____
x^2 + 6x
-4x - 24
_____
x^2 + 2x - 24

Try to be careful with the signs, though! Remember, I was multiplying something by negative 4 there, so the signs I get have to match. If both bits are positive, or both bits are negative, their product is positive - they're negative if they have different signs.

You can actually try this method with normal multiplication too. As some practice, try multiplying 24 * 76 like I did above, but rewrite 24 as 20 + 4, and rewrite 76 as 70 + 6. Then try multiplying them in the way that I multiplied the binomials.

It should look something like this:

Spoiler

Code:

20 + 4
70 + 6
______
1400 + 280
120 + 24
______
1400 + 280 + 120 + 24

Combining terms together:
1400 + 400 + 24
1800 + 24
1824
(Or you can just use a calculator; either works.)

It works, right? (You can try this with smaller numbers too, and in any combination you want. Even something as 8 times 7, rewritten as 5 + 3 times 4 + 3, will work this way.)


Multiplying w/FOIL Method

Another way of doing this is called the FOIL method, which is a slightly different way of doing what we just did. FOIL is an acronym for "First, Outer, Inner, Last", referencing the order in which you multiply things before adding them together. Let's use our example from before:

(x + 6)(x - 4)

FOIL method has you multiply the terms in this order:
First: The two terms that are first in each parentheses.
Outer: The two terms that are on the outside.
Inner: The two terms that are on the inside.
Last: The two terms that are last in each parentheses.

We should get the same answer as we did before, but let's try it.
First: The first term in each parentheses is an x. x times x is x2, so that's the first thing.
Outer: On the far left there's an x, and on the far right there's a -4 (don't forget the sign!). x times -4 is -4x.
Inner: These are the two on the inside, next to each other. There's a 6, and there's an x; multiply those and you get 6x.
Last: The last term in the first group is a 6, and in the other it's a -4; 6 times -4 is -24.

Now we add them all up. We have x2, -4x, 6x, and -24. If you sum those up, we get x2 + 2x - 24. Same answer as before, right? It's just a different way of doing the same thing. Use whichever way is more comfortable to you to write out and process, as you will arrive at the same result.

Do note, FOIL works best when both terms are binomials, as we just did. If we have any trinomials, and/or you have to multiply more than 2 things together, then either use the first method, or modify the FOIL method you use to account for all the terms. (The first method is preferred, though, since there might be a lot of terms to deal with.)

Oh, and here are a few handy identities for multiplying binomials together:

(a + 2 = a2 + 2ab + b2
(a - 2 = a2 - 2ab + b2
(a + * (a - = a2 - b2

where a and b are both positive terms. The first two are called 'perfect squares'; the third one is called a 'difference of squares'.

You can actually try this out for yourself with various combinations; for example, try finding out what (x + 6)2, (x - 6)2, and (x + 6)(x - 6) will become when you use the first method or the FOIL method.

The spoiler below has the answers, but make sure to do the work yourself first; don't peek!

Spoiler

You should get x2 + 12x + 36, x2 - 12x + 36, and x2 - 36, respectively.

Basic Polynomial Factoring

Now that we have multiplying down, let's get to factoring - the opposite! ...and what you actually wanted to see in the first place. I'm going to start with a sample problem:

Factor x2 + 10x - 24.

Notice it's in the form ax2 + bx + c. I'm going to start off with examples where a is 1, and show a few tricks you can use to start factoring.
The first thing you want to do is identify what a, b, and c are. If a is 1, it's a bit simpler. If it's not 1...I'll get to that in a bit. So, looking at this, we have:

a = 1
b = 10
c = -24 (don't forget to mark the signs!)

When we factor this, we're going to factor out all the binomials as much as we can, like this: (x + _) * (x + _). For those blanks, there are 2 important rules to follow, so remember these:

:: The two numbers have to multiply together to equal c.
:: The two numbers have to add together to equal b.

Let's call those blanks s and t, for clarity, so we're going to have (x + s) * (x + t). Some courses call this the Blanks Method, so you may know it under that name.
A lot of factoring involves trial and error, and finding out what works and what doesn't. However, there are a few tricks you can use to help you:

:: If c is positive, then s and t will be either both positive or both negative. (A negative times a negative is a positive.)
:: If b and c are both positive, then s and t will both be positive. (Remember, s and t add up to b, so if c is positive, they have the same sign, and if b is positive, then they're both positive.)
:: If b is positive, but c is negative, then s and t will both be negative. (Same as the previous.)

:: If c is negative, then s and t will be opposite signs (one positive, one negative).
:: If b is positive, and c is negative, then the larger out of s and t will be positive. (When I say 'larger' in this case, it means 'furthest away from zero'. For example, -10 would be 'larger' than 2 in this example.)
:: If b is negative, and c is also negative, then the larger out of s and t will be negative. (Same as the previous. If you want, you can just say 'furthest away from zero' instead of 'larger' if it confuses you.)

:: If b is 0 and c is negative, then you have a difference of squares, so you can factor it that way. (For example, if you have x2 - 49, then that is broken up into x + 7 and x - 7.)
:: If b is 0 and c is positive, then you cannot factor it - it is irreducible. (There are a few other cases in which you have something irreducible and b isn't 0, but I will not cover those yet.)

Most of the time, s and t will be integers, so you won't have to think too hard about it. If your integers don't work, you might have to use the quadratic formula (which will be discussed another time). The 'trial and error' process I mentioned earlier will be finding the factors of c - that is, finding numbers that, when multiplied together, equal c.

You are going to want to start closer to the 'middle' and work your way outward. I'll explain. First, the factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24. If you look at the middle, that's 4 and 6. (You can usually just eyeball it, but writing out the factors also helps.)

Remember, we are dealing with -24 here, so when picking s and t, one's positive and one's negative - and since b is positive (10), the furthest away from zero is positive. I'll just say that s is the positive one and t is negative for now.

Do 6 and 4 work? It might, since 6 and 4 do add up to 10, but t is going to be negative, and 6 and -4 add up to 2. That's not going to work. Now, if c were positive 24 (i.e. x2 + 10x + 24), this would work, since s and t would both be positive. So this won't work.

Do 8 and 3 work? 8 and -3 add up to 5, and we're trying to aim for a target number of 10, so that won't work either. Time to go further out again.

Do 12 and 2 work? Let's see, 12 and -2 add up to 10...so that means this DOES work! 12 and -2 multiply together to equal -24 (which is c), and they add up to 10 (which is . We have succeeded in factoring! We write out the answer like so:

x2 + 10x - 24 = (x + 12)(x - 2)

:: When c is negative, and you write out the factors for c, you want to pick 2 factors (ignoring the signs for a moment) whose DIFFERENCE equals b. Remember, above, the difference between 12 and 2 was 10. We knew that 12 (the larger one) was the positive one because b was positive. If b was negative instead, the 12 would be the negative one.
:: When c is positive, and you write out the factors for c, you want to pick 2 factors (ignoring the signs for a moment) whose SUM equals b. In the above example, if c were 24 instead of -24, 6 and 4 would have worked, since the sum of 6 and 4 is 10. Then they'd be either both positive or both negative, and since b was positive, they were both positive.

Factoring is not as intimidating as it looks, but it does require a bit of practice. Here, I'll give you some examples to work with. Remember, figure out what a, b, and c are before you begin, and use the rules above. To make it easy, a is going to be 1 in all these cases, and I'll give you all the factors for c.

:: x2 + 6x + 9 (The factors of 9 are 1, 3, and 9.)
:: x2 - 6x - 16 (The factors of 16 are 1, 2, 4, 8, and 16.)
:: x2 + 3x - 40 (The factors of 40 are 1, 2, 4, 5, 8, 10, 20, and 40.)
:: x2 - 17x + 60 (The factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60.)

To check your work open the spoiler below, but try not to cheat!

Spoiler

:: x2 + 6x + 9 = (x + 3)(x + 3)
:: x2 - 6x - 16 = (x - 8)(x + 2)
:: x2 + 3x - 40 = (x + 8)(x - 5)
:: x2 - 17x + 60 = (x - 12)(x - 5)

Polynomial Factoring when a is not equal to 1

But what happens when a isn't 1? You'll be relieved to know that the method is basically the same as above, just with a few extra steps. The same principles apply. Here's the example I'm going to use:

Factor 3x2 + 19x + 20.

:: First, make sure that you can't factor a out of the entire equation, and end up getting a quadratic where a is 1 again. For example, if you have 4x2 + 16x + 12, you can actually factor a 4 out of everything, so you end up with 4 * (x2 + 4x + 3), which can be factored easier.
:: Next, write out the factors of a. a will usually not be a very large number so the list will be short.
:: Unlike with c, it's actually recommended that you start at the more extreme ends and work inward. If a is prime you won't have this problem, but if it's not you may have to resort to trial and error.

Just so I don't lose you somewhere, I'll show you what I'll do next. The factors of 3 are just 1 and 3, so instead of just x and x leading the binomials, there will be a 3x and a 1x. So I have this:

(3x + s)(x + t)

a = 3
b = 19
c = 20

Like before, s and t are going to multiply to equal c. However, with b it's a little different. While they're still going to add up to b - somewhat - based on the coefficients of the x's, s and t will be counted multiple times. Some courses call this the Modified Blanks Method, since while it works the same way, some numbers are counted multiple times.

Remember how we did the FOIL method? Well, think of the Outer and Inner groups as two pairs, in which you'll be counting them multiple times. In this case, the 3x is paired with t (the Outer pair), and the x is paired with s (the Inner pair). The s will only be counted once, but the t will be counted 3 times (since it's paired with 3x). So, when you're trying to find b, in this case it's s + 3t = 19.

Also, the rule about s and t and which is further than zero doesn't directly carry over. Instead, these 'suggestions' are substituted in their place:

:: If b is positive, and c is negative, then the one out of s and t that is counted more often is usually (but not always) the positive one.
:: If b is negative, and c is also negative, then the one out of s and t that is counted more often is usually (but not always) the negative one.

These are NOT hard-and-fast rules, and do not always apply all the time. For example, you could count a 2 three times, but you could also count a -10 only once and b would be negative. That is why I called them 'suggestions'. The exceptions to these suggestions usually surface the closer b is to 0.

I'm probably going too fast, so I'm going to do this example, and a few others, to illustrate.

First, the factors of 20: 1, 2, 4, 5, 10, and 20. Like before I'm going to start in the middle and work outward. Thankfully, c and b are both positive, so all the numbers will be positive.

Will 4 and 5 work? Let's see. I'm going to have to count one of them 3 times, and get the two to add to 19. What if I count 4 three times? (4*3) + 5 = 17, so I'm close, but that's not it. What if I count 5 three times? (5*3) + 4 = 19. That works! Remember, in this case, the 3x is in the Outer pair, so that means t is counted 3 times, so t is the 5. Thus we have:

3x2 + 19x + 20 = (3x + 4)(x + 5)

You can double-check any factoring you do using the FOIL method, just to make sure your answer works. Remember, you only count multiple times for adding up to b. How you factor and choose numbers for c is unaffected.

This is undoubtedly the hardest part of factoring (when a isn't 1), so I'm going to do a few more examples for you, so you can get the hang of how to do this in multiple situations.

I just did a problem where b and c were both positive. The following examples I'm going to do will have one or both of b or c be negative. I'll condense these in spoiler tags, but feel free to peek at these!

An example where b is positive and c is negative:

Spoiler

Factor 4x2 + 17x - 15.

First things first, let's see if we can factor anything out of a. No, we can't (4 is 2*2, and neither 17 nor 15 are even). So now let's identify a, b, and c, as well as the factors of a and c:

a = 4
b = 17
c = -15

Factors of a: 1, 2, 4
Factors of c: 1, 3, 5, 15

Notice that a isn't prime, so there is a bit of guesswork involved: we could either have (4x + s)(x + t), or we could have (2x + s)(2x + t). However, you can eyeball which one is correct right away, actually. Notice that b is 17. If we went with the second choice, that means we each count s and t twice. If we did that, no matter which numbers we picked, we'd get an even number. And 17 isn't even, so that isn't going to work. So we're going to go with the first choice:

(4x + s)(x + t)

Remember, c is negative, so s and t will be opposite signs. Since b is positive, t is most likely positive too, as we're counting it 4 times. As always we're going to start in the middle and go outward.

Let's start with 3 and 5. There are actually four different combinations we could do with these two numbers (s could be 3, -3, 5, or -5, and t is whatever makes it multiply to -15), but let's start easy. Let's start with t being 5, and s being -3. t is counted 4 times, so (4*5) + (-3) = 17. That actually works.

4x2 + 17x - 15 = (4x - 3)(x + 5)

This doesn't always happen, as you will usually need to try all of the various combinations to see what works. As you can imagine, this is a lot more tedious than if a is 1, since the order of s and t actually does matter (whereas if a is 1, they can usually be interchanged and the order doesn't matter).

An example where b is negative and c is positive:

Spoiler

Factor 5x2 - 19x + 18.

We can't factor the 5 out, so it's time to get to work. Just like before, I'm going to identify each one, then factor a and c:

a = 5
b = -19
c = 18

Factors of a: 1, 5
Factors of c: 1, 2, 3, 6, 9, 18

Since c is positive, yet b is negative, it means that both s and t are going to be negative as well. And, since a is a prime number, there's only one way the two can be paired - with a 5x and a 1x - so our setup becomes this:

(5x + s)(x + t)

t is going to be counted 5 times, and remember, s and t are both going to be negative.

Our first pair of factors is 6 and 3. If we count the 3 five times, we get (-3*5) + (-6) = -21. That's not b, and if we switch them around we get -33 instead, so 6 and 3 is not going to be our pair.

Next we have 9 and 2. I'm not going to count the 9 five times as it's pretty obvious it will overshoot b (-47, to those curious), so let's count the 2 five times and see what we get: (-2*5) + (-9) = -19. We have found our pair, so our answer becomes:

5x2 - 19x + 18 = (5x - 9)(x - 2)

Remember, the one that was counted multiple times was t, not s, so that is why the -2 is in the second parentheses. (Having s counted multiple times is uncommon, and only happens when a is nonprime and you don't use 1 as a factor of a.)

An example where b is negative and c is also negative:

Spoiler

Factor 6x2 - 7x - 20.

Generally, problems where a isn't 1, and where b and c are both negative, are the hardest out of the factoring problems. And, once again, we can't factor anything out of a (while 6 and 20 are both divisible by 2, 7 isn't), so...well, you know the drill by now.

a = 6
b = -7
c = -20

Factors of a: 1, 2, 3, 6
Factors of c: 1, 2, 4, 5, 10, 20

We're going to start with the ends and work in with a, and start with the middle and work out with c. Since there are two combinations we can use for a's factors, we may have to try all the possibilities. (Remember, if you see a combination that adds up to b, you can stop.)

So, first:
(6x + s)(x + t)

As there are a lot of possibilities I'm not going to use as many words this time, but I will show my work. Remember, s and t are going to have opposite signs, and the goal is -7.
4 & 5:
(6x - 4)(x + 5) : 6*5 - 4 = 26
(6x + 4)(x - 5) : 6*-5 + 4 = -26
(6x - 5)(x + 4) : 6*4 - 5 = 19
(6x + 5)(x - 4) : 6*-4 + 5 = -19

You'll notice that if I switch the signs on the two pairs, the answers are opposites. That's not a mistake at all, so if you find an answer that simply has the opposite sign we want, just switch the signs. (This really only comes into play when c is negative, since s and t will have different signs.)
Oh well, none of those are -7, so let's move on.

2 & 10:
(6x - 2)(x + 10) : 6*10 - 2 = 58
(6x + 2)(x - 10) : 6*-10 + 2 = -58
(6x - 10)(x + 2) : 6*2 - 10 = 2
(6x + 10)(x - 2) : 6*-2 + 10 = -2

1 & 20:
(6x - 1)(x + 20) : 6*20 - 1 = 119 (whoa)
(6x + 1)(x - 20) : 6*-20 + 1 = -119
(6x - 20)(x + 1) : 6*1 - 20 = -14
(6x + 20)(x - 1) : 6*-1 + 20 = 14

You can probably eyeball 1 and 20 and deduce it won't apply here, since the numbers will be too different to make them sum to -7; regardless I included them for completeness. As none of the 6x/1x combos work, it's time to move on to 3x/2x instead.

As a note, if b is somewhat close to 0, you might actually want to work inward on the a factors too. I prefer to work outwards in on a, but there's nothing wrong with working inwards out either. All you need to do is just keep trying until you find a combination that works, then you can stop.

(3x + s)(2x + t)

4 & 5:
(3x - 4)(2x + 5) : 3*5 - 2*4 = 7
(3x + 4)(2x - 5) : 3*-5 + 2*4 = -7 (Found it!)
(3x - 5)(2x + 4) : 3*4 - 2*5 = 2
(3x + 5)(2x - 4) : 3*-4 + 2*5 = -2

See, if you start off a problem with a different method, you can get the answer a lot sooner - as I've just demonstrated. I don't even need to do all the rest, because we have a -7 right there:

6x2 - 7x - 20 = (3x + 4)(2x - 5)

Some problems can be really short or really long, depending on how you start them off. In my case I chose a longer way. But maybe you already figured it out before you got here because you chose the short way. Either way you'll arrive at the same answer - the only difference here is time.

Here are some examples for you to work on, to get into practice. I'll include the answers below...but try not to peek. (Oh, and you won't be able to factor a number out of a in any of these, so you can skip that step.)

:: 4x2 + 12x + 5
:: 5x2 + 6x - 8
:: 3x2 - 11x + 6
:: 6x2 - 5x - 6

Spoiler

:: 4x2 + 12x + 5 = (2x + 5)(2x + 1)
:: 5x2 + 6x - 8 = (5x - 4)(x + 2)
:: 3x2 - 11x + 6 = (3x - 2)(x - 3)
:: 6x2 - 5x - 6 = (3x + 2)(2x - 3)

Polynomial Long Division

And now I'm going to conclude this tutorial with a subject very often hated - polynomial long division. As if normal long division wasn't painful enough, now we have to do it with x's involved!? Thankfully, it is not as big of a monster as you might think, but when long dividing, you need to keep one very important thing in mind: do NOT confuse positive and negative signs.

Division is still the reverse of multiplication, even with polynomials, and is still approached much the same way. The only real difference is how you will start subtracting terms from the polynomial at large. And, most of the time, it's actually approached the same way as factoring! To illustrate, I'll actually give you an example.

Factor x3 + 5x2 - 12x - 36. Hint: One of the factors is x + 2.

First of all, if you're given a polynomial in which the highest degree term (that is, the highest exponent on an x in the problem) is more than 2, they will give you some kind of hint as to what one of the factors is, or put it so that you can factor out an x. (For example, x3 + 10x2 + 25x can have an x factored out of it, so it becomes x(x2 + 10x + 25), the latter of which you can factor easily.)

For the setup, start by writing it out like you would any long division problem, like so:

Code:

         _________________________
   x + 2 | x^3 + 5x^2 - 12x - 36

Look at the leading x term of your divisor (it's x + 2, so just x), as well as the leading x term of your dividend that you're going to deal with (so the first one is x3). Divide the latter by the former: x3 / x = x2. That's going to be the first term of your quotient:

Code:

          x^2
        _________________________
  x + 2 | x^3 + 5x^2 - 12x - 36

Now, to figure out what goes below, multiply that quotient term you just put by the divisor, so here, x2 * (x + 2) = x3 + 2x2. Write that below:

Code:

          x^2
         _________________________
  x + 2 | x^3 + 5x^2 - 12x - 36
          x^3 + 2x^2

This part is crucial, don't forget this. In fact, if you mess up here, there's a chance your whole answer will be wrong, so remember this.
Remember, with division you are going to be subtracting down. Thus, the signs are vitally important here! With this bit they're all positive so it's simple, but later on in this problem there'll be one different. So, I subtract all that, and bring the next term down:

Code:

          x^2
         _________________________
  x + 2 | x^3 + 5x^2 - 12x - 36
          x^3 + 2x^2
          __________
                3x^2 - 12x

Notice the x3 term has gone to 0. You should do that each time you go to the next term: when you divide that bit and subtract, your leading term should go to 0, leaving you with the next term.
Here our leading term is 3x2. So, what do we put up top this time? As before, we divide that bit by the leading term in the divisor (the divisor is x + 2, so the leading term is just x) to get: 3x2/x = 3x.

Code:

          x^2 + 3x
         _________________________
  x + 2 | x^3 + 5x^2 - 12x - 36
          x^3 + 2x^2
          __________
                3x^2 - 12x

Multiply that 3x by the divisor (x + 2) and prepare to subtract...

Code:

          x^2 + 3x
         _________________________
  x + 2 | x^3 + 5x^2 - 12x - 36
          x^3 + 2x^2
          __________
                3x^2 - 12x
                3x^2 + 6x

Now, this is where the problem with long division lies, and where a lot of people will be mixed up. Remember, you're subtracting - so the 3x2 bit will go to 0, but what about the other bit? Subtracting a positive number is the same as adding a negative number (as subtraction is the opposite of addition) so you won't end up with a -6x, but rather, you will end up with a -18x, since -12 minus 6 is -18.

Code:

          x^2 + 3x
         _________________________
  x + 2 | x^3 + 5x^2 - 12x - 36
          x^3 + 2x^2
          __________
                3x^2 - 12x
                3x^2 + 6x
                __________
                     - 18x - 36

Getting the signs right, and not mixing up addition and subtraction, is the hardest part of polynomial long division. I always recommend practicing things to help you get better at them, though with long division I implore you to take extra special care.

Now, we have a -18x. What would the last part be? If you think 18, you're almost there but you're a little off. Remember, -18x is negative, so when dividing by x there, you'd end up with -18. And when you multiply that through...

Code:

          x^2 + 3x   - 18
         _________________________
  x + 2 | x^3 + 5x^2 - 12x - 36
          x^3 + 2x^2
          __________
                3x^2 - 12x
                3x^2 + 6x
                __________
                     - 18x - 36
                     - 18x - 36
                     __________
                              0

A lot of textbooks, I've noticed, will shift the quotient over, so that all the x2s are lined up, all the x's are lined up, and so on. I don't do that, because I find it to be confusing, and it doesn't match up with the subtraction and division very well.

Okay, so we've finished one step of our problem. We've taken one factor out of x3 + 5x2 - 12x - 36, and we now know it's (x + 2)(x2 + 3x - 18). Can you factor that last bit? We've done examples like that before, so see if you can do that before scrolling down.

...
...
...
...

I hope you cheaters didn't peek. Anyway, you can break that second bit into (x - 6) and (x + 3). So, put all together, our final answer is...

x3 + 5x2 - 12x - 36 = (x + 2) (x - 3) (x + 6)

But sometimes division problems aren't always really clean, and they end up being a little messy at the end. Here's an example:

Divide 12x3 + 4x2 - 61x - 63 by 2x + 3.

As always, set it up like a division problem. Here's what it's going to look like first:

Code:

          _________________________
   2x + 3 | 12x^3 + 4x^2 - 61x - 63

This time, the leading term in the divisor is a 2x, so finding the quotient bits will be a bit tricky. Not by much, though, so we just need to divide by 2 on top of that. The first one is going to be 6x2, as you may have guessed (12x3/2x):

Code:

            6x^2
          _________________________
   2x + 3 | 12x^3 + 4x^2 - 61x - 63
            12x^3 + 18x^2
            ____________
                  - 14x^2 -61x

Recall, 4 minus 18 is -14, so that's why we end up with that. The next one is -14x2/2x, or -7x:

Code:

            6x^2 - 7x
          _________________________
   2x + 3 | 12x^3 + 4x^2 - 61x - 63
            12x^3 + 18x^2
            ____________
                  - 14x^2 -61x
                  - 14x^2 -21x
                  ____________
                          -40x - 63

Subtracting a negative number is the same as adding a positive number, so it became -61 + 21, which is -40. We're almost done, don't fret! -40x is next, so -40x/2x is -20...

Code:

            6x^2 - 7x   - 20
          _________________________
   2x + 3 | 12x^3 + 4x^2 - 61x - 63
            12x^3 + 18x^2
            ____________
                  - 14x^2 -61x
                  - 14x^2 -21x
                  ____________
                          -40x - 63
                          -40x - 60
                          _________
                                 -3

And now we run into a problem - this problem has a remainder. We can't really divide after this point because there's nothing left to divide, so what do we do? You may be surprised, but we handle it much like we would any other division problem with a remainder.

Usually when you have a remainder, you usually write the quotient out and then have an "R #" on the end, meaning Remainder - or you can write it out as a decimal or fraction. Depending on the course you're taking, some will want you to write it as an R remainder, while others will want you to write it out as a fraction. I'm going to do the latter. To write the remainder out as a fraction, simply take the last number you have (in this case, -3) and put it as a numerator, while you put the divisor (in this case, 2x + 3) as the denominator: -3/(2x + 3).

So when you put that all together, our final answer is going to be:

(12x3 + 4x2 - 61x - 63) / (2x + 3) = 6x2 - 7x - 20 - 3/(2x + 3).
(It subtracted at the end, because the remainder is negative. If the remainder were positive, you'd put a plus sign at the end.)

That...is a lot to take in. To help you hone your craft, I've prepared a few examples for you to try on your own. But since long division is so tricky, I'll give you a few hints:

:: All of the dividends are going to be degree 3. (That is, the highest x term will be an x3.) The x3 is also just going to be a 1x3 - no 3s or 6s or any larger numbers in these examples. The divisors will also be simple.
:: I'll ask you to divide one polynomial by another. Just do the long division out first.
:: If you get a remainder, just stop there. Write out the quotient with the remainder as your answer.
:: If you don't get a remainder, then factor it out completely, and that'll be your answer.


Sounds good? Okay, here are some practice problems. Remember, no peeking yet!
Oh, and a calculator is fair game if you need to use one.

:: Divide x3 + 3x2 - 34x + 48 by x - 3.
:: Divide x3 - 19x2 + 120x - 254 by x - 6.
:: Divide x3 + 2x2 - 29x - 30 by x + 6.
:: Divide x3 + 25x2 + 208x + 585 by x + 9.

Spoiler

:: x3 + 3x2 - 34x + 48 = (x - 2)(x - 3)(x + 8)
:: (x3 - 19x2 + 120x - 254)/(x - 6) = x2 - 13x + 42 - 2/(x - 6)
:: x3 + 2x2 - 29x - 30 = (x - 5)(x + 1)(x + 6)
:: (x3 + 25x2 + 208x + 585)/(x + 9) = x2 + 16x + 64 + 9/(x + 9)

Conclusion

In the event that you have a quadratic equation, and you cannot factor it with the methods discussed so far, no matter what you do, or no matter how hard you try, then it may be a job for either completing the square or the quadratic formula. But that is a tutorial for another time - your brains may be fried enough at this point and I don't wish to exacerbate that. For now, just practice! The quadratic formula bit will have a few new things to cover, but a lot of it is just built upon factoring here.

Next tutorial planned: Solving equations via factoring, and the quadratic formula

 

Solving equations via factoring, and the quadratic formula
Requested by @[member=Azure]

Introduction

At this point I am going to begin by assuming that you already have a decent grasp on how to factor polynomials. If you don't, I suggest you practice it, possibly by reading the factoring tutorial above - then come back here.

So we know how to factor now, that's wonderful. But what can we do with this knowledge? More often than not they are used to solve word problems or fractions involving binomials. Those kinds of problems come up a lot, however, so it is worth knowing what to do with your factoring knowledge. I plan to start this off first by teaching you how to do the setup, and then transitioning into word problems. I will also briefly go over completing the square, and finish with the quadratic formula.


Solving when a Polynomial is Equal to Zero

This kind of factoring problem is by far the most common, and the one in which a lot of word problems will go with. It involves a polynomial of degree 2 (that is, the highest exponent on an x term will be x2) being equal to a certain amount. Let's start with an example problem, and I'll show you how I do it:

Solve the following equation: x2 - 8x = 3x + 26.

The first thing we need to do before anything else is make sure to place all of the terms on one side of the equation. The side of the equation you'll choose is the one where you will end up with a positive x2 when you move everything over. In that case, it's going to be the left side, since the positive x2 is already there.

So let's move everything over. Here's what we start with:

x2 - 8x = 3x + 26

Subtract 3x from both sides of the equation (remember, anything you do to one side of the equation, you must do the other):

x2 - 11x = 26

Subtract 26 from both sides:

x2 - 11x - 26 = 0

When you finally have a polynomial that's equal to zero, now you can get to solving it. You'll notice that the left side looks suspiciously like it can be factored - and that's the intent; you are going to factor the left side of the equation. And, since a is 1 here, it's easy enough to use the Blanks Method. (Remember, c and b are both negative.)

See if you can factor it yourself before we move on here, then we'll go to the next step.

...
...
...
...

You should have factored it so that s and t are -13 and 2. So, our expression becomes this:

(x - 13)(x + 2) = 0

Now, this is the important bit. Remember that any number multiplied by 0 is 0 itself, so for the above equations, only 1 of those has to equal 0 to make the whole thing go to 0.

If (x - 13) is the one that equals 0, then x can equal 13.
If (x + 2) is the one that equals 0, then x can equal -2.

So, which one is it? You may be surprised to hear this, but both of those numbers are valid solutions. x can be either 13 or -2, and either way the equation will equal 0. To double-check, you can plug those numbers back into the original equation, and they should be correct.

When x = 13:

x2 - 8x = 3x + 26
(13)2 - 8(13) = 3(13) + 26
169 - 104 = 39 + 26
65 = 65


When x = -2:

x2 - 8x = 3x + 26
(-2)2 - 8(-2) = 3(-2) + 26
4 + 16 = -6 + 26
20 = 20

(This works just as well as if you use the 'equals 0' form, as it is equivalent.)

See, that wasn't too difficult, right? Now that you have the hang of factoring down, solving equations with factoring should be a breeze.

In regards to the above problem having 2 solutions, that is actually a property inherent to polynomial equations: if you have a polynomial of degree k (that is, the highest exponent on an x term is xk), and it is equal to 0, then that equation will have, at most, k solutions. For example, an ax2 + bx + c = 0 equation will have at most 2 solutions, while an ax3 + bx2 + cx + d = 0 equation will have at most 3 solutions.

As a rule of thumb, the solutions will usually all be different, unless a perfect square (or perfect cube, etc.) is involved. For example, the equation x2 + 4x + 4 = 0 has only 1 solution, as it's a perfect square. (The solution is -2, for those curious, as it's x + 2 squared.) Also, you may come across a problem that appears to have solutions, but they don't appear to be numbers. For example, the equation x2 + 25 = 0 has 2 different solutions, but neither of them are real numbers; when such a problem appears, you simply write there is 'no real solution'. (The two solutions are 5i and -5i, which are imaginary numbers, but I am not going to cover those at this time.)

Here's another example problem of solving by factoring:

Solve the following equation: 3x3 + 15x2 = 150x.

When you get to equations like these, be careful. The leading term is not an x2, but an x3. Up until this point, and in my factoring tutorial, I have only discussed factoring degree 2 polynomials, and I didn't touch on degree 3 ones. This is because, in algebra, there is standardly no quick way to factor degree 3 polynomials, unless they can be somehow reduced or re-written to a form where you can deal with it as a degree 2.

And this one can, and I'll show you how, with a new step.

Now, after you isolate all of the terms on one side of the equation (again, it'll be the one where the highest degree term is positive, so the left again), there's something you'll want to check for. First, you want to see if there is a term, either an integer or a variable, that you can factor out of all terms in the equation. Once we isolate the terms to one side, we get this:

3x3 + 15x2 - 150x = 0

Can we isolate any terms? Yes, there's two, actually. The first one is a 3; all of the terms there are evenly divisible by 3. So, we can factor a 3 out and get this:

3(x3 + 5x2 - 50x) = 0

The other one is an x (notice that each term contains at least one x), so we can factor out an x from all terms as well:

3x(x2 + 5x - 50) = 0

Now, look at the term inside the parentheses - it's a degree 2 polynomial, which can be factored, and thankfully, it's one of the easier varieties too. After we factor that using the Blanks Method, we get:

3x(x + 10)(x - 5) = 0

As a note, when you have an integer factored out (like the 3 here), it is usually safe to divide both sides of the equation by that integer, to make it disappear, and it won't change the solutions. (0 divided by 3 is still 0, after all.) It is NOT SAFE, however, to divide by x! Remember, we don't know what numbers x is yet, and there's a possibility that x could be 0 - and dividing by zero is, as we all know, a big no-no and a way to make your calculator explode.

I'm not going to divide the 3 out for this equation, but you can if you want - it doesn't make too much of a difference. Now, for the solution - remember, any of those terms can equal 0 and it will drive the whole thing to 0. We have three separate terms here: 3x, x + 10, and x - 5. We can take each one separately and set them equal to 0:

3x = 0
x + 10 = 0
x - 5 = 0

By solving these three mini-equations, we now get three different solutions: 0, -10, and 5. (You will usually be asked to write them in ascending order, though, so write it as x = -10, 0, 5.) All three of these solutions are valid - and this shouldn't come as a shock; remember, we started with a degree 3 polynomial, so there will be at most 3 solutions.

As before, here are a few practice problems for you to try out. I will have a few in which you may have to factor something out, but it is nothing new that we haven't covered. Answers are in the spoiler tag when you're done.
Hint: When you get a factor like (3x - 4), you would still solve the normal way, just expressing your answer as a fraction (for example, x = 3/4 for that one).

:: Solve x2 - 3x - 10 = 0.
:: Solve x2 + 72 = -24x - x2.
:: Solve x3 + x2 - 12x = 0.
:: Solve 2x3 = 3x - x3.

Spoiler

:: (x - 5)(x + 2) = 0; x = -2, 5
:: 2(x + 6)(x + 6) = 0; x = -6
:: x(x - 3)(x + 4) = 0; x = -4, 0, 3
:: 3x(x - 1)(x + 1) = 0; x = -1, 0, 1

Factoring Sums and Differences of Cubes

Before I move to the next bit of this tutorial, I would like to mention that there actually is a way to factor certain cubes, but it's highly specific. It only applies to sums of cubes and differences of cubes (similar to a difference of squares). The identities for such are below:

a2 - b2 = (a + (a -
a3 + b3 = (a + (a2 - ab + b2)
a3 - b3 = (a - (a2 + ab + b2)

I posted the difference of squares there as well because, although we've already covered it, it bears repeating. The sums and differences of cubes won't come up too often, but you want to be aware if they do. They're also really weird to memorize, and not quite as intuitive as the difference of squares, but I'll tell you how I remembered it:

The first factor is always a with b. The sign is the same as whether it's a sum or difference - for example, if it's a3 + b3, the first factor is a + b.
The second factor is a2, ab, and b2. (It's NOT 2ab - that's a perfect square). The first sign is always the opposite of the one in the first factor - I did a + b before, so it's going to be a2 - ab to start off. Lastly, the b2 at the end is ALWAYS positive, no matter whether it's a sum or difference, so I finish with + b2 - altogether, that means the second factor is a2 - ab + b2.

Something can be a square or a cube without needing to have a 2 or a 3 in the exponent; the exponent can just be a multiple of 2 or 3 as well. For example, x4 - 16 is a difference of squares: x4 is x2 squared, while 16 is 4 squared. Thus:

x4 - 16 = (x2 + 4)(x2 - 4)

And, after that, you can actually factor the second one even further, because x2 - 4 is ALSO a difference of squares:

x4 - 16 = (x2 + 4)(x + 2)(x - 2)

The same kind of principle applies for other things, such as x6 + 27y9 (a sum of cubes), 64s12 - 125t6 (a difference of cubes), and 2x7 - 8xy4 (a difference of squares, as soon as you factor the 2x out).

Yes, I know this bit should have technically gone in the factoring tutorial before, but I didn't think of it until now. If you want, though, you can try and factor the examples I just gave, as some practice problems. Don't worry - you won't have to solve for anything.
Hint: start by identifying what each factor is a cube of - that is, identify a and b. These are a bit tricky, but keep a level head and you should be able to handle them.

:: Factor the sum of cubes x6 + 27y9.
:: Factor the difference of cubes 64s12 - 125t6.
:: Factor 2x7 - 8xy4, which includes a sum of squares.

The perfect squares of integers, up to 102, are: 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100.
The perfect cubes of integers, up to 103, are: 1, 8, 27, 64, 125, 216, 343, 512, 729, and 1000.

Spoiler

:: x6 + 27y9 = (x2 + 3y3)(x4 - 3x2y3 + 9y6)
:: 64s12 - 125t6 = (4s4 - 5t2)(16s8 + 20s4t2 + 25t4)
:: 2x7 - 8xy4 = 2x(x3 + 2y2)(x3 - 2y2)

Solving by Completing the Square, and Radical Solutions

Completing the square is an alternative method to solving quadratic equations. Rather than moving everything all terms over to one side of the equation, so you have something equal to 0, you move everything over except for a positive integer, which you leave on the other side. Then you complete the square, and take square roots, to solve the equation.

This method is a bit trickier to do than isolating everything to one side - and, in my opinion, more impractical - but some courses require that you know it, so I'm going to cover it here for completeness. It's also very easy to get lost as well, so I'll go over the steps slowly. Here's an example we can work on:

Solve by completing the square: 2x2 = 8x + 64.

The first step in solving an equation by completing the square is, as mentioned before, to bring all the terms over to one side, except for an integer. Usually this just means bringing all the xs over to one side, as you only need to move those. So we'll subtract 4x from both sides, and get this:

2x2 - 8x = 64

The next step is to divide out any integers so that we are just left with an x2, rather than a 2x2 or a 3x2 or so on. In this case we need to divide by 2, so we divide everything in the equation by 2:

x2 - 4x = 32

This is essentially the same as factoring out the 2 like we did before, and we could divide it out so that it doesn't affect the outcome of the equation. When completing the square, though, you may end up with situations in which, after you divide the integer out, you get fractions. This is possible. It makes the solution a bit messier later on, but it's still accurate. (I'll get to an example like that shortly.)

Next bit is the tricky part. We want to get the left side into the form of ax2 + bx + c, and do so that it is a perfect square, i.e. (x + s)2. We have the ax2 and the bx, but we don't have the c yet. To find c, do the following:

:: Find out what b equals here. (In this problem, it's -4.) Mentally note that number.
:: Halve it (-2 for this problem), so it becomes b/2. Mentally note this number as well - you'll need it later!
:: Square it (4 for this problem), so it becomes (b/2)2. This number is what your c is.
:: Add that number to both sides of the equation.

We'll do a bit more after this as well, but let's go slowly. As expressed above, in this particular example, c is 4 to make the left side a perfect square, so we are going to add 4 to both sides of the equation:

x2 - 4x + 4 = 36

This act of adding is called 'completing the square', which is where this procedure comes from.
Do note: if, at this point, the right side is a negative number, this will NOT work - later you need to take the square root of both sides, and under normal circumstances you can't take the square root of a negative number. I won't have any examples where that does happen, and it's very rare that it will, but be aware of this restriction.

The next thing we're going to do is to 'compress' the left side, so it is some expression squared. We know that's a perfect square, so you can factor it using the Blanks Method, OR take note of something you did a little while ago: remember when you found b/2? That is actually what s is for the equation. You found b/2 as -2, and sure enough, when you compress it, x2 - 4x + 4 becomes (x - 2)2. Neat, no?

(x - 2)2 = 36

After that, we're going to take the square root of both sides. When you do that, there's something important you need to remember. When squaring a number, if that number is negative, when you multiply it by itself, it becomes positive, so the equation x2 = 25 actually has two solutions: 5 and -5, as Martin Prince of The Simpsons might readily tell you. That principle does apply here too. The left side is pretty simple as it just reduces to x - 2, but there can be two solutions to that, so to compensate, what we do is, when taking the square root of the integer (on the right side), we put a plus-minus sign in front of it:

x - 2 = ±6

Stated another way, if x - 2 squared is 36, then that means x - 2 can be one of two numbers: 6 or -6. Again, this goes with the idea that a degree 2 equation can have up to 2 solutions.

Now we need to isolate x...get it by itself. This part is simple, but easy to mess up. You're going to add 2 to both sides of that equation there so you get x by itself, but when adding it to the right side, you're going to put it in FRONT of the plus-minus sign, like so:

x = 2 ± 6

DO NOT DO THIS:

x = ± 6 + 2
(While this technically is not wrong, it's improper syntax, and it looks a lot messier. If evaluated correctly it'll lead to the same result, but it's not very clean-looking, so don't do it.)

So, now we have this:
x = 2 ± 6

Since this is technically two numbers/solutions, we need to split this equation in half and evaluate each one separately. 2 + 6 is 8, and 2 - 6 is -4, so the two solutions for x are -4 and 8. (You can plug those back into the original equation to double-check as well; they will come out correct.)

Now, here's an example of a problem where you get fractions...and something else weird as well, which I'll get to.

Solve by completing the square: 3x2 = 78 - 8x.

Just like before, we move all but the integer term to one side of the equation. We do this by adding 8x to both sides:

3x2 + 8x = 78

Since there is a 3 in front of the x2, we need to divide both sides by 3, so the x2 is left by itself. When we do so, this happens:

x2 + (8/3)x = 26

Notice that b, the coefficient for x, is a fraction. We can still work with this; it's just going to be a bit messier. We need to find out what c is in this case; we find this by halving b and then squaring it, resulting in 16/9. We add that number to both sides of the equation:

x2 + (8/3)x + 16/9 = 27 7/9

As mixed numbers are not appropriate to use in this scenario, we convert it to a fraction:

x2 + (8/3)x + 16/9 = 250/9

Additionally, check to see if you can reduce the fraction, writing it in lowest terms. However, 250 and 9 have no common factors, so we leave it as is. Now we compress the left side into the perfect square form; remembering from before that half of b was 4/3, that is what the s in the (x + s)2 will become:

(x + 4/3)2 = 250/9

The next step is to take the square root of both sides, remembering to insert the plus-minus sign on the right side:

x + 4/3 = ±√(250/9)

You will notice that the right side is not a perfect square. That is the other half of this problem I had to cover - it is possible that, when completing the square, you will end up with a radical on the right side of the equation. This actually happens fairly often, though it might not always look as messy as my example here.

However, as you probably remember when dealing with square roots, we want to see if we can clean up the square root, and extract any perfect squares from it, to drag them outside the radical. Thankfully 250 is a neat number, and it separates into 25 * 10, so that radical can be rewritten as 5 * √(10). 9 is a perfect square as is, so that reduces to 3. Our equation now looks like this:

x + 4/3 = ±5√(10)/3

Next, we isolate x, by subtracting 4/3 from both sides of the equation:

x = -4/3 ± 5√(10)/3

And lastly, we separate the plus-minus sign out so that x's two distinct solutions are listed:

x = (-4 + 5√(10))/3
x = (-4 - 5√(10))/3

(As the denominators for both terms were the same, they were combined into a single fraction.)

So there you have it - completing the square. It's a lot more annoying to work with than setting one side equal to 0, and a lot messier too (especially with the square roots), but it does have one distinct advantage over the former method: you can use it when you can't factor the polynomial completely using the methods you just learned. In fact, completing the square is essentially the precursor to another factoring method...which I'll discuss at the very end.

But before that, let's work with a few practice problems on completing the square before moving along. Remember, the answers are in the spoiler tags, but don't peek until you're done!

Solve each of these by completing the square:
:: x2 - 6x = 55
:: 2x2 = -20x + 48
:: 3x2 + 12x = 84
:: x2 - 15x = 16

Spoiler

:: x = -5, 11
:: x = -12, 2
:: x = -2 - 4√2, -2 + 4√2
:: x = -1, 16

Lowest Common Denominators, and Polynomial Fractions

When dealing with fractions that have binomials in the denominator (e.g. 2/(x + 4)), you will usually have to find the lowest common denominator of that expression, if you want to add them together. (When multiplying you don't have to worry about that; you can just factor out everything and start cancelling like factors.)

The lowest common denominator, or LCD (out of a set of denominators) is the lowest denominator possible that has all factors in common out of that set - otherwise known as the least common multiple of denominators. I'll start with how to find lowest common denominators for real numbers, then I'll move on to how to do that with binomials. Here's my example:

Find the lowest common denominator out of 10, 12, and 18.

To find the lowest common denominator out of a group of numbers, first break down each number into its prime factors. This will let you see what kind of factors will be needed to find your LCD.

If you don't remember how to find prime factors, it helps to use a tree. Start dividing your number by prime numbers until you can't divide anymore - start at 2, then 3, then 5, then 7, and so on. Usually you won't need to go above 7, but in some large number cases, you might. So, let's factor these three. I'll use the trees to do so.

Code:

2|10
  __
   5

2|12
2|6
  _
  3

2|18
3|9
  _
  3

When written out horizontally, we have:

10 = 2*5
12 = 2*2*3
18 = 2*3*3

Next, we count out how many different kinds of prime factors we see here. There are only three in this case: 2, 3, and 5. (These numbers are by far the most common when doing these kinds of problems.) After that, we want to see where each of those prime factors is represented the most, so we know how many of these prime factors we need in the LCD.

Let's start with 2. 10 and 18 each have one, but 12 has two. So, whatever number we're going to use for the LCD, it's going to need two 2s in it.
Now for 3. 10 doesn't have any, 12 has one, and 18 has two. Like before, that means that our LCD is going to have two 3s in its prime factors.
Lastly for 5, only 10 has a 5 in it, but it tells us that our LCD is going to have one 5.

Listing those together, we have two 2s, two 3s, and one 5. The LCD is going to be the product of all of those prime factors, or stated another way, when the LCD is broken down into prime factors, it's going to be 2*2*3*3*5. Doing that in reverse, if you multiply that all out, you get the number 180. That tells you that 180 is going to be the least common denominator of those three numbers.

So if you have a problem like this:

Add: 5/12 + 7/18 + 3/10

It means that we are going to have to make all of the denominators 180 before we can add them together. We do that by multiplying the top and bottom of each fraction individually by whatever number is missing from the LCD.

You can do that one of two ways: either dividing the LCD by the current denominator, or counting which prime factors are missing from that denominator. (Either way gives you the same answer.) I'll solve this problem using both methods.

The first term is 5/12; the denominator is 12. If I use the first method, I just divide 180 by 12, and get 15. If I use the second method, I compare prime factors: 180 is 2*2*3*3*5, while 12 is 2*2*3. There is a 3 and a 5 missing from 12, so the number to multiply be will be 3*5, or 15. Either way I get the same result. When I multiply top and bottom by 15, the first term becomes 75/180.

The second term has 18 in the denominator. The first method tells me that 180/18 = 10, so 10 is my missing number. The second method compares 2*2*3*3*5 (180) with 2*3*3 (18). There's a 2 and a 5 missing from 18, and 2*5 = 10. Multiplying 7/18 by 10/10 gives 70/180.

The third term is 3/10. With the first method, dividing 180 by 10 gives 18; with the second method, comparing the prime factors 2*2*3*3*5 with 2*5 shows me there is a 2 and two 3s missing, so 2*3*3 = 18 - the result is the same either way. Multiplying top and bottom by 18 gives 54/180.

Now my equation becomes:

75/180 + 70/180 + 54/180

Those all have the same denominator, so I can combine the fractions and add them together, like so:

(75+70+54)/180 = 199/180

The last step is to see if numerator and denominator have any common factors now, to see if they can be cancelled. After a quick check, they don't share any common factors (199 is prime, actually), so we can stop the problem right there and give 199/180 as the answer.

Now that we've done an example with numbers you're more familiar with, let's move on to LCDs with binomials.

Add: 2/(2x + 6) + 4/(2x - 6) + 6/(x2 - 9)

When dealing with the LCDs of binomials, you need to factor each term as much as is possible, including integer factors. Similar to the above, you will list the factors of each one side by side, and then determine what factors your LCD is going to need. In this case, we have the following:

2x + 6 = 2(x + 3)
2x - 6 = 2(x - 3)
x2 - 9 = (x + 3)(x - 3)

Now, note how many different kinds of factors you have for this set. For this problem, you have three: 2, x + 3, and x - 3. We'll count each one separately.

As far as 2 goes, the first two terms have one, but the last one doesn't. Either way, we are only going to need one 2 for our LCD, as nothing has multiple 2s.
Now for x + 3. For that factor, the first and last ones each have one, and the second one doesn't. Thus, we only need one x + 3.
Similarly, for x - 3, the first one doesn't have any, and the second and third each have only one, so our LCD only needs one x - 3.

Put that all together, and your LCD for these three is going to be 2(x + 3)(x - 3). You could rewrite it as 2x2 - 18, but I recommend you don't do that until you're finished. Just as we were checking for factors before after we finished our last LCD problem, we are going to be doing the exact same thing with this one.

After multiplying the top and bottom of each fraction by whatever term is missing, you will get the following numerators:

First: 2x - 6
Second: 4x + 12
Third: 12

The sum of those three is 6x + 18, so your expression at this point, after you sum them up, will look like this:

Code:

    6x + 18
_______________
2(x + 3)(x - 3)

Done, right? Well, not yet. As I mentioned earlier, the last step is checking for any common factors between the numerator and denominator...and, as it turns out, there is one. The numerator can be factored out as 6(x + 3), so it becomes this:

Code:

    6(x + 3)
_______________
2(x + 3)(x - 3)

As you can cancel like factors, you can cancel the x + 3, and the expression becomes this:

Code:

   6
________
2(x - 3)

...and after that the 2 in the denominator can be cancelled with the 6 in the numerator, thus leaving the final answer as 3/(x - 3). Ta-da!

But that isn't the only thing we can do with lowest common denominators, actually. In addition to adding expressions together as a sum, you can also use them to solve equations as well - and even multiply two expressions together that initially look radically different. I'll go over each example below, as well as how to solve each one.


Solving an Equation by Adding LCDs

Spoiler

Solve:

Code:

   2         6            8
_______ + _______ = _____________
 x + 5    x^2 - 9   x^2 + 2x - 15

When trying to solve an equation, the steps are slightly different, but most of the same rules and procedures still apply. The first step is still the same: to list out the factors of all the denominators separately.

x + 5 = x + 5
x2 - 9 = (x + 3)(x - 3)
x2 + 2x - 15 = (x + 5)(x - 3)

If you look at all of the common factors and determine the LCD, you will determine that it is (x + 5)(x + 3)(x - 3) - which you do not need to multiply out, as the factors will start cancelling anyway. There is one important thing you must be aware of at this point, however:

Just like with any other fraction, the denominator is not permitted to be 0, or else the number is undefined. Thus, we look at the LCD and see what possible values of x would make the LCD 0. Remember, if just one bit equals 0, the whole LCD goes to 0. The numbers that would result in that would be -5, -3, and 3. Thus, if any of the solutions that we obtain are any of those numbers, they must be rejected, since they will result in undefined numbers when plugged back into the original equation.

The next thing we are going to is multiply both sides by the LCD - or, multiply the top and bottom of each term separately by the LCD. Do whichever way is more comfortable for you - either way you're going to get the same result. (Keep the LCD as the separate factors - don't combine it together into x3 + 5x2 - 9x - 45, because that expression is NOT going to help you!) After resolving that, your problem should look like this (I'll simplify it as I go along, going in small steps):

2(x + 3)(x - 3) + 6(x + 5) = 8(x + 3)
2(x2 - 9) + 6x + 30 = 8x + 24
2x2 - 18 + 6x + 30 = 8x + 24
2x2 + 6x + 12 = 8x + 24
2x2 - 2x + 12 = 24
2x2 - 2x - 12 = 0
x2 - x - 6 = 0
(x - 3)(x + 2) = 0
x = -2, 3

So, after simplifying everything, we appear to have two solutions for our problem: -2 and 3. Or do we? Remember what I said earlier - you cannot accept a solution that makes the LCD (or any denominator, for that matter) go to 0. As we established before, 3 is one of the numbers that makes the LCD go to 0, so we reject that solution. Thus our problem only has one solution: -2.

To double-check, substitute that number back into the original equation, and it should work. (Notice that you can't do that with 3; you'll get 2/8 + 6/0 = 8/0, which is basically nonsense. With -2 you'll get 2/3 - 6/5 = -8/15, which is correct.)

Multiplying Polynomial Fractions

Spoiler

Multiply:

Code:

2x^2 + 18x + 40   x^2 + 10x + 21
_______________ * ______________
 x^2 + 8x + 15       2x + 14

Looks grisly, right? What if I told you the answer is x + 4? You might not believe me right away, as it looks so big and scary, but trust me, that's the answer. I know. I wrote the problem.

Multiplying polynomial fractions is actually easier than adding them together, because you entirely skip the step of trying to find an LCD, and go right to cancelling out factors. The difference is that, with the multiplication problems, you cancel out a lot more factors.

What you do with multiplication problems is you separate out all terms in the equation into factors. All of them. Every single one, just like you did when trying to find LCDs. All of the degree 2 polynomials can be factored using the Blanks Method (you can factor a 2 out of the top left one), so this isn't too tricky.

After a bit of factoring, our problem now looks like this:

Code:

2(x + 4)(x + 5)   (x + 3)(x + 7)
_______________ * _____________
(x + 5)(x + 3)       2(x + 7)

That's a little easier to look at than the thing we started with, right? Now for the fun part: cancelling. If there is a common factor in any numerator with any denominator, it gets cancelled. Since you're multiplying together, you can cancel out across into other fractions, since they are the same separate as they are combined (that is not the case with addition). There is an x + 5, an x + 3, an x + 7, and a 2 that exist in both the numerator and denominator, one copy of each.
(Note, you have to cancel in pairs. If there were two x + 7s in the denominator and only one in the numerator, for example, you could cancel one pair, but the other x + 7 in the denominator can't be cancelled.)

After all the cancelling is done, you should end up with just x + 4...just like I said it would. Problems like these look scary on the surface, but they're actually quite fun, since it's basically breaking them up into factors and then cancelling away!

Solving an Equation by Multiplying Fractions

Spoiler

Solve:

Code:

x^2 + 10x + 25   x^2 + 11x + 28
______________ * ______________ = 0
     x + 7           x + 5

A little similar to what we did before, though this time, it's put in the form of an equation. Just like when solving with an addition problem with fractions, we have to reject any solution that would make any denominator go to 0. In this case, the numbers are -7 and -5, so if we get that as a solution, we reject it.

Now, after that, factor out all terms in the numerators and denominators, like before. We don't have to do that with the denominators this time, so that's half of the work we don't have to do, but after factoring the numerators, we get:

Code:

(x + 5)(x + 5)   (x + 4)(x + 7)
______________ * ______________ = 0
   (x + 7)          (x + 5)

Now we can start cancelling. There is an x + 5 pair, as well as an x + 7 pair, that we can cancel. Once we do that, we get the following:

(x + 5) * (x + 4) = 0

From this we get two solutions, -5 and -4. However, -5 is one solution that makes a denominator go to 0, so we have to reject that one, leaving us with just -4 as our answer.

And, of course, there are a few examples for you to try out on your own. The answers are hidden below, but do the work yourself before you peek!

Code:

:: Add:

     5             6           1
___________ - ____________ + _____
x^2 + x - 6   x^2 + 6x + 9   x + 3

Code:

:: Solve:

  3       2        7
_____ - _____ = ________
x - 7   x + 7   x^2 - 49

Code:

:: Multiply:

x^2 + 10x + 24      3x - 12
______________ * _____________
   x^2 - 16      x^2 + 2x - 24

Code:

:: Solve:

(x + 7)(x - 7)   (x + 6)(x - 6)
______________ * _____________
    (x + 6)         (x + 7)

Spoiler

:: (x2 + 21)/[(x + 3)(x + 3)(x - 2)]
:: x = -28
:: 3/(x - 4)
:: x = 6, 7

Word Problem Examples

Here are a few examples I've seen of word problems used in algebra courses, that are solved using factoring. It's essentially taking what we learned with factoring, and applying it to life situations.

If you have any examples you've seen but aren't here, and want to have them added, let me know! I will be glad to enter them into this guide. Your name will be mentioned above any examples you've contributed. If I get enough of these, I may move them into a separate tutorial.

All examples in here will be concealed in spoiler tags, to save space. There will be no example problems for you to try in this section, either, so you get off scot-free. Lucky.

As with all word problems, try to visualize the scenario in your mind, and mentally draw it, so you will know how to reproduce it with math.


Combined workload:
(Not actually a factoring example, but included here anyway)

Spoiler

An office building has two printers they use for high-duty print jobs. One of them is an older model that they've had for some time, and it takes about 5 hours for it to finish a day's worth of printing. The other printer is much newer and more efficient, and it only takes 3 hours to finish a day's worth of printing. How much time would it take to finish a day's worth of printing, if both printers are used at once?

This one is a bit tricky as the answer isn't all that obvious, but here's the setup used.
Old Printer takes 5 hours to finish a day's printing.
New Printer takes 3 hours to finish a day's printing.
Old and New Printers together take an unknown amount of hours to finish a day's printing. We'll call it x.

Now let's express that another way - how much can a printer get done in 1 hour? If it takes Old Printer 5 hours to finish 1 day's worth, then in one hour, Old Printer has done 1/5 day's worth of printing. That makes sense, right? Similarly, since New Printer takes 3 hours to finish a day's worth, in one hour it has done 1/3 day's worth of printing. And, if they work together, they get done in x hours, so in 1 hour, they get 1/x day's worth of printing done.

This becomes our setup. We put together the amount of work each printer can do in 1 hour:

1/5 + 1/3 = 1/x

Stated another way, Old Printer's hour + New Printer's hour = Combined hour.

As this is a problem with least common denominators, we find the LCD of the three terms. Noting all the factors, the LCD is 15x, so we multiply both sides of the equation by 15x (or individually multiply each term's top and bottom by 15x, which is the same, and also works).

The problem becomes this (I'll simplify as I go):

15x(1/5 + 1/3) = 15x(1/x)
15x(1/5) + 15x(1/3) = 15x(1/x)
3x + 5x = 15
8x = 15
x = 15/8

Thus, when the printers work together, it takes them 15/8 hours (a little under 2 hours) to get a day's worth of printing done. This makes sense, as the answer should be less than either of the printers alone, since they work together here. You can substitute it into the original problem as well:

1/5 + 1/3 = 8/15

...and it will turn out correct.

Moving in a river current:
(Not actually a factoring example, but included here anyway)

Spoiler

Randy is riding in a boat on a river. When the current is flowing, it takes Randy 3 hours to go downstream a certain distance, but it takes him 7 hours to go upstream that same amount of distance. The current of the river is 6 miles per hour. How fast does Randy row?

When you are in a river, and you're moving with the current, your speed is increased by that of the current (since you're moving with it), but when you're going against the current, you're hindered by it, so its speed counteracts your own, thus reducing your speed by that of the current.

We don't know how long the distance Randy traveled, but it doesn't matter: we know the time at which he traveled, as well as the speed, and we know that either way it amounts to the same amount of distance. (Distance equals rate multiplied by time; if you go 50 miles per hour for a half-hour, you've gone 25 miles.)

We want to find out Randy's rowing speed, so let's call it x miles per hour. When Randy's moving with the current, the current adds to his speed, so he would be moving at x + 6 miles per hour then. However, when he's moving against it, the current counteracts him, so he'd move at x - 6 miles per hour instead.

When moving downstream you're with the current, when moving upstream you're against it. So it takes 3 hours for Randy at x + 6 mph to get some place, but 7 hours at x - 6 mph to get there. Here's our setup:

3 (x + 6) = 7 (x - 6)

It's fairly easy to solve:

3x + 15 = 7x - 42
3x + 57 = 7x
57 = 4x
x = 57/4

Thus, we know that Randy's rowing speed is 57/4 mph, or 16.25 mph (he's a pretty strong rower). This answer makes sense, because Randy, at the very least, needs to have a rowing speed greater than 6 mph, or he would be completely unable to go against the current.

Two people moving in different directions:

Spoiler

Nadia and Ernie are both at an intersection, riding on bicycles. They both start off at the same time, with Nadia traveling north and Ernie traveling east. After a while, they stop. Ernie has traveled 7 kilometers further than Nadia, and the two of them are 17 kilometers apart from each other. How far did Nadia travel?

Here you have two people who start at the same point and go in different directions - north and east. The distance they are apart isn't via the routes they just took; it's the direct distance, if you drew a line between the two. Drawing that out, you actually get a right triangle.

So let's draw this out. Nadia traveled some number of kilometers away; let's call that 'x'. She traveled x kilometers away. (You can use whatever letter you like.) Ernie traveled 7 kilometers further, so he traveled x + 7 kilometers. And they are exactly 17 kilometers away from each other. This gives you a right triangle with one leg as x (Nadia), one leg as x + 7 (Ernie), and the hypotenuse as 17 (the distance between).

This you have to solve with the Pythagorean theorem: a2 + b2 = c2. With the numbers we have above, we'll plug those in:

a = x
b = x + 7
c = 17

If you write that out, this is what you get (I will also simplify the expression as I go along):

x2 + (x + 7)2 = (17)2
x2 + x2 + 14x + 49 = 289
2x2 + 14x + 49 = 289
2x2 + 14x - 240 = 0
x2 + 7x - 120 = 0

This equation can be solved by factoring:

(x + 15)(x - 8) = 0
x = -15, 8

So, after solving that, Nadia either traveled -15 kilometers or 8 kilometers. However, though we have two different solutions here, one of them isn't valid...and I'm sure you've noticed the reason why: you can't travel a negative amount of distance in this case. (We aren't dealing with vectors.) So, we reject the -15 answer, and arrive at our final solution that Nadia traveled 8 kilometers. (Ernie traveled 15, which is 7 more than 8.)

The hardest part to any word problem is the setup, but here are some examples of word problems I've seen that may help you decipher bits of your own homework. I hope this is of help to you.

And now that we've gone over that, it's time to wrap up this section...with the quadratic formula.


The Quadratic Formula

Look upon the quadratic formula, mortals, and despair!

Seriously, though, while the quadratic formula may look mean and nasty at first glance, it does have a very nice perk to it: it will ALWAYS work. Always. No exceptions. Any quadratic equation can be solved simply by plugging in a, b, and c into the quadratic formula, and reading the solutions. Don't forget to get the signs correct on a, b, and c!

The quadratic formula is as follows: If you have a quadratic equation, i.e. of the form ax2 + bx + c = 0 (where a is not equal to 0), then the solutions are:

Code:

x =  -b ± √(b^2 - 4ac)
     _________________
            2a

Ick. No wonder everyone hates this. Usually the only way out with this on a test is just to memorize it. The bit inside the square root is the hardest part to memorize (the rest is kinda simple, right?), but don't fret: the quadratic formula is intended as a last resort. Only use it when nothing else works.

It's essentially the same thing as completing the square, combined with moving all the terms to one side.

But how did we get from ax2 + bx + c = 0 to that...that thing?

I'll show you.

(Some courses require you to derive the quadratic formula as a bonus problem. Not all courses require this, but if you need to know, here it is. You can skip this if you want, but I recommend you read the part about the 'discriminant'.)

So let's start off with our quadratic equation, left alone. And, believe it or not, we are actually going to derive the proof by completing the square - which we just did not too long ago!

ax2 + bx + c = 0

Remember what the first step was in completing the square? It was to get all of the x-terms to one side, and the integer term on the other side. We're going to do just that here; since c is the integer term here, we move it over by subtracting both sides by c:

ax2 + bx = -c

Next, we need to reduce the x2 to a form by itself. a is an integer, so we need to divide both sides by a in order to leave the x2 by itself.
Note: although we don't know exactly what number a is, the inherent restriction of a quadratic equation is that a can't equal 0. Since that means dividing by a will never mean dividing by 0, this is permissible.
Dividing both sides by a gives this:

x2 + (b/a)x = -c/a

(The first two steps can be done in either order. It doesn't matter which step you do first: subtracting c or dividing a, as you will end up at the same result.)

Now is where we can complete the square. Note the coefficient of x (which is b/a). Halve it, giving you b/(2a), then square it, giving you b2/(4a2). Add this number to both sides of the equation:

Code:

      b      b^2       c    b^2
x^2 + _ x + ______ = - _ + ______
      a      4a^2      a    4a^2

Compress the left side so it appears as a perfect square. Remember that the halved coefficient was b/(2a), so that's what our perfect square will have:

Code:

     b        c    b^2
(x + __ )^2 = - + ______
     2a       a    4a^2

At this point, you might remember that the right side there needs to be non-negative, or the equation will have no real solutions - which is correct. However, when deriving the formula, we don't need to worry about this just yet - it'll become more obvious in the finished product.

Next is to take the least common denominator of the right side, to make it all one large fraction. The least common denominator is 4a2, so to make the -c/a have that denominator, we multiply top and bottom by 4a, so the top becomes -4ac. Combining that into one fraction, we get:

Code:

     b        b^2 - 4ac
(x + __ )^2 = _________
     2a          4a^2

Notice the numerator is b2 - 4ac - the same term that was under the radical in the quadratic formula at the end. We must be getting close. After we finish the derivation, I would like to note something special about that term.

But we need to finish completing the square. At this point, we need to take the square root of both sides. The left side is easy to take the root of, but the right side is less so. The denominator, 4a2, is a perfect square, so that bit becomes 2a, but the numerator, b2 - 4ac, is not, so a radical goes over it. Don't forget the plus-minus sign either!

Code:

    b    ±√(b^2 - 4ac)
x + __ = _____________
    2a        2a

Lastly, subtract -b/(2a) from both sides, so that x is left by itself. Since the denominator is 2a, the two can be combined into a single fraction...which leads to us finishing our derivation.

Code:

x =  -b ± √(b^2 - 4ac)
     _________________
             2a

Nifty, isn't it? Turns out completing the square IS good for something after all.
And now that our derivation is finished, I must talk a bit about that term under the radical there.

That term, b2 - 4ac, is called the 'discriminant', and it lies under the radical in the quadratic formula. The discriminant is rather important, because it essentially determines how many solutions the equation in question will have.
Unlike a term like 4a2, which is always positive, b2 - 4ac, the discriminant, has no such restriction: it could be positive, negative, or even zero. Depending on that, it will actually tell you how many solutions there are to the problem:

:: If b2 - 4ac < 0, there are no real solutions. The 2 solutions to this problem are both imaginary numbers, and both different.
:: If b2 - 4ac = 0, there is exactly 1 solution. The original equation is a perfect square.
:: If b2 - 4ac > 0, there are 2 real solutions, both of which are different numbers.
:: Additionally, if b2 - 4ac is a perfect square, that means the equation can be factored normally too. (You will get the same answers either way.)

Note that the discriminant doesn't tell you what the solutions are exactly: just how many. However, it is recommended that, if you have to use the quadratic formula, you work out what the discriminant is first, since that's the hardest part of the formula. If it's negative, you can usually just stop; if it's 0, then your single answer is easy (x = -b/(2a)); if it's positive, that might require a bit more work.

A reminder, you only need to use the quadratic formula if you cannot factor the equation any other way. It works very similarly to completing the square, but there are less steps to do overall - all you need to do is plug in a, b, and c, and simplify your expression when necessary. It isn't even necessary that a be positive - or that all of them be integers! However, it is required that they all be rational numbers (i.e. numbers that can be expressed as a fraction of 2 integers), and that a is not 0.

Here are a few examples below, of problems involving the quadratic formula. You can work on plugging in the numbers any way you choose, but I will begin by working on the discriminant, and going from there.

Solve: 2x2 - 8x + 7 = 0.

The first thing you will do with the quadratic formula, if it hasn't been done already, is to move all terms to one side of the equation, so that one side equals 0. Once you do that, identify a, b, and c:

a = 2
b = -8 (remember the sign)
c = 7

Now, evaluate the discriminant, b2 - 4ac, to see how many solutions we're dealing with:

(-8)2 - 4(2)(7) = 8

The discriminant is positive, and also not a perfect square, so we're dealing with 2 solutions, neither of which we can factor the normal way. When you actually put the radical in, though, √(8) can be rewritten as 2 * √(2), so I'm going to do that. Plugging the values into the quadratic formula produces this:

Code:

x =  -(-8) ± 2 * √(2)
     ________________
            2(2)

Note, you can cancel common factors out of the quadratic formula, but when you do so, you have to do so from THREE areas: the denominator, the numerator left of the plus-minus sign, and the numerator right of the plus-minus sign. All of those areas need to have a common factor if you want to remove it. I'll demonstrate:

Code:

x =  8 ± 2 * √(2)
     ____________
           4

The common factor in this case is 2, so I cancel a 2 out of the 4, the 8, and the 2 √(2):

Code:

x =  4 ± √(2)
     ________
        2

We can even split up the fraction so only one bit has a denominator:

x = 2 ± √(2)/2

And lastly, we get both solutions by evaluating the plus-minus sign:

x = 2 + √(2)/2
x = 2 - √(2)/2

Long and tedious? Yes. At least it only involves plugging numbers in; the formula does most of the work for you. Let's move on to a few other examples.

Solve: 9x2 - 12x + 4 = 0.

As usual I'm going to start with noting what a, b, and c are, then evaluating the discriminant:

a = 9
b = -12
c = 4

(-12)2 - 4(9)(4) = 0

That was easy! Since that means the √(b2 - 4ac) term now becomes √(0), that term disappears entirely, so all we are left with is:

x = -b/2a
Plug those numbers in and we get x = 12/18, which reduces to x = 2/3. Ta-da! Now for our last example.

Solve: 2x2 - 5x + 4 = 0.

As usual I'm going to start with noting what a, b, and c are, then evaluating the discriminant:

a = 2
b = -5
c = 4

(-5)2 - 4(2)(4) = -7

The discriminant is negative here, so that means there are no real solutions. If your course deals with imaginary numbers, you would finish evaluating it anyway (your solutions would have a √(7)*i in them), but otherwise, you would simply note that there are no real solutions, and stop right there.

As always there are a few practice problems below for you to try, to use the quadratic formula. The only real hard part to using the quadratic formula is memorizing it to begin with, and reducing the square roots and fractions when possible. (You can actually reduce the square roots at the end of your solution instead of in the middle - either works, as you'll get the same answers.) Simply plug in the numbers and solve, but don't peek! If the equation has no real solutions, just note that and you can stop.

Solve all of these using the quadratic formula:
:: 3x2 - 17x - 6 = 0
:: 5x2 + 8x - 10 = 0
:: 9x2 - 6x + 1 = 0
:: 4x2 + 15x + 20 = 0

Spoiler

:: b2 - 4ac = 361; x = -1/3, 6
:: b2 - 4ac = 264; x = (-4 - √(66))/5, (-4 + √(66))/5
:: b2 - 4ac = 0; x = 1/3
:: b2 - 4ac = -95; no real solutions

Conclusion

That's about all for factoring for now, actually! Problems like these take many shapes and variants, but in each case the same principles apply. All that differs is what kind of steps to take and what kind of formulas to use, but at this stage, you should be very well equipped to tackle any kind of factoring problem that comes your way!

Next tutorial planned: Piecewise functions

 

Piecewise functions
Requested by @[member="Azure"]

(Note: This tutorial on piecewise functions, like all my others, will be mostly text-based. As I believe that a subject like this is best illustrated with images as well, I will try to find some images to accompany it, and where I cannot find adequate ones, I will describe them as best I can.)

Introduction

A function is a set of points plotted on a graph, usually connected by a line or curve, across its given domain. Usually they're expressed as functions of another variable, with f(x) (or y) being the most common: f(x) = 2x + 3, f(x) = sin x, and even f(x) = 3 are all examples of functions - you plug x into the right side of the equation (input) to get another number on the left side (output).

A piecewise function is like that too, except it's comprised of 2 or more 'sub-functions', which are just like any ordinary functions, except only defined for a specific portion of its domain. Each sub-function is given separately in an ordered list, along with which x values it will correspond to.


Reading piecewise functions

The most well-known example of a piecewise function is the absolute value function, y = |x|, shown below:



As you can tell by the graph above, the absolute value function will always return a positive number; if your x is positive, then your y will be the same as that, but if your x is negative, the function will return the positive form of that number.

When expressed in piecewise function terminology, the function looks like this:



If the image doesn't show up very well:

Code:

        -x, if x < 0
|x| = {  x, if x ≥ 0

(Usually, the left side is written as f(x) instead, to avoid confusion.)

This is the common form for most piecewise functions, and it actually functions similarly to a lookup table. When you're given the formula for a piecewise function, and you want to find its value at a certain x-value, use the following steps:

:: First, locate the x-value that you're trying to find the output for.
:: Look to see which sub-function's domain that x-value applies to.
:: Simply substitute the x in to that part of the piecewise function to get your variable.

Let's try this with an example problem here, with an example piecewise function:

Code:

         x^2, if x < 0
f(x) = { 0.5x, if 0 ≤ x < 6
         2x - 9, if x ≥ 6

Given the above piecewise function, find f(-2), f(4), and f(8).

It seems we have our piecewise function here above. So far the problem is just asking us to plug in 3 x-values and give the corresponding y-values.

Now, first note that this piecewise function is comprised of 3 separate sections, which have domains of x < 0, 0 ≤ x < 6, and x ≥ 6. Notice that there are no overlaps whatsoever between those three domains (which I will discuss in short after this problem.)

The first one we need to find is f(-2). Let's follow the example steps. The x-value that we're dealing with is -2. That falls into the first domain, x < 0, so over that stretch, the sub-function is f(x) = x2. When we plug in -2 to that, we get (-2)2 = 4, so that means that f(-2) = 4.

Next we have f(4). The x-value 4 is in the second domain of 0 ≤ x < 6 (4 is between 0 and 6), so the sub-function to plug it into is the one for that domain, 0.5x. Plugging 4 into that gives 0.5(4) = 2, so f(4) = 2.

The last one is f(8). Our x-value now is 8, and 8 falls into the final domain of x ≥ 6, which is matched up with the sub-function 2x - 9. When we plug 8 in, we get 2(8) - 9 = 7, so f(8) = 7. Not too bad, right?
This method can also be used when given the actual graph of a piecewise function, when the axes are labeled. In that case, it isn't much different - find the x-value on the x-axis, and see what y-value it corresponds to. Whether or not it's graphed out, or you just have the raw formula, you can come to the correct solutions either way.

But there is one more part to this example that I'd like to illustrate.

Find the domain and the range of the above piecewise function.

To find the domain and range of a piecewise function, you need to look at the respective domains of all of its sub-functions, then find the ranges of those sub-functions within that area. While the domain bit of the problem is easier, the range bit is a bit harder. I'll illustrate.

First, the domain of the entire piecewise function is going to be the sum of all of its sub-domains. Easier said than done, but bear with me. As before, we still have the 3 sections of x < 0, 0 ≤ x < 6, and x ≥ 6. If you plot that out, you will find that, between those 3, they include all real numbers (the first domain includes every negative number, the second one includes 0 and all positive numbers up to but excluding 6, and the third one includes 6 and every positive number beyond it), so the domain of this function will be (-∞, ∞), or you can write it as ℝ, i.e. all real numbers.

Similar to the domain, the range of the piecewise function is going to be the sum of all of its sub-ranges. This one is a bit tricky, as it requires knowledge on how each sub-function behaves (like how it increases and decreases, for example), so this bit is usually easier done when having a graph as an assist. Let's go over these functions one at a time.

First, we have f(x) = x2. That function is a parabola, that has its lowest point at 0, and then continuously increases in either direction. When x < 0 for it, it will start at the origin (though not include it) and then continuously increase off to the left forever. So, in this case, the range of the first sub-function will be (0, ∞) - notice that 0 has a parenthesis next to it, because the sub-domain is x < 0 - it does not actually include 0.

The next one is 0.5x. This one is a bit easy as it's linear, continuously increasing at a rate of 0.5, within its sub-domain. There are no weird tricks in between, so to find the range, for this, you just need to look at the points on either end - that is, find f(0) and f(6). f(0) is 0, and f(6) is 3, so the second bit is going to only have y-values around there. Its range will therefore be [0, 3) - again, 3 has a parenthesis, because in the second domain, 6 is not included.

The last one is 2x - 9, over its subdomain x ≥ 6. You can handle this just like the last one; at its starting point, f(6), it's going to be 3, and will then continuously increase from there on out at a rate of 2. The range for the third sub-function will thus be [3, ∞).

Lastly, we just sum the ranges together and see what we get. We have:
(0, ∞)
[0, 3)
[3, ∞)

When all of that overlaps, the range of the entire piecewise function becomes [0, ∞). Note that this isn't the range of the first sub-function - that didn't include 0; the entire function will. Finding the range is harder for some functions than it is for others, but usually what you need to know is the endpoints on its sub-domain, plus how the function itself will behave within it (e.g. if it has any points that are higher or lower than on the ends).

Before going on, there are a few important rules about piecewise functions to consider.

:: The sub-functions and sub-domains of a piecewise function can use either an equality (i.e. an equals sign, =), or any or all of the inequalities (<, >, ≤, ≥, ≮, ≯, ≠), since piecewise functions are defined over domains, though they can also be defined over specific points. (Note that ≮ and ≯ are equivalent to ≥ and ≤ respectively, so you won't see those very often.) There are restrictions, however:

:: When establishing the sub-domains and sub-ranges, and combining them together, each x-value must produce exactly 1 distinct y-value, and there is no overlap permissible. If the domains will overlap, and multiple sub-functions will overlap the same x-value, they all have to produce the same y-value from that x-value. For example:

Code:

f(x) = { x^2, for x < 3
         2x, for x > 0

...would not be permissible, because for every value within the interval (0,3), the function will produce 2 different y-values from each x-value (with the coincidental exception of f(2)), so this is not a valid piecewise function. However:

Code:

f(x) = { x^2, for x ≤ 2
         2x, for x ≥ 2

...this would work, because although both sub-domains claim x = 2, each sub-function will return the same y-value at f(2). (Writing sub-functions like this isn't always good practice, unless it's known for sure that they will overlap, and only at the 1 specific point in which they would change over.) The one piecewise function in the example did have overlaps at the changing points, and thus all of the < and > could be replaced with ≤ and ≥ and yet remain unchanged, but this will not always be the case.

:: It is not required for a piecewise function to be continuous (i.e. an entirely solid line with no breaks) over its entire domain. For example, take a look at this:



As you can see the function is continuous everywhere, except for that one point, in which it suddenly skips up a bit. This is still a valid piecewise function; just so long as the domains are clearly defined, and there is no overlap of having multiple ys to one x, it is acceptable. The gaps do not necessarily have to be at the changing points, either.

:: (This applies mostly for people dealing with calculus, and this won't come up very often Most piecewise functions - though not all of them - are not differentiable at the changing points, since there is a sharp change and there will not be a tangent line. When you differentiate a piecewise function, its derivative would also be a piecewise function. For instance, consider our example problem above:

Code:

         x^2, if x < 0
f(x) = { 0.5x, if 0 ≤ x < 6
         2x - 9, if x ≥ 6

Its derivative would be this:

Code:

          2x, if x < 0
f'(x) = { 0.5, if 0 < x < 6
          2, if x > 6

Notice that the numbers 0 and 6 are no longer in the domain. The function will also skip at each of those points (the original did not), but it's still a piecewise function. (If you aren't in calculus, don't worry about this section. It's not too important.)

To wrap up this part of the section, here are some practice problems for you to work with. In each example, first identify if the function in question is a valid piecewise function (you want to look to see if an x-value can correspond to multiple y-values). If it isn't, you can stop. If it is one, solve for the outputs it gives you (if possible), and - if you want - try and figure out its domain and range as well.

I'll include the answers below, but don't peek until you've worked on it for yourself.

Code:

         x - 5, if x < 0
f(x) = { 5, if x = 0
         x + 5, if x > 0

Find f(-2), f(0), and f(2).

Code:

f(x) = { 3x, if x = 0
         x, if x < -2

Find f(-2), f(0), and f(2).

Code:

         2x, if x < 2
f(x) = { x + 2, if x > 1

Find f(-2), f(0), and f(2).

Code:

         0.5x + 5, if x ≤ -4
         -2x - 5, if -4 ≤ x ≤ -2
f(x) = { 0.5x, if -2 ≤ x ≤ 2
         -2x + 5, if 2 ≤ x ≤ 4
         0.5x - 5, if x ≥ 4

Find f(-3), f(-1), f(1), and f(3).

Spoiler

:: f(-2) = -7; f(0) = 5; f(2) = 7; domain = ℝ; range = (-∞, -5) U [5, ∞)
:: f(-2) doesn't exist; f(0) = 0; f(2) doesn't exist; domain = (-∞, -2) U {0}; range = same as domain
:: Not a valid piecewise function
:: f(-3) = 1; f(-1) = -0.5; f(1) = 0.5; f(3) = 1; domain = ℝ; range = ℝ

Constructing piecewise functions

This section will have a few examples on how to take certain word problems that can be translated into piecewise functions to solve. There won't be sample problems for you to try at the end, but it will cover a few common examples, and show you how to set them up, and go from there.

(If any of you have examples of word problems on piecewise functions you'd like to contribute that aren't covered here, let me know and I'll add the problem in to your credit.)


Cooking temperature:

Spoiler

A certain brand of homemade cookie mix involves one step of its recipe placing the cookie dough on a pan, and heating it in an oven for 350 degrees Fahrenheit for 20 minutes, before removing them to cool. However, it goes on to say that, at altitudes 1000 metres above sea level or more, it should be heated at 300 degrees Fahrenheit for 20 minutes instead. What will the piecewise function for this cookie baking look like?

Let's consider the variables here, and see what changes. Normally we're going to be baking these cookies at 350 degrees Fahrenheit, but at altitudes at least 1000 metres, we'll bake them at 300 degrees F instead. (The baking time is the same, so that doesn't change.) The thing that we can flexibly change based on our own circumstances is the altitude; we can be at any point from -88 to 8800 metres above sea level, depending on our location. That will essentially become our x-value. The thing that changes based upon the altitude is the temperature, our y-value. In essence, altitude is the independent variable, and baking time is the dependent variable.

Now we know which is which. Baking time is normally 350 degrees F, but it's 300 degrees F at 1000 metres or more. Let's start with the 1000 bit, because it's the anomaly here. Now, what sign are we going to use? Are we going to use a >, or a ≥? Note the language of the question: you use the baking temperature at 300 degrees F if you're at altitudes of 1000 metres above or more, so that is going to include the 1000 metres. Thus, the sign we'd be using for this part is a ≥, as below:

Code:

f(x) = { 300, if x ≥ 1000

Note that we didn't use x here in the input, because no matter what x is, as long as it's at least 1000, the output is going to be 300. We've covered that much. But what about the other part? As it said, the normal baking temperature is given as 350 degrees, so that is assumed to be essentially 'everywhere else'. Now, what sign are we going to use? Are we going to use a <, or a ≤? Remember that there can't be any overlap between x-values, and note that if we put ≤, then 1000 would give us two y-values, which would be incorrect. Another way to look at it is to view what the opposite of ≥ is (i.e. ≱). If it's not greater than, and it's not equal to, the only thing left is 'less than', so we'd be using a < in this case:

Code:

         350, if x < 1000
f(x) = { 300, if x ≥ 1000

And that's the piecewise function for our cookie baking! If you actually plot it out, it will look like two straight lines, one with a white hole at its right endpoint, before skipping down to the second line. This is one of those examples in which it's fairly straight-forward and there are no xs to input, but you still have to keep the domains in mind when constructing the sub-functions.

Rental car fare:

Spoiler

A car rental company loans out its cars to customers, and it charges based on how many kilometres the renter drives in that car. There is a flat $1000 fee required to rent the car, and after that it charges 20 cents per kilometre driven. After 1200 kilometres, it starts charging 30 cents per kilometre for any amount driven after that point, and after 3000 kilometres (if anyone reaches that point!), the company charges 40 cents per kilometre after that.
Construct a piecewise function that represents the fee required based on kilometres driven. How much will a customer pay if he drives 800 kilometres before returning the car? if he drives 1500 kilometres?

You can tell this is a piecewise function because the fare keeps increasing the more the car is driven. So, let's look at what our independent (x) and dependent (y) variables would be, first of all. As indicated in the problem, the fee charged is based upon the kilometres driven, so for this, our independent variable (x) is the kilometres, and our dependent variable (y) is the fee charged by the rental company.

So let's see what's what. Right off the bat we know that we're going to have to pay $1000 regardless of how much we drive, so that's going to be part of our first sub-function. After that, we are going to be charged 20 cents, or $0.20, per kilometre driven. So if x is our kilometres, we're going to have this as the first part of the piecewise function:

Code:

f(x) = { 1000 + 0.20x, if __________

...since there's the flat rate of 1000, which increases by 0.20 per kilometre driven. 1000 doesn't have an x by it because it isn't directly based on kilometres, but the 0.20 does. Now, how much does this go up to? The first turning point is going to be at 1200 kilometres, so this sub-function will apply up to that point. Will we be using a < or a ≤ sign? Look at the language of the problem: the turning point is after 1200 kilometres, so when we're at 1200 kilometres, we're still being charged according to the first bit. So we would use a ≤ here, as 1200 is included in the first bit:

Code:

f(x) = { 1000 + 0.20x, if 0 ≤ x ≤ 1200

The next bit has us being charged 30 cents per every kilometre after 1200. What would you think we'd use for this piece? If you guessed 1000 + 0.20x + 0.30x, you'd be wrong, since our rate is going to be different. Rather, we need to do two things. The first is to figure out how much money we would normally pay up to that point, i.e. how much money we would pay at exactly 1200. We do this by substituting 1200 in for x in the first equation and solving, which gives us 1240. So, we're going to get something that looks like this:

Code:

         1000 + 0.20x, if 0 ≤ x ≤ 1200
f(x) = { 1240 + _______, if 1200 < x ≤ 3000

(The language for the 3000 turning point is the same as before, hence the bound.)
Now, what goes in the blank space? If you thought it was 0.30x, you'd be wrong. Why is that? Note that the rental company only charges extra for the kilometres beyond 1200 - the first 1200 is still being charged by the lower rate. So what do we do here? We only have an x...or do we? Since we only need to account for the kilometres after 1200, we can actually use an (x - 1200) expression in the problem. We do this so we only are accounting for what is past 1200, which will be charged for the higher rate. So, our next piece will look like this:

Code:

         1000 + 0.20x, if 0 ≤ x ≤ 1200
f(x) = { 1240 + 0.30(x - 1200), if 1200 < x ≤ 3000

which, when you multiply it out (0.30 * 1200 = 360, and 1240 - 360 = 880), becomes:

Code:

         1000 + 0.20x, if 0 ≤ x ≤ 1200
f(x) = { 880 + 0.30x, if 1200 < x ≤ 3000

Okay, one last part, then we can start plugging numbers in. Now we need to figure out how much we're being charged at the highest rate of all, beyond 3000 kilometres. Just like before, we figure out how much we're being charged at exactly 3000 for the first bit, so we plug 3000 into x into the second equation (not the first one, since that wouldn't be the correct sub-domain) to get an answer of 1780:

Code:

         1000 + 0.20x, if 0 ≤ x ≤ 1200
f(x) = { 880 + 0.30x, if 1200 < x ≤ 3000
         1780 + ________, if x > 3000

And, just like the last bracket, rather than starting at x, we start at (x - 3000) since we only charge past that point. Charging that at the higher rate of 0.40 per kilometre, and putting that into the function, we get:

Code:

         1000 + 0.20x, if 0 ≤ x ≤ 1200
f(x) = { 880 + 0.30x, if 1200 < x ≤ 3000
         1780 + 0.40(x - 3000), if x > 3000

which, when you multiply it out, like before, becomes:

Code:

         1000 + 0.20x, if 0 ≤ x ≤ 1200
f(x) = { 880 + 0.30x, if 1200 < x ≤ 3000
         580 + 0.40x, if x > 3000

And that becomes our finalised piecewise function. Hooray!
Now for the easy part. We just need to know how much we'd be charged at 800 or at 1500 kilometres. Simply find which subdomain those two numbers go into, and solve appropriately, then we're done.

800 is in the first sub-domain:
1000 + 0.20(800) = 1160

and 1500 is in the second sub-domain:
880 + 0.30(1500) = 1330

So, if we rented a car from this company and drove it 800 kilometres in all, we'd have to pay $1160 upon returning it, but if we drove it for 1500 kilometres, we'd have to pay $1330 upon returning it instead.

Toy production:

Spoiler

A children's toy company is planning on putting out a brand new model of toy on the market soon. They are trying to figure out how many units they want to put on the shelves of a certain store to try it out. Because of supply and demand, if they put less units on the shelves, they can charge more for each one, while if they put more units out, they have to charge less for each one. The company is trying to find a balance between the two: charging more per unit, or selling more units overall. Put simply, they want to maximise profit.

Each toy costs $0.50 to produce, and readying the machine that mass-produces them incurs a cost of $100 - these numbers are fixed and are a flat rate, and won't change based on number of units sold.

The company's marketing department has made some forecasts, and has calculated how much they can sell the toys for each, based on how many they put on the shelves:

100 or less - $1.80
101 to 300 - $1.50
301 to 500 - $1.20
501 to 1000 - $0.90
1001 to 1500 - $0.60

The company will not produce more than 1500 units for the initial run. How many units should the toy company produce in order to maximise profit?

Wow. Information overload. Let's tackle these bits one thing at a time. Before we get too far, for those who are not aware, profit is equal to revenue minus cost. (Revenue is how much the company receives when something is sold.)

Let's start with the piecewise bit first, since that is going to have a lot of parts to it. Thankfully each piece is fairly simple, and doesn't really have extra numbers affixed to it. The problem is even kind enough to provide us with the numbers for the sub-domains as well.

(Note: since we are dealing with whole numbers of units for this problem, the phrases 301 ≤ x ≤ 500 and 300 < x ≤ 500 are essentially identical. We aren't dealing with fractions, and neither of those domains include 300. Writing them in the former method is more accurate to the problem, while writing them in the latter method allows you to visualise it as a continuous graph. Either is acceptable. I'll be writing them in the former method, just so I don't confuse you.

Okay, so we can convert the revenue into this:

Code:

         1.80x, if 0 < x ≤ 100
         1.50x, if 101 ≤ x ≤ 300
f(x) = { 1.20x, if 301 ≤ x ≤ 500
         0.90x, if 501 ≤ x ≤ 1000
         0.60x, if 1001 ≤ x ≤ 1500

The reason the first one is a < instead of a ≤ is because, if you aren't producing any toys, you won't turn the machine on.

Don't start plugging in numbers yet, because we have only accounted for the revenue here. We still have yet to account for the cost, which we'll need to figure out the profit. Now, what does the problem tell us? It takes the company $0.50 per unit for each toy, but also, activating the machine also costs $100, and they need to do both of those things to produce the toys. The cost is actually not a piecewise function at all; it's expressed as f(x) = 0.50x + 100 (the cost per unit plus the flat rate to activate the machine).

The next step is to derive the profit. Profit is revenue minus cost, so to get the function for profit, we need to take the revenue function, and subtract the cost function from it - and, in the case of a piecewise function, you have to subtract it from each sub-function. Thankfully the cost isn't piecewise so the subtraction will be consistent throughout, but if the cost were also piecewise, you'd have to subtract appropriately, with the given sub-domains. (In fact, if the sub-domains of the two don't line up, your result will have even more sub-functions than before. Not fun.)

So for this, we'll have to do (1.80x) - (0.50x + 100) for the first piece, (1.50x) - (0.50x + 100) for the second piece, and so on. When we do that, we'll get this:

Code:

         1.30x - 100, if 0 < x ≤ 100
         1.00x - 100, if 101 ≤ x ≤ 300
f(x) = { 0.70x - 100, if 301 ≤ x ≤ 500
         0.40x - 100, if 501 ≤ x ≤ 1000
         0.10x - 100, if 1001 ≤ x ≤ 1500

Now, how do we find the point at which profit is maximised? You can graph each bit out if you want, but the easiest way to find out is to calculate the endpoints. As all of these functions is linear and increasing, the answer will be at the rightmost endpoint of one of the sub-domains. This kind of function is an example of what is called a 'ceiling function'. (If it were decreasing instead, the answer would be at one of the leftmost endpoints. That would be an example of a 'floor function'.)

So let's plug some numbers in. The rightmost endpoints are 100, 300, 500, 1000, and 1500. We know it's going to be one of those, so let's plug in the numbers for the appropriate subdomains, and see what we get:

1.30(100) - 100 = 30
1.00(300) - 100 = 200
0.70(500) - 100 = 250
0.40(1000) - 100 = 300
0.10(1500) - 100 = 50

You could try to plug in the numbers for the left endpoints as well, but they're going to be lower than anything else in the subdomain, so those answers won't work. Nonetheless, I'll include them here for comparison's sake:

1.30(1) - 100 = -98.7
1.00(101) - 100 = 1
0.70(301) - 100 = 110.7
0.40(501) - 100 = 100.4
0.10(1001) - 100 = 0.1

See how they are all substantially less than the right endpoints; you have profits here of $1, $0.10, and even a loss of $98.70! (If you want to be fun and graph or plug the numbers in yourself, you'll see that the company has to make at least 77 units if it wants to make any profit at all.)

The last thing we need to do is just compare the numbers and see which one is highest. We see that the highest result number here is 300, which happens when the toy company produces and sells 1000 units. Thus, our final answer for this problem is that the toy company should produce 1000 units (and sell them at $0.40 apiece) in order to maximise profit. A bit of a hassle overall, and toying with bits and pieces, but the problem is now finished!

Conclusion

The piecewise function tutorial isn't particularly long (yet), largely because it's simply a step up from existing functions - the only difference is that you're taking bits and pieces of various other functions and putting them together, which can be very tricky to draw and figure out - especially when you need to figure out the domain or range of the new piecewise function, then it gets messy. Some people will do better visualising piecewise functions by actually graphing them out; others will be fine simply by being given the formula and going from there. I hope this tutorial was helpful to you!

 

Linear & exponential growth and decay
Requested by @[member="Azure"]


Introduction

When people speak of growth and decay, they usually speak of an object with a quantity that can increase or decrease over the course of time. A few common examples of this would be money in a bank account, population of a species in an environment, or the half-life of a radioactive isotope. Math courses love using these as examples, but probably because they're the easiest examples to use to demonstrate a point. Growth is the steady increase of a quantity of some object over time, whereas decay is a steady decrease of such.
Commonly these equations can be expressed graphically, though you don't need a graphing calculator or to draw the graph either, as the equations are somewhat simple. You will need a calculator all the same, though, just because of the complexity of the equations. As alluded to above in the title, there are two main types of growth/decay patterns: linear and exponential. (There are other growth models as well, but these are two that increase, or decrease, at a predictable rate.) I'll begin with the simpler linear growth model at first, then move on to the exponential model.

Note: the formulae that I use are essentially the same as those taught commonly in textbooks. A main difference you might see is that I might use different letters for the variables, e.g. some textbooks use k for rate instead of r, and/or use P for quantity instead of A. Either way, as long as the numbers you plug in are correct, they will operate the same way - the letters chosen don't matter too much.


Linear growth model

The linear growth model is appropriate for quantities and objects that increase or decrease at a constant rate. No matter at what time you will see the thing, the rate of change will always be the same. It's called the linear growth model because, when it's expressed as a graph over the passage of time, it will be a line.

A linear growth model standardly has this form:

A(t) = A0 * (1 + rt)

These variables stand for the following:
A(t) is the population of the object at the time 't'. It is sometimes written just as A.
A0 is the starting amount/quantity of whatever object you're analysing. It is either pronounced 'A-zero' or 'A-naught'.
r is the rate of change of the object's amount/quantity, expressed as a decimal. It will always gain (or lose) quantity at the same constant rate.
t is how much time has elapsed since you've started observing the object.

There are two important notes to keep in mind, though - one about the variable r, the other about t.

First, r is always expressed as a decimal. If the problem gives r to you as a percentage, convert it to a decimal to get your figure. Additionally, based on r's value, it will tell you what kind of model you're dealing with:
:: If r is positive, then you're dealing with linear growth. The slope of the line will be positive. For example, if something is increasing at a linear rate of 5% growth, then r will be 0.05.
:: If r is negative, then you're dealing with linear decay. The slope of the line will be negative. For example, if something is decreasing at a linear rate of 3% decay, then r will be -0.03.
:: If r is 0, there is no change at all. You're dealing with a horizontal line; the slope will be 0. (To see for yourself, plug in 0 into r in the above equation. A(t) will always be equal to A0 and there will be no change at all.) And, if there's no change, it is neither growth nor decay.
:: The slope of a linear growth or decay line will always be A0 times r.

A tricky part about growth and decay problems (both linear and exponential) is converting r into decimal form. Some formulae will have you enter the rate in as 1.05 or 0.97, but this is uncommon; the "1 +" portion is usually separate, as it is in my example above.

Second, concerning t... t can be measured in any unit of time (minutes, hours, days, weeks), but you must establish clearly what unit of time that t is counting. For example, if t is being counted in minutes, but you want to find the population after 2 hours, then t would be 120, not 2, because there are 120 minutes in 2 hours. Similar conversions have to be made if you're trying to convert one value of time into another. Here are some common ones (along with a few not-so-common ones), ranging from smaller to larger:

60 seconds = 1 minute
60 minutes = 1 hour
3600 seconds = 1 hour
24 hours = 1440 minutes = 86400 seconds = 1 day
7 days = 168 hours = 1 week
1 month = 30 days*
1 month = 4 weeks*
3 months = 1 quarter
365 days = 52 weeks = 12 months = 4 quarters = 1 year
10 years = 1 decade
100 years = 10 decades = 1 century
1000 years = 100 decades = 10 centuries = 1 millennium

*When converting days or weeks into months, it depends on the context of the problem. Standardly, if it wants you to convert months into days or weeks, or vice versa, go with what kind of unit is actually given in the problem. See if it says one of these things, or its reverse:

:: "If [population] grows [this much] in a week, how much does it grow in a month?" - Use the 1 month = 4 weeks conversion factor.
:: "If [population] grows [this much] in X days, how much does it grow in a month?" - Use the 1 month = 30 days conversion factor.

Similar discretion must be used with converting weeks or months into years, but this is less common and also more accurate than the above. However, never ever use 4 weeks = 30 days as a conversion factor. When converting between weeks and days, always use 7 days = 1 week.

But you didn't come here to read about unit conversion (or did you?), so I will simply take the above information and move on to some examples. However, before I do, know that there are a few things in common to any growth or decay problem: when it gives you the formula, i.e. A(t) = A0 * (1 + rt) , it will only leave 1 of those values unknown, and give you all the others. (That is, out of A(t), A0, r, or t, it will only leave out one of those. It's usually t, but it can be A(t). Very rarely will it be r or A0.)

Now, let's tackle a problem and see how we go about this.

A farmer has a corn field that currently has 400 fully grown corn plants in it, and all are productive. He wants to plant more corn in the field but he doesn't want to overdo it. He determines to plant an amount each year equal to 8 percent of how many corn plants he has right now. How many corn plants will he have after 5 years (provided none of them die)? How long will it take him to have double the number he has now? (Round to 2 decimal places where appropriate.)

I was really tempted to have a problem about rabbits, but I'll save one of those for later. For now, this example. How do we know it's a linear growth problem? Look at the language used: it says each year, he'll plant new corn equal to 8 percent of what he has right now - not 8 percent of whatever he starts that year with. That means that the number of new corn plants he'll get will be constant. Thus, this is a linear growth equation.

Let's start with our example formula.

A(t) = A0 * (1 + rt)

How many of these do we already know, and which ones will we have to figure out? We know what A0 is; it's however many we're starting out with, and that's given as 400. We also know what r is: it's 8 percent, so expressed as a decimal, that's 0.08. We know this is linear growth, so it's positive. (If it were decay, it'd be -0.08.) Also note that it's 8 percent per year, so that tells us that t is going to be measured in years. So we can plug those in...

A(t) = 400 * (1 + 0.08t)

This example has 2 problems: one of them gives us t and we need to find A(t); the other gives us A(t) and we need to find t. Let's do the first one.

How many corn plants will he have after 5 years (provided none of them die)?

The time it gives us is already in years, so we don't need to convert it to some other number. t = 5 (years), and we need to find out how many corn plants he'll have after 5 years. So we just plug in the 5...

A(5) = 400 * (1 + 0.08*5)
A(5) = 400 * (1 + 0.40)
A(5) = 400 * (1.4)
A(5) = 560

Thus, the farmer will have 560 corn plants after 5 years. Not too bad, right? The other one is a bit different, but solved in a similar way.

How long will it take him to have double the number he has now? (Round to 2 decimal places where appropriate.)

In this case we're given the output, but not how much time it takes to get there. He has 400 corn plants right now at the start, that much we know, and double that is 800. So we plug that 800 in, and solve for the variable we don't know, namely t:

800 = 400 * (1 + 0.08t)
(Divide both sides by 400)
2 = 1 + 0.08t
1 = 0.08t
t = 1/0.08
t = 12.5

So, it will take 12.5 years for the original stock of 400 corn plants to double in number. You might have noticed a pattern there, but if not, I'll just say it: in a linear growth, the amount of time it will take for the original to double in size is always 1/r. You don't even need to know the original quantity either. I'll demonstrate:

A(t) = A0 * (1 + rt)

So we're back where we started, and I want to find out how long it would take a population to double in size. If A0 is how much we start with, then double that will be 2 * A0. Now if I plug that in...

2 * A0 = A0 * (1 + rt)

...and then divide both sides by A0 (which is permissible, since it's always a positive number):

2 = 1 + rt

...then the rest becomes easy.

1 = rt
t = 1/r

A similar principle can actually be used for linear decay functions, in which you want to see how long it takes for something to halve in size:

0.5 * A0 = A0 * (1 + rt)
0.5 = 1 + rt
-0.5 = rt
t = -0.5/r

Now, you might think that's a negative number, but remember, in linear decay models, r is a negative number. Thus you get a negative over a negative, which becomes a positive. Those are two pretty handy shortcuts to have on hand, right? Exponential growth and decay have shortcuts like that as well!

But I'm getting ahead of myself. Let's do a linear decay example.

Unlike the farmer we just talked about, the other farmer down the road decided to use all of his available acreage to plant potato plants, and he managed to plant a whopping 900 potato plants. Things are going well for him, until he realizes he's basically sucked the soil dry of nutrients, and now his plants are dying out along with the soil. He's going to lose 6 percent of the amount of crops he has right now, each year. How many potato plants will he have remaining after 4 years? How many years until he only has half as many? How many years until he runs out? (Round to 2 decimal places where appropriate.)

As anyone who is familiar with planting crops will know, you have to rotate the fields every so often to keep the nutrients fresh, as certain crops will draw in certain nutrients from the soil while giving them other ones, and it's different for each crop. (If memory serves, the rotating crop system is to have a third of the field with 1 crop, another third with a different crop, and the last third fallow, and to change this every year. Don't quote me on that, though. I don't know the details.)

In any case this potato farmer is foolish and he's now paying the price for it. Now, time to start plugging in numbers again.

A(t) = A0 * (1 + rt)

A0 is our starting amount, which in this case is 900 plants. Our rate is given in the problem as losing 6 percent each year (hinting that t is going to be measured in years), so r is going to be -0.06 (remember, it's decay, we're losing things, so r is negative). And the rate is 6 percent of whatever he has right now, so it's linear decay.

A(t) = 900 * (1 - 0.06t)

Now we can tackle these problems one at a time, by plugging in the appropriate number.

How many potato plants will he have remaining after 4 years?

t is measured in years so we don't need to do any extra conversions, thankfully. So, after 4 years, what do we have? We plug in the 4 and...

A(4) = 900 * (1 - 0.06*4)
A(4) = 900 * (1 - 0.24)
A(4) = 900 * (0.76)
A(4) = 684

After 4 years, he'll only have 684 of his original 900 plants left. Not good for him.

How many years until he only has half as many?

I'm going to solve this part of the problem in two ways - the first way I'm going to use our little shortcut we discussed earlier, and the second way I'm going to plug in all the other numbers and go that way. Now, first, our little shortcut:

t = -0.5/r

For this, all we need to know is the rate, which is -0.06. So we just plug that in...

t = -0.5/-0.06 = 8.33333...

Therefore it would take 8 years and 4 months (1/3 year) for him to lose half his plants at that rate. Now for the old-fashioned way...

A(t) = 900 * (1 - 0.06t)

Half of our initial amount is 450 - half of 900. That's going to be the output, but we don't know how much time it will take. We'll plug in 450 for A(t), and we should end up with the same number as we did with the shortcut. Let's see...

450 = 900 * (1 - 0.06t)
(Divide both sides by 900)
0.5 = 1 - 0.06t
-0.5 = -0.06t
t = -0.5/-0.06
t = 8.33333...

Yep, exact same result. Note that particular shortcut only works with halving, but it will always work and produce the correct result. (Other fraction sizes have different shortcuts, though they are derived similarly - for example, to derive your shortcut for how much time it takes to quarter the original amount, take the original formula and substitute 0.25 * A0 in for A(t), then solve for t.)

How many years until he runs out?

As a note, before I solve this, running out of something is standardly only possible with linear equations, and doesn't even come up very often. It is not possible for something to run out entirely that has an exponential decay function (but I'll get to that later) - though it can become extremely small.

Now, in this case, he's run out of all of his plants (provided he just sits on them and does nothing to improve his situation - foolish man), so his A(t) at whatever time has become 0. We can plug that in...

0 = 900 * (1 - 0.06t)
(Divide both sides by 900)
0 = 1 - 0.06t
-1 = -0.06t
t = -1/-0.06
t = 16.6666...

Thus it will take 16 years and 8 months (2/3 year) for all of his potato plants to die out. (Notice that is exactly double the time it takes for the crop amount to halve. That is not a coincidence; it is inherent to the linear decay model.)

Similarly, the shortcut to see when something runs out entirely for linear decay is t = -1/r. It's essentially the opposite of how much it takes something to double...which makes sense, because when something is doubled, you're gaining 100% of it, but when you lose all of something, you're losing 100% of it.

Nothing too strenuous, right? You just start with some numbers and plug in some other ones to find the missing ones. (Well, that was fantastically vague of me.) As a note, non-compounded interest also follows a linear growth curve with much the same formula (they might include an n part as well to convert units into years, but it's essentially the same). Most interest problems, however, are compound interest...which I'll get to later.

Before we move on to the exponential growth and decay models, let's give you some examples to work with. As always I'll include the answers in a spoiler tag below, but please don't peek until you're done!

:: Starting amount is 1500; growth rate is 4% per week. How much does it have after 2 months? How long does it take to reach 2700?
:: Starting amount is 3000; decay rate is 2% per day. How much does it have after a week? How long does it take to halve in size?
:: Starting amount is 2400; growth rate is 6% per month. How much does it have after one quarter? How long does it take to double in size?
:: Starting amount is 5000; decay rate is 1% per hour. How much does it have after a day? How long does it take to reach 2000?

Spoiler

Yeah, I threw in a new twist; you have to convert the time into a different unit for each one!
:: 1980; 20 weeks
:: 2580; 25 days
:: 2832; 16.6666... months
:: 3800; 60 days

Introduction to exponential growth and decay

So now that we're done with linear growth and decay, exponential growth and decay is probably a lot scarier, right? Well, not necessarily. Exponential growth and decay models are still functions, with formulas structured like before, except the rate of change is no longer constant; the rate itself steadily rises in intensity. With exponential growth, it becomes incrementally larger, whereas with exponential decay it becomes incrementally smaller. All of the formulas for various types of exponential growth and decay have one thing in common: the t (that is, the units of time we're measuring) is found in the exponent of the basic formula.

If you want a picture of what exponential growth looks like, compare the functions f(x) = x2 and f(x) = 2x, starting at 0. The latter one, 2x, is an exponential function, and if you plot the first few integers of each one, starting at x = 0:

x2: 0, 1, 4, 9, 16, 25, 36
2x: 1, 2, 4, 8, 16, 32, 64

...notice that, while x2 initially has the lead, 2x eventually far surpasses it (unlike the parabola's steadily increasing growth, the exponential function doubles every time), and just keeps going beyond that. While we won't deal with anything quite as drastic, that should give you an idea of what kind of growth you're dealing with.

Exponential decay is a bit smoother, and not quite as drastic, as linear decay. One important thing to note with exponential decay is that, no matter how small of a quantity you get, it will never reach exactly 0. (It will come very close, though not actually be 0.) If you want a picture of exponential decay, compare the functions f(x) = (1/x)2 and f(x) = (1/2)x. I'll take the first few integers of each, starting at x = 1, and round to 4 places for each one:

(1/x)2: 1, 0.25, 0.1111, 0.0625, 0.04, 0.0278, 0.0204
(1/2)x: 0.5, 0.25, 0.125, 0.0625, 0.0313, 0.0156, 0.0078

The trend here is similar - initially the first function has the lead with its steadily increasing decay, but it's quickly overtaken by the exponential function, which halves every time. This should give you an idea of what to expect - exponential growth and decay is incremental, but quite quick. (The prior examples had bases of 2 for growth and 0.5 for decay; normally they won't be much more than 1.2 or much less than 0.8, but this is just an example to show how exponential growth rate outspeeds other kinds of growth rates.)

I'm going to go over three often-used formulas for exponential growth and decay: basic growth/decay models, compound interest, and continuous growth/decay models. Each of those will smoothly transition into the next one, building off what was already covered. Also, note that while basic and continuous models can exhibit growth or decay, compound interest is a growth model only. It's actually a specific application of the basic model that's used in the business world...but I'll get to that later.


Basic exponential growth/decay models

Just like linear growth and decay, exponential growth and decay follows its own distinct formula. The basic formula for exponential growth or decay is as follows:

A(t) = A0 * (1 + r)t

where A(t), A0, r, and t mean exactly the same things that they meant for linear growth/decay (current quantity at time t, initial quantity, rate of change as a decimal, and amount of time since the start respectively). The restrictions for r still apply as well: if r is positive it's growth; if r is negative it's decay; and if r is 0 there's no change at all.

You've probably noticed that there is only one difference between the linear formula and the exponential one: where the t is placed. In the linear it's multiplied by r, while in the exponential, it's actually the exponential term itself. But what does that difference actually mean? What does it equate to, and how does it change things? I think the best way to show the differences is via an example.

A farmer has a corn field that currently has 400 fully grown corn plants in it, and all are productive. He wants to plant more corn in the field but he doesn't want to overdo it. He determines to plant an amount each year equal to 8 percent of how many corn plants he has at the start of that year's growing season. How many corn plants will he have after 5 years (provided none of them die)? How long will it take him to have double the number he has now? (Round to 2 decimal places where appropriate.)

Wait. Isn't this the exact same example we had to demonstrate the linear model? Yes, it is. The only difference is the bold text you see - instead of planting 8 percent based on how much the farmer had at the very start, he's planting 8 percent based on how much he started that year with - which is always going to be more than the previous year. That change alone means that we're dealing with an exponential problem.

So, let's get to work. For our old frame of reference, in the linear problem, he had 560 plants after 5 years, and it took him 12.5 years to double up to 800 plants. Let's see how those numbers change, shall we? Oh, and before I forget, to solve any exponential problem, you will need, at minimum, a scientific calculator. This is because your calculator needs to have at least 2 specific buttons:

:: The ^ or yx button. (Those are 2 different names for the same thing.) It lets you raise numbers to specific exponents. You can get away with this on a basic calculator simply by multiplying the same number by itself multiple times (by hitting the * button repeatedly), though this will not work if you have to deal with fractional exponents. (Fractional exponents don't come up very often, though.) Doing this on a basic calculator is also much more tedious.
:: Either the log (logarithm base 10) or the ln (natural logarithm, base e) button. One of these is absolutely mandatory for any exponential problem that requires you to solve for t, e.g. "How much time does it take for [object] to reach [quantity]?". Basic calculators do not have either of these buttons. The ln button is usually preferred, as its numbers are easier to work with. The log button can work in some instances, as the two have similar properties, but it will not work with continuous models, which will come later. Thus, for consistency, ln will be used throughout this tutorial.

Now then. Get your scientific calculator and try to follow along with me.

How many corn plants will he have after 5 years (provided none of them die)?

Let's look at our starting formula, and start plugging the appropriate numbers in.

A(t) = A0 * (1 + r)t

Our initial amount is still 400 corn plants, so that becomes our A0. Our rate is still 8 percent, or 0.08, so that becomes our r. The only difference is we're using the exponential model instead of linear. Let's plug these numbers in:

A(t) = 400 * (1 + 0.08)t
Simplify the parentheses:
A(t) = 400 * (1.08)t

We need to find A(t) when t is 5 years, so...

A(5) = 400 * (1.08)5
A(5) = 587.73

So the farmer has 587.73 corn plants after 5 years. That's really weird to imagine, so imagine that he has 587 complete plants, and one that's three-quarters fully grown. (You'll pretty much always end up with decimals when doing exponential problems, and will have to round. What the decimals mean in context depends on the problem, though you probably won't have to worry about that too much.)

Notice in the linear problem he had 560 after 5 years. With the exponential model he has 587. Not a drastic amount more, but it's something. If you graphed the lines out, you'd see a straight line going upward at a constant rate (the linear model), and another line that starts at the same place but creeps up and up above it a little faster, in a steady curve (the exponential model).

How long will it take him to have double the number he has now?

Same kind of deal as before:

A(t) = 400 * (1.08)t

We're given the ending amount of 800, and we need to find how much time it takes for the guy to reach that point. You've probably predicted that, since we got more crops in 5 years with the exponential model, it'll probably take us less time to double our crops. Well, let's see if that prediction is correct. First we plug in the ending amount:

800 = 400 * (1.08)t
2 = (1.08)t

...now we run into a problem. We need to solve for t but it's stuck in the exponent. How do we extract it? The solution: logarithms. You might remember logarithms of various bases (log5 25 = 2), or how to combine them together (log a + log b = log ab), and how they can be used to manipulate exponents. We're going to do that exact same thing here.

When dealing with exponential equations, there is one rule of logarithms you need to remember:

log (ab) = b * log a

...and when converting an exponent out of (or into) an expression, this is the rule you use. This is also the rule we're going to use to extract that t out of the exponent. Note that the logarithm can be any base (not just base 10), so for simplicity's sake, any time I am going to use a logarithm in a problem, it is going to be ln, the natural logarithm (base e).

Returning back to our problem...

2 = (1.08)t

The next step to do is to take the natural logarithm of both sides of the equation, like so:

ln 2 = ln (1.08t)

Note the rule above. The next thing we're going to do is to extract that t out of the exponent, and bring it outside:

ln 2 = t * ln 1.08

Finally, divide both sides of the equation by ln 1.08. (Remember that ln is a function that is defined for all positive numbers, so this is permissible. However, you cannot divide both sides of an equation by ln 1, because ln 1 = 0. But you'd never get ln 1 in these equations because it means nothing would change.) In any case...

t = ln 2/ln 1.08

And that's why you need the scientific calculator, to get you the actual number. When rounding to 2 decimal places, we get:

t = 9.01

That means it will take the farmer just a smidgen over 9 years to double his initial corn crop count, if he plants his crops exponentially. That is a massive difference - it takes him 3.5 entire years less to double his crop than he would do so planting at a linear rate. Every little bit adds up, and this is a prime example of that.

But will exponential decay exhibit a similar pattern? Let's find out with another example.

Unlike the farmer we just talked about, the other farmer down the road decided to use all of his available acreage to plant potato plants, and he managed to plant a whopping 900 potato plants. Things are going well for him, until he realizes his crops have caught a blight, and are starting to die out. He's going to lose 6 percent of his remaining crops each year. How many potato plants will he have remaining after 4 years? How many years until he only has half as many? How many years until he runs out? (Round to 2 decimal places where appropriate.)

I must really like to torment this guy. Thankfully he's fictional, so he feels no pain. Now instead of barren soil, we're dealing with a potato blight (which, as people aware of history may remember, caused a huge exodus from Ireland in the mid-19th century - not fun). And he's going to lose 6 percent of what's left of his crops each year if he doesn't do something.

So let's start with our basic exponential model again, and plug in some numbers.

A(t) = A0 * (1 + r)t

Our starting amount is still 900, and our rate is still -0.06 (negative, remember, as it's decay). We plug those in.

A(t) = 900 * (0.94)t

With linear decay, the guy had 684 plants after 4 years, and it took him 8.3333... years to lose half his crops, and 16.6666... years to lose all of them. Let's see how those numbers change.

How many potato plants will he have remaining after 4 years?

t is 4 years. So all that needs to be done is plug in t, and solve with a calculator:

A(4) = 900 * (0.94)4
A(4) = 702.67

So after 4 years, he has 702.67 plants left...he actually has more remaining under exponential decay than he does linear.

How many years until he only has half as many?

You might be thinking that, since in 4 years he didn't lose as many plants as he did from the linear decay, it will actually take more time for him to lose half his crops. Let's see how that turns out. Half of his initial amount is 450, so let's go from there...

450 = 900 * (0.94)t
0.5 = (0.94)t

Take the natural logarithm again...

ln 0.5 = ln (0.94t)

...then extract the exponent.

ln 0.5 = t * ln 0.94

Finally divide both sides by ln 0.94.

t = ln 0.5/ln 0.94
t = 11.20

So it takes 11.20 years for him to lose half his crops if he's losing them exponentially. That's almost 3 years more time than if he lost them linearly. But what does this mean? In both growth and decay, the exponential function ended up with higher numbers than the linear one. And the growth rate for exponential functions is incremental...and in much the same ways. Just as exponential growth functions have their growth rate steadily increase, exponential decay functions have their decay rate steadily decrease. When you graph them out, the linear decay will shoot down at a constant rate, but the exponential decay will always be a little bit above it, the decay rate gradually slowing down bit by bit.

How many years until he runs out?

Did you already guess it was going to be double the previous result of 11.20, i.e. 22.40? That's actually the amount of time it takes the farmer to have a quarter of his original crop amount, i.e. it's been halved again. So how long does it take to get him to 0? You might be surprised at the answer. Let's work it out and see what we get. So our final amount is 0. That means...

0 = 900 * (0.94)t
0 = (0.94)t
ln 0 = ln (0.94t)
ln 0 = t * ln 0.94
t = ln 0/ln 0.94

...right?
No, not right. A few of you reading this probably already noticed where I made the intentional mistake:

ln 0 = ln (0.94t)

I made it right here.

Recall what a logarithm is for a moment: it's the exponent that you can raise a number (the logarithm's base) to, to get the expression inside - essentially, exponents in reverse. For example, log5 25 is 2, because you can raise 5 to the power 2, and get 25. Now, look at ln 0. ln is the same as loge, though in this case that doesn't matter too much. What power could I raise e (or any other number) to, in order to get 0?

If you answered that there is no such number, you're right. If you said negative infinity, you need to take a break from your calculus homework - an infinite limit is not the same thing as actually using infinity as a number. As surmised, ln 0 (or log 0 with a log of any base) is undefined; when using an exponential decay function, you will never actually reach a quantity of 0. I did say and hint at this before, but there's the proof. Realistically the farmer in this example will run out of potato plants eventually (it happens after 110 years that you get numbers of less than 1 plant, for those curious), as we're dealing with whole numbers, but there are some quantities in the real world that might change exponentially that never reach 0. One such example is temperature (when measured on the Kelvin or Rankine scales), but there are other kinds.

Okay, let's slow down. That's the end of this section; I'll go into compound interest next, which is just a variant on the basic exponential growth model. (It doesn't deal with decay.) Before moving on I have a few examples for you to try. Like before, try not to peek at the answers before you're done! Round to 2 decimal places for each answer.

:: Starting amount is 3300; growth rate is 3.3% per day. How much does it have after a fortnight? How long does it take to reach 7000?
:: Starting amount is 2700; decay rate is 4.5% per hour. How much does it have after a day? How long does it take to halve in size?
:: Starting amount is 5000; growth rate is 2.4% per month. How much does it have after two quarters? How long does it take to double in size?
:: Starting amount is 4500; decay rate is 1.5% per minute. How much does it have after a half-hour? How long does it take to reach 2500?

I thought about using the same examples as I did in the linear example just for you to see how they're different, but if you really want to see, go back and do those yourself to see how they change.

Spoiler

:: 5198.98; 23.16 days (A fortnight is two weeks, or 14 days.)
:: 894.22; 15.05 hours
:: 5764.61; 29.23 months
:: 2859.56; 38.89 minutes

Compound interest

Compound interest is a specific application of the basic exponential growth formula, calculated specifically for calculating how much interest is earned on money deposited in a bank or similar institution. The formula to calculate compound interest is almost identical to the basic exponential growth formula:

A(t) = A0 * (1 + r/n)nt

This doesn't actually calculate the interest; it calculates the total saved investment. If you want to calculate just the interest earned, subtract A0 from your final result. There are a few new rules to keep in mind for the compound interest model:

:: r is always positive. This is a model for exponential growth, not exponential decay.

:: There is a new variable, n, to keep in mind. n is how many of a certain period of time is in a year, based on how often the investment's interest is compounded. For example, if the interest is compounded monthly, then n is 12, because there are 12 months in a year. The values that are possible for you to see for n are as follows:
n = 0.5: Compounded biannually
n = 1: Compounded annually (In this case, there is no difference from the basic exponential growth model)
n = 2: Compounded semiannually
n = 3: Compounded quarterly
n = 6: Compounded bimonthly
n = 12: Compounded monthly
n = 52: Compounded weekly
n = 365: Compounded daily
It is extremely rare that you will see examples of n that are not one of the above. (You may have heard of or seen something called 'compounded continuously'. That will be covered later, as it is a special case of compound interest.)

:: t is always measured in years. If the problem wants to find interest after a period of time that isn't years with no remainder, it must be converted. For example, 18 months is 1.5 years. Use the conversion tables to convert values into years:
60 seconds = 1 minute
60 minutes = 1 hour
3600 seconds = 1 hour
24 hours = 1440 minutes = 86400 seconds = 1 day
7 days = 168 hours = 1 week
1 month = 30 days*
1 month = 4 weeks*
3 months = 1 quarter
365 days = 52 weeks = 12 months = 4 quarters = 1 year
10 years = 1 decade
100 years = 10 decades = 1 century
1000 years = 100 decades = 10 centuries = 1 millennium
*Remember, be careful when converting weeks or days into months.

:: Because t is always measured in years, A(t) is how much total money is in the investment after t years.

Really the only thing that changes is that we have an n now in the equation, and we'll always have to convert t into years whenever it comes up. Nothing too drastic or new, right? That's the idea - I'm trying to introduce things as slowly and as incrementally as possible, so nothing comes as too much of a shock here.

Let's illustrate compound interest with an example.

I need to buy a new computer, as my old one is on its way out and is succumbing to age. I think I'll need about $1000 to buy a computer, but I want to see if I can earn that much money in interest - I have some money but I want to save it for emergencies. So, I go to the bank and deposit $5000 in a new account that earns 5% interest, compounded monthly. How much interest will I have made after 1 year? How long until I earn enough interest to buy that computer?

Calculating compound interest is a bit tricky. While the 5% interest is intended to be the amount over the entire year, the amount in interest gets periodically paid and recalculated (compounded) at each of the time periods. In this case we have 5% calculated monthly. Since the 5% interest is over the whole year, in the first month we gain interest equal to one-twelfth of 5% of my initial investment of $5000. Then, in the second month, we still gain interest equal to one-twelfth of 5%, but it's on the $5000 plus the amount I got in interest in the previous month. And so on. That's how compound interest works, mechanically.

Mathematically, though, it's just a matter of plugging our numbers in. First, let's see how much money in interest I make a year:

A(t) = A0 * (1 + r/n)nt

A(t) is what we're trying to find out, so I can't plug in a number for that.
A0 is my initial investment of $5000, so that goes there.
r is my rate of interest, which is 5%, or 0.05.
n is how often it's compounded in a year...it's monthly, so 12 times.
t is how many years have passed, so just 1 in this case. I plug all of this in...

A(1) = 5000 * (1 + 0.05/12)12

And a bit later, after I plug that into the calculator:

A(1) = 5255.81

So after 1 year, my initial investment has now become $5255.81. That isn't the answer though; it asked how much interest I made. Don't worry; just subtract the initial amount (A0, 5000) from the answer. So, in 1 year, my investment made $255.81 in interest. That's a little over a quarter of the way to my computer!

Now, before I figure out how many months it takes to earn all the interest I need, I'd like to take a moment and talk about something called the 'real interest rate', sometimes called the effective interest rate. (That is probably the more common name.) The real (effective) interest rate is the interest rate you actually earn in a year when you take the compounding into account. Its formula is:

R = (1 + r/n)n - 1

Basically it's your actual interest rate after 1 year (t = 1). In this case we can do the equation pretty quick; it's (1 + 0.05/12)12 - 1, which comes out to 0.0512, or 5.12%, if I round to 4 places. The less often the interest is compounded, the lesser the real rate will be, and the closer it will be to the advertised interest rate. The change is extremely slight (for example, if you take compounded monthly out to more decimal places, it's 5.1162%, whereas 5% compounded daily has a real rate of 5.1267%), but when you deal with larger and larger investments and periods of time, it all adds up.

Now for the 'fun' bit - when I can actually buy my computer. (As of the time of this writing, I really do need a new computer.)

A(t) = 5000 * (1 + 0.05/12)12t

A0, r, and n are the same, so I plugged those in. We're trying to solve for t in this case (how many years it would take), so we leave that alone. What about A(t)? We do know that, but if you say it's 1000, you missed a detail. Remember, A(t) is the total amount of the investment, not just the interest. So, in this case, A(t) is 6000. Sound good? Okay, time to solve. And, just like with the previous exponential growth/decay problems, you take the exact same steps.

6000 = 5000 * (1 + 0.05/12)12t

1.2 = (1 + 0.05/12)12t
ln 1.2 = ln ((1 + 0.05/12)12t)
ln 1.2 = 12t * ln (1 + 0.05/12)
ln 1.2/12 = t * ln (1 + 0.05/12)
t = ln 1.2/(12 * ln (1 + 0.05/12))

Ick! That is going to be nasty to enter into a calculator. Unfortunately, results like this are fairly common, so you'd best get used to them. When we enter all that...

t = 3.6540

So it takes me 3.6540 years to gather enough interest to buy my laptop. I'd like to convert that into months if I can, so I multiply by 12 to give me 43.8483 months (I multiplied before rounding off), so that's technically rounded up to 44. Of course, by then my computer might have died. Maybe I should have just gone ahead and bought it now...or made a bigger starting investment.

Are all compound interest problems like this, you ask? Yes, they are. Not much new to learn besides that. Not too different, right? Let's test this with a few examples - as before I'll include the answers but try not to peek. I won't torture you too much, though - this will just be two problems each with two parts (so it's technically four like always, but with repeats).

:: An initial investment of $7500 is deposited in an account that earns 3% interest compounded quarterly. How much money does it earn in 2 years? How long does it take to earn $1000 in interest, in quarters?
:: What's the real interest rate of that account (3% interest compounded quarterly)?
:: An initial investment of $2000 is deposited in an account that earns 7% interest compounded weekly. How much money does it earn in 3 years? How long does it take to earn $500 in interest, in weeks?
:: What's the real interest rate of that account (7% interest compounded weekly)?

Round any money amounts you get to 2 decimal places, and any other answer you get to 4 decimal places. Remember, it's asking how much money earned, not how much money total.

Spoiler

:: $461.99; 16.7509 quarters (4.1877 years)
:: 3.0339%
:: $467.01; 165.8753 weeks (3.1899 years)
:: 7.2458%

Compound interest, compounded continuously

Compounded interest that is 'compounded continuously' is a special case of the compound interest formula. Here it is again for reminder:

A(t) = A0 * (1 + r/n)nt

n = 0.5: Compounded biannually
n = 1: Compounded annually
n = 2: Compounded semiannually
n = 3: Compounded quarterly
n = 6: Compounded bimonthly
n = 12: Compounded monthly
n = 52: Compounded weekly
n = 365: Compounded daily

'Compounded continuously' is what happens when n is equal to infinity - that is, the money is compounded an infinite number of times within the year. Note that this application is entirely theoretical, so it doesn't have any applications in the real world (as far as I'm aware). Regardless, they teach it to mess with you. It is mostly used as an illustration for a theoretical infinite, if you compound interest more and more times.

The formula for compounded interest that is compounded continuously is this:

A(t) = A0 * ert

...in which e is a constant, specifically, the base of the natural logarithm (ln is the same thing as loge). It's possible to derive this formula yourself by using calculus, though this is normally covered in algebra so I won't confuse you. I will hint, however, that for the function f(x) = (1 + 1/x)x, its limit as x approaches infinity is e.

You might think that having an irrational number here will make things harder. It won't. It'll actually make them easier, because for your final calculation, you should only have one ln in it to evaluate, not multiple. I'll show you with an example.

Let's use my computer example again. $5000, at 5%, compounded continuously. How much money do I earn after a year? How much time does it take for me to earn $1000 in interest?

Before, I earned $255.81 in interest within a year, and it took me 3.6540 years to earn $1000 interest. How will the numbers look here? Since it's being compounded more and more times, I should probably earn more interest in a year, and it'll take me less time to earn all the interest - not by a drastic amount, but possibly enough to matter.

So let's see.

A(t) = 5000 * e0.06t

This is our starting setup. Let's see how much interest I earn in a year. Remember, t is still in years.

A(1) = 5000 * e0.06

On a scientific calculator, use the e^ button to perform this equation. We get...

A(1) = 5309.18

So I earned $309.18 when my interest is compounded continuously. That's about $50 more than monthly. That's not a huge amount but every little bit helps. Now, how long does it take me to earn the whole $1000 in interest?

6000 = 5000 * e0.06t
1.2 = e0.06t
ln 1.2 = ln e0.06t

Now, this is the fun part. ex and ln x are inverse functions of one another:

eln x = x
ln ex = x

...which means that, when you apply the natural logarithm (ln) to both sides, where the e is, they will cancel each other out. That means the right side of your equation looks like this:

ln 1.2 = 0.06t

That looks MUCH neater! It means our solution won't have to plug as many things into the calculator.

t = ln 1.2/0.06
t = 3.0387

So with continuously compounded interest, it only takes a little over 3 years (36.4643 months, rounded up to 37) to get all the interest I need. A difference of 7 months may seem small if your scale is larger, but again, the difference is as substantial as you want to make it.

I don't have any example problems for continuously compounded interest, because I'm going to use it as a pivot to proceed into my final point, which I shall start immediately.

Continuous exponential growth and decay

Continuous exponential growth and decay models are models in which the growth rate isn't really tied to a period of time - rather it's divided into an infinitely large number of small pieces, all stacked together - kinda like compound interest. Rather than something growing at 6% per year, it's said to grow continuously at 6% per year. (It doesn't necessarily have to be years either.) The formula for continuous growth and decay is this:

A(t) = A0 * ert

...wait, isn't this the exact same as the continuous compounded interest formula?
Yes it is. A(t) and A0 are still the same, though r and t are a bit different:

:: t doesn't necessarily have to be in years. However, like any growth/decay problem, the units of time must be expressed, and converted if need be. For example, if t is measured in hours, then growth after 90 minutes would mean t = 1.5.
:: r follows the same rules for growth and decay as usual: if r is positive it's growth; if r is negative it's decay; if r is 0 there's no change at all.

So we're basically back to using our normal growth/decay model as always, just that we have e as a base now. This means that we might actually have neat-looking answers for continuous growth/decay problems, won't we? Yes we will! And unlike continuous interest, continuous growth actually does have a few real-world examples, which I'm going to illustrate.

Let's try this out with an example or two - and this time, I'm not going to use the farmers again.

There's been a pest infestation lately, and farmers are starting to lose their crops to rabbits. They don't know where they came from or how they got here, but they've found a rabbit warren nearby, and its denizens are beginning to multiply like...well, rabbits. There were 50 rabbits here initially, and they're growing at a continuous rate of 8% per month. How many rabbits will there be after 3 months? How long will it take for the warren population to double in size? Round to 2 decimal places where appropriate.

Okay, I half-lied. There are still farmers but the problem is based around the rabbits. No Fibonacci nonsense here though, thankfully. So let's see, what do we know?

A(t) = A0 * ert

We started with 50 rabbits, so that's A0. They're increasing continuously at a rate of 8% per month too, so r = 0.08 (positive, because growth). t will thus also be measured in months.

A(t) = 50 * e0.08t

How many rabbits are there after 3 months? Since t is still measured in months, we don't need to do any unit conversion, so we can just plug the figures in:

A(3) = 50 * e0.08*3
A(3) = 50 * e0.24
A(3) = 63.56

So there's 63.56 rabbits after 3 months. They reproduce quickly. (Imagine the .56 as a rabbit that's only halfway to being fully grown, if you need to conceptualize it.) I'm not sure if rabbits actually reproduce that fast in real life, but it doesn't take them long to reach maturity, so this could be an accurate figure.

Now how long will it take for them to double in size, i.e. to 100? Similarly, we plug the 100 in and solve for t:

100 = 50 * e0.08t
2 = e0.08t
ln 2 = ln e0.08t
ln 2 = 0.08t
t = ln 2/0.08
t = 8.66

It will take only 8.66 months for that warren of rabbits to double in size. That is a pretty important pest problem.

While I'm on the topic, similar to linear growth/decay shortcuts to see how long it takes for something to double or halve in size, continuous models have one too, and it's pretty simple. The time it takes for an initial quantity under continuous growth to double in size is ln 2/r. Similarly, the time it takes for an initial quantity under continuous decay to halve in size is ln 0.5/r, or -ln 2/r. (ln 0.5 = -ln 2; they're the same number, so either way works. For consistency I will use the ln 0.5 shortcut in this tutorial.)

If you want to try out the farmer examples from before and see how they work under continuous growth or decay, you are welcome to it, but I will not cover that here, for those who are curious to find out for themselves to see how they compare.

How about an exponential decay problem? Let's try one of those.

The half-life of carbon-14, a radioactive isotope of carbon, is 5730 years. How long will it take a 20-gram sample of carbon-14 to decay to 12 grams?

...what. That wasn't very much information; where do we begin? This problem is indeed solvable, but before I begin, let's start with a few definitions.

First of all: half-life. A half-life is the amount of time it takes for a quantity of radioactive substance to decay into half of the original amount. For example, above, for that 20-gram sample of carbon-14 to decay to 10 grams would take exactly 5730 years. (So you should probably anticipate that to decay into 12 grams would take less time.)

A radioactive isotope is a form of an element that is prone to decay, and gives off radiation in various forms to do so (alpha, beta, and gamma particles usually, though there are other exotic kinds - but I'll save that explanation for a chemistry course). Normally, carbon exists as carbon-12 (the number 12 being its atomic weight), and the two extra neutrons that contribute to the weight of carbon-14 make it unstable, and prone to radioactive decay. Radioactive substances do exhibit continuous decay, just as certain animal populations exhibit continuous growth - so these models do have real-life applications.

But how are we going to solve this decay problem if we don't have much to work on? Well, I'll show you.

A(t) = A0 * ert

They give us figures for A0 and t initially, but what they don't give us is r, so we actually need to solve for r first. As it takes 5730 years for a sample of carbon-14 to halve in size, t will be thus measured in years, and will be 5730 for our example. Now, as for A(5730), you can plug in 10 grams (half of 20) if you want, but whatever the case it's going to be half of the original, so you can just as easily plug in 0.5 * A0. For the sake of this illustration, that's what I'm going to do:

0.5 * A0 = A0 * e5730r

Divide both sides by A0, and solve for r...

0.5 = e5730r
ln 0.5 = ln e5730r
ln 0.5 = 5730r
r = ln 0.5/5730

If you plug that into a calculator, you get -0.000120968 (to 10 places). That's a pretty small number. It also should not surprise you; that tells you that carbon-14 decays continuously at a rate of 0.012097% per year, and it's been established that it takes 5730 years for a sample to halve in quantity.

How long will it take a 20-gram sample of carbon-14 to decay to 12 grams?

Now, while I could plug that number in for r in the next part of the problem, I'm actually going to leave it as (ln 0.5/5730) - the exact figure - so that my answer will be more accurate. You can put in -0.000120968 if you wish, though; our answers may be different but only at a few decimal places down the line.

We actually have all the numbers we need now and we just need to solve for t:

12 = 20 * et * (ln 0.5/5730)

Here goes.

0.6 = et * (ln 0.5/5730)
ln 0.6 = ln et * (ln 0.5/5730)
ln 0.6 = t * (ln 0.5/5730)
t = ln 0.6/(ln 0.5/5730)

Or, rewritten:

t = 5730 * ln 0.6/ln 0.5
t = 4222.81

So it will take 4222.81 years for that sample of carbon-14 to decay from 20 grams to 12. You'll notice that the ln from r remained in the solution - and that's normal; you'll probably have 2 ln expressions to deal with if you have to end up solving for r yourself (such as in the case of half-life problems like these), and you don't round r prematurely. Otherwise it'll usually just be one.

So is that really all there is to continuous growth and decay? Yes, actually. Most of it is knowing the rules of logarithms, and knowing where to plug in which numbers. Once you get past that bit, the rest can be rather fun! However, you will always end up with solutions that look messy (as the ln function is used) so you will have to round pretty much every time.

And, as is custom, I have a few examples for you to try. Solve them as you always have, just plugging in numbers where appropriate, and don't feel afraid to use your shortcuts (ln 2/r for something to double in quantity, ln 0.5/r for something to halve) if necessary.

Round whatever answers you get to 4 decimal places.

:: Starting amount is 120; continuous growth rate is 4% per week. How much does it have after a fortnight? How long does it take to reach 180?
:: Starting amount is 450; continuous decay rate is 5% per day. How much does it have after a week? How long does it take to halve in size?
:: Starting amount is 2000; continuous growth rate is 2% per month. How much does it have after half a year? How long does it take to double in size?
:: The half-life of strontium-90 is 28.79 years. How long will it take a 5-gram sample of strontium-90 to reach 1 gram?

Spoiler

:: 129.9944; 10.1366 weeks (remember, a fortnight is 2 weeks)
:: 317.1096; 13.8629 days
:: 2254.9937; 34.6574 months
:: r = ln 0.5/28.79, or roughly -0.0241; 66.8483 years

Conclusion

And that covers all of the main kinds of growth and decay models that are covered in algebra courses! I probably make it sound easier than it actually is, though most of it is just plugging in numbers to where the appropriate variables go, converting t into the appropriate unit of time, and then going from there. Oh, and natural logarithms come up everywhere in the exponential models. In any case, if you think I missed anything in my explanations or they were somewhat incomplete, let me know so we can talk and I can address your concerns personally. As always I do hope this tutorial was of use to you, and helped you to more easily bear through your algebra homework!

 

Integration (anti-derivatives) and basic techniques, plus a few common integrals
Requested (indirectly) by @[member="Megarai111"]



Introduction

Integrals are what shy many people away from calculus, and for fairly good reason - it's doing differentiation in reverse. In fact, at some points, because of that integration may seem frustrating or even impossible because you can't find the right path back. While I certainly won't claim that integration is easier than differentiation (differentiation is probably the easier of the two), I can certainly help in assisting your learning.

So first off, what is an integral and why do people need to know them? Outside of physics I haven't found a practical application for integrals yet (though for derivatives there's plenty). But I'm still looking. Yet, that doesn't mean I can't explain them.

Like a derivative is defined as the slope of a line tangent to any particular point on a curve (usually expressed as an entire function), an integral is defined as the area underneath the curve, between the x-axis and the curve itself. When the curve is above the x-axis, the area is denoted as positive. When it's below the x-axis, the area there is usually denoted as negative.

Integrals are usually begun by talking of approximating the area underneath the curve via a series of rectangles, each with the same width. The thinner you make the rectangles, the more there will be, and the closer they will approximate the area underneath the curve. The integral itself basically happens when you take that approximation to its end, and with 'infinite rectangles' you will have the exact area under the curve.

But how do you calculate the area of infinite rectangles? And why am I mentioning this? The answer is only so I can get that explanation out of the way, so I can move on to the heart of the matter.


The basic definition of an integral

To begin, let me begin by introducing integral notation:

∫ f(x) dx

This is usually how integrals are presented. I will go over each part of this one at a time.

First let's start with the symbol on the left. That symbol there is called the integral symbol, and is the notation used to signal that you're integrating a function. The symbol can be either by itself, in which case it's called an 'indefinite integral', or there can be numbers above and below it, in which case it's called a 'definite integral'. More on that later.

f(x) is your basic function of x. It means the same thing it's always meant, so there's nothing too scary here. The 'dx' at the end means that x is the independent variable in question. If the function were a function of t instead of x, then the expression at the end would be a dt, i.e. ∫ f(t) dt, since t is the independent variable. If you see any other letters in there besides whatever the 'd_' denotes, those are treated as constants, not variables.

Now, to give you the equation that gives you the mathematical definition of an integral:

Suppose there's a function f(x), and its derivative is g(x) - that is, f'(x) = g(x). If that is so, then:

f(x) = ∫ g(x) dx

This is what I alluded to before when I mentioned that integration is doing differentiation in reverse. If g(x) is the derivative of f(x), then f(x) is the anti-derivative of g(x). Usually, the terms 'integral' and 'anti-derivative' are used interchangeably; however, while they are not exactly the same thing, the context is still understood. (The main difference is because "taking the integral gives you the anti-derivative", but that's just terminology mumbo-jumbo. Don't worry about it too much.)

So let's say that f(x) = x3. That would mean that g(x) = 3x2, right? So, if I take the derivative of x3, the result is what I'd expect:

d/dx (x3) = 3x2

However, when I take the integral of 3x2, something unusual happens:

∫ 3x2 dx = x3 + C

"Wait a minute there," you ask. "Where did that C come from? And what is C, anyway?" Excellent questions - which I will explain immediately.
Recall from differentiation that the derivative of any constant is 0. When you have a line like y = 3, or y = -2, it's perfectly flat, it has no slope, so if it were added on to another function, it would just move it up or down on the graph, and not affect its slope or shape. So, you can have the function x3, along with the functions x3 + 1, x3 - 2, x3 + 5, or x3 - 1000. And when you take the derivative of each of those functions, they all end up to be the same: 3x2. That's because the shape of the graph didn't change when appending the constant to the end - it just moved it up or down.

That's why the C is there. The C is something called the 'integration constant'. It's not any specific number, but rather it's an arbitrary figure to show that any of those possible anti-derivatives could satisfy the condition when you take the integral. So when you integrate 3x2, the anti-derivative could be x3 + 1, or x3 - 2, or x3 + 5...and so on. That's why a "+ C" is placed on the end. When dealing with indefinite integrals, adding a "+ C" on the end will be the very last step you do. It's mostly convention, but it is important.


Integration rules

Just like how differentiation has its own set of rules (like the Power Rule), integration has its own set of rules as well. Since integration is doing differentiation in reverse, a lot of these will look very similar to rules you've already encountered in the past. There are a lot of integration techniques, but for now I'll cover the basic ones, and move onto more complicated ones at a later time.

Remember to always put a +C on the end as your final step!

First, some basic stuff:

∫ f(x) + g(x) dx = ∫ f(x) dx + ∫ g(x) dx

Just like derivatives, you can break up integrals if they're being added together, and you can integrate each part separately. You can 'shortcut' this step by integrating each term added or subtracted, without breaking it up into smaller terms - you'll get the same result at the end. And even if you break it up into smaller terms, you will still only have 1 +C at the end.

∫ A * f(x) dx = A * ∫ f(x) dx
(where A is any constant)

And, just like derivatives, you can remove constants while integrating, and place them back again when you're done.

∫ A dx = Ax + C
(where A is any constant)

The integral of a constant equals that constant times x.

Next, the Power Rule for Integration:

∫ xn dx = xn+1/ (n+1) + C
where n ≠ -1

This is essentially the Power Rule for Differentiation backwards. Nothing too weird. To integrate a power term, increase the power by 1, then divide the function by the new power - so, the integral of x4 would be x5/5 +C. This even applies if the power is 0 - which was shown directly above, as ∫ A dx.

If the power is -1, though, then the integral is different:

∫ 1/x dx = ln |x| + C

This is the only power for which that exception applies, however. For all other powers, even ones that aren't integers, the above rule applies.

Other than polynomial integrals (which you can derive with the Power Rule), here are a few more integral formulas you may find useful:

Basic exponential and logarithmic integrals:
∫ ln x dx = x ln x - x + C
∫ 1/x dx = ln |x| + C
∫ ex = ex + C
∫ Ax dx = Ax / ln A + C
(where A is any constant, except for 1 or 0)

Basic trigonometric integrals:
∫ sin x dx = -cos x + C
∫ cos x dx = sin x + C
∫ tan x dx = ln |sec x| + C

|These| are absolute value expressions, in case you were curious. They are mandatory for those particular integrals.

Depending on the course you're taking, you may only be given one or the other of these categories to practice on. Most of the time, when starting out, you will deal with polynomial integrals. However, for an example, why don't we work on a problem together?

Find the integral: ∫ 2x3 + 6x2 + 4x + 7 dx.

This might look really complicated, but we'll do it step by step. For this problem, we'll want to keep 3 things in mind: the Power Rule for Integration; the fact we can separate integrals added together; and the fact we can extract constants to put them back later.

So, in a sense, we'll be integrating 4 terms here. (I will be explaining this problem with all of the necessary steps, and separating everything out as much as I can. In practice, the more you do this, you will end up mentally skipping the steps for breaking up the integrals via adding, and taking out the constants via multiplying. You'll just end up doing them by habit eventually.)

First, let's break up the integral into the 4 separate functions, then work on each one at a time:

∫ 2x3 dx + ∫ 6x2 dx + ∫ 4x dx + ∫ 7 dx

Let's start with the first one, 2x3. First, let's extract the 2 from it, so we are just dealing with the x3 term:

2 ∫ x3 dx

Via the Power Rule for Integration, to integrate that, we increase the power by 1, then divide the function by the new power. The new power would be 4 (as it's 3+1), so the integral would be x4/4:

2 (x4/4)

Now we can bring the constant back in. Multiplying the two together, and cancelling factors, it becomes x4/2. One down, three to go! Now our original expression becomes:

x4/2 + ∫ 6x2 dx + ∫ 4x dx + ∫ 7 dx

You may be thinking that we'll handle the other three terms in almost the exact same way. If you are thinking that, you would be absolutely right. When we have 6x2, we can take the 6 off to the side:

6 ∫ x2 dx

We integrate x2 with the Power Rule for Integration, giving us x3/3:

6 (x3/3)

Then we multiply the two together and cancel factors:

2x3

Putting that back to the original term gives us:

x4/2 + 2x3 + ∫ 4x dx + ∫ 7 dx

And thus we'll do the last two the same way. (In fact, you can do this with virtually any indefinite integral for polynomials. There may be more - or less - steps, but the process is the same.) Doing the same with the other two gives:

x4/2 + 2x3 + 2x2 + 7x

Then as our final step, we have to add a "+ C" on the end, to conclude the problem. Putting that all together, we get:

∫ 2x3 + 6x2 + 4x + 7 dx = x4/2 + 2x3 + 2x2 + 7x + C

Not too tricky, right? Let's try another example, and see if that has any success.

Find the integral: ∫ 2ex - 3 sin x dx.

Instead of using the Power Rule for this one, we'll end up using the other formulas instead, for the other functions. The two we'll need for this are:

∫ ex = ex + C
∫ sin x dx = -cos x + C

Handling the first one is, thankfully, simple. We can take the 2 out and integrate it, leaving us with 2ex again as the first term, since ex is the only function whose integral (and derivative) is itself.
Now for the second one. The integral of sin x is -cos x, so that second term becomes -3 (-cos x). Multiplying that out makes the negative become a positive, so putting those two things together:

∫ 2ex - 3 sin x dx = 2ex + 3 cos x + C

And as always, the last step is to add the "+ C" onto the end.

With the close of this section, I'll put some practice problems for you to work with. Again, I'll have the answers below in the spoiler tag, but please try to work on them on your own, and then use the spoiler to check your work when you're done.

:: Find the integral: ∫ 3x2 + 6x + 2 dx.
:: Find the integral: ∫ 6x3 + 12x + 7/x dx.
:: Find the integral: ∫ 2x4 + 6ex + 8x4 dx.
:: Find the integral: ∫ 5x + 2 sin x - 3 ln x dx.

Spoiler

:: x3 + 3x2 + 2x + C
:: 3x4/2 + 6x2 + 7 ln |x| + C
:: 2x5 + 6ex + C
:: (5x / ln 5) - 2 cos x + 3x ln x - 3x + C

Definite integrals

So we've been calculating indefinite integrals and taking anti-derivatives and finding out how the functions line up with one another. That's all well and good, but "aren't integrals supposed to be areas under the curve?" you ask. "How is a function an area? Isn't an area a number?" Indeed they are, and that is what definite integrals are for.

Definite integrals are computed just like indefinite integrals, except they have what is called 'bounds' on them. These are displayed as numbers above and below the integral symbol, with the bottom number being the 'lower bound', and the top number being the 'upper bound', like so:


b
∫ f(x) dx
a
(Sometimes the a and b are written directly above; sometimes they're written at the top and to the right.)

This basically represents the area between the curve f(x) and the x-axis, between x=a and x=b. So if a were 3, and b were 5, that would be calculating the area under the curve between the x-values 3 and 5. Generally, the function also has to exist everywhere on that interval as well, and doesn't have vertical asymptotes there.

Calculating the area isn't too tricky. Taking the precept from before:

f(x) = ∫ g(x) dx

where g(x) = f'(x), taking the definite integral instead results in:


b
∫ g(x) dx = f(b) - f(a)
a
Thus, to find the area under a curve from a to b, you would have to take the anti-derivative of the function, plug in the bound figures, and then take the difference. That's a lot of confusing words, so it may be easier to show you how it works in practice:

Evaluate:


5
∫ 3x^2 dx
2
Solving definite integrals only has 1 additional step to indefinite integrals, and you don't have to deal with the +C at the end either. (Technically speaking, the +Cs cancel each other out, so you don't need to worry about it.)

So first, let's integrate 3x2. Via the Power Rule and multiplying the constant, we find the integral of 3x2 is x3. After we do that, we write the problem like this:


5
[x^3]
2
The bracket with the numbers above and below it is also a special kind of notation used in integrals, meaning you're about to evaluate the integral between those two numbers. Essentially, that there is the same as writing out (5)3 - (2)3. Subtracting those numbers, we find the answer is 117.

That means that the area under the graph of 3x2, between the x-axis, and the lines x=2 and x=5, is 117 (square units). Not too strenuous, right? Admittedly, this example is a lot more simple than most, so there weren't very many steps to perform.

Oh, and there's also one other thing about definite integrals. Remember this formula:


b
∫ g(x) dx = f(b) - f(a)
a
Now, let's say there's a number 'c', that is between b and a. Knowing that, you can split the integral up like this:


b c b
∫ g(x) dx = ∫ g(x) dx + ∫ g(x) dx
a a c
Instead of going from a to b, you can go from a to c, and then c to b, and you'll get the same result. Neat, right? You usually won't have to do this unless the function in question is a piecewise function, and you have to break it up into smaller bits. Those examples are rare, but for a test, why don't we work on one?

Evaluate:


4
∫ f(x) dx
0
where f(x) is:


x^2, if x < 2
f(x) = { 2x - 9, if x >= 2
We can tell right off the bat here that the function is going to jump (or skip) at x = 2, so we can't just go all the way through and evaluate it. We'll need to split the integral up and evaluate each piece separately. The break happens at x=2, so the first thing we're going to do is break the integral up there:


2 4
∫ f(x) dx + ∫ f(x) dx
0 2
From there, we can substitute the pieces of the function in directly, since on the parts where we broke them up, the function doesn't change between pieces at all. So for the first integral, we can put in x2, and for the second integral we can put in 2x - 9:


2 4
∫ x^2 dx + ∫ 2x - 9 dx
0 2
After that we take each integral separately. Thankfully, neither of these are too hard, so the integrals become easy:


2 4
[x^3/3] + [x^2 - 9x]
0 2
Remember that each of those brackets means you're evaluating the function at the upper bound, and then subtracting from it the value of the function at the lower bound. So for the first one, it would be (2)3/3 - (0)3/3, and the second bracket is done similarly. So, with that in mind:

[23/3 - 03/3] + [(42 - 9*4) - (22 - 9*2)]
[8/3 - 0] + [(16 - 36) - (4 - 18)]
8/3 + [-20 + 14]
8/3 - 6
-10/3

That tells us that the area under the curve between x=0 and x=4 is -10/3 (square units). You may find it rather surprising that the area is a negative number - after all, isn't area supposed to be positive? The reason why the area is negative is because there are portions of the graph that are below the x-axis. Any area above the x-axis is positive, while area below is negative. When summed up, if the area below is more, it causes the result to be negative.

There are also a few other rules about definite integrals to keep in mind. Here they are below:


b a
- ∫ f(x) dx = ∫ f(x) dx
a b
Multiplying the integral by -1 has the same effect as switching the lower and upper bounds of integration. This may not have much practical application, but if you run into a situation in which your lower bound is a higher number than your upper bound (rare, but it can happen), that is a good way to switch the bounds.


a
∫ f(x) dx = 0
a
If you go from a to a, you go nowhere, so sensibly, the area under the curve in that case would be 0.

The last two rules concern odd and even functions. Remember, an odd function is a function that is symmetric about the origin (0,0), and an even function is a function that is symmetric about the y-axis.

Now, if o(x) is an odd function, and e(x) is an even function, then these things are true:


a
∫ o(x) dx = 0
-a

and:

a a
∫ e(x) dx = 2 * ∫ e(x) dx
-a 0
So if you see either an odd or an even function crossing 0 like that evenly on both sides, you can make the work a little easier on yourself by breaking the integral in half, or just resolving it to 0. Just make sure the function is actually odd or even, before you do this. (Most functions are neither.)

Now then, with that in mind, let's throw a few practice problems for you to test your skills on definite integrals. As usual the answers will be in the spoiler tags below, but again, do it on your own first before checking your answers.

Evaluate the following definite integrals:


6
∫ x^3 + 4x dx
1

Code:

3
∫ e^x + 2x^2 dx
0

Code:

 2
 ∫ 4x^3 dx
-2

Code:

 4
 ∫ 4x^4 dx
-4

Spoiler

:: 393.75
:: e3 + 18
:: 0 (it's an odd function)
:: 819.2 (it's an even function)

Conclusion

Thus concludes the basics of integration. There are many more techniques to integration beyond this, but this should give you a basic framework and idea of what integration is, what it means, how it works, and most importantly, how to perform it. In another tutorial I'll probably speak of other integration techniques - including my personal favourite, U-substitution - but for now, just let your brain digest this material.

Next tutorial planned: Either differentiation, or U-substitution

 

If I knew how to do that, I certainly would. Unfortunately, I've only taken single-variable differential and integral calculus - I never moved on to vectors. (I've never taken physics either, as I wasn't able to find a teacher who could explain it to me very well.)

I apologise for being unable to help you in this regard.

 

@[member="Azure"]
I wasn't able to find any of my old geometry textbook or notes at all (sadly), so all of what I'm about to tell you will be based largely off memory, plus a little bit of help from Google. (This may come in helpful later, but the first link contains the basic building blocks.)

The starting point in geometry is first to define what a postulate is to begin with. A postulate, sometimes called an axiom, is something that's assumed to be true by default. Postulates become the starting points for all future proofs that you will end up deriving. Things such as the definitions of line segments, rays, lines, and planes, as well as Euclid's Postulates (see the first link), fit into this group. To use a loose analogy, postulates are much like atoms - they can't really be subdivided any further, and from those small things, the rest of the field is built up.

Proofs, otherwise called theorems, are concepts and statements that are derived from postulates or other theorems. From what I recall when I studied geometry, it is largely about proof derivation. One practice that I never did then (and recommend now) is that, each time you learn or derive a new proof/theorem, note or write it down somewhere, as you'll most likely use it to derive another proof. Geometry is like math in this regard, in that the new things you learn are derived from the old things you already know, and you build up from there.

In the case of proofs, statements made about them generally have to be true in all cases that it allows. If it says something about an object, it needs to be true for all cases of that object. For example, the Base Angle Theorem (second link) states that "If two sides of a triangle are congruent, the angles opposite these sides are congruent". That means this applies to all triangles no matter what they look like, as long as the triangle fits those criteria.

As for basic postulates, the first link I gave has a very nice list of those - and I also found a nice picture explaining them visually. I'll start by explaining the ones they list as point-line-plane postulates (since you most likely have heard the first list - but if you need help with those I can repeat them):

Spoiler

Remember, these are postulates as well - they are assumed to be true without further need for proof. Postulates serve as the foundational statements of geometry.

Unique Line Assumption: A line is straight, drawn connecting two points, and extending infinitely in both directions. If you only have exactly 2 points, you will only be able to draw 1 unique line between the two. (You could draw multiple, but they would be congruent to the initial line in every respect.) You could draw multiple line segments, but not multiple lines (since lines are infinite length in both directions, and line segments are finite).

Dimension Assumption: A plane is a 2-dimensional surface, extending infinitely out in its directions of length and width. However, real life exists in 3 dimensions, and planes won't account for the 3rd dimension, height. Similarly, a line is only 1-dimensional (length only), and thus won't account for width (the 2nd dimension) or height. You will end up with points in a plane not on a line (separated by the 2nd dimension), and you will end up with lines not on a plane (separated by the 3rd dimension). They will be on a different plane or line, though - just not the 'given' one.

Number Line Assumption: This deals with numerical measurements, especially when corresponding to distance. When dealing with geometric figures, distance isn't always given - it may not be given at all. The Number Line Assumption basically means we can choose to assign figures or numbers to the distances of lines how we please, just so long as our measurements remain consistent. This also deals with the first part of the Distance Assumption too.

Distance Assumption: The distance between 2 points on the same line will always be the same between those 2 points; the number isn't going to change if the points (or line) don't change position.
Points have no dimensions at all, so when 2 points are brought together, it introduces the 1st dimension, length, to connect them together. So, if those 2 points are on the same plane (surface), the line that connects them is on that plane as well. (Also goes back to the Unique Line Assumption.)
Non-colinear simply means that the 3 points aren't all on the same line. (2 points for sure will be on the same line, as above, but the 3rd one will be somewhere else.) At most, there will be only 1 plane that has all 3 of those points in it. You can conceptualise this as well by being familiar with shapes: imagine the 3 points with lines between them, and it becomes a triangle, the simplest polygon. However, you couldn't have 3 points not be on the same plane; the simplest polyhedron, the tetrahedron, has 4 vertices.

Remember that a plane is just a flat surface; it doesn't mean it needs to be parallel with the ground! (Say that points 1 and 2 were on the floor, and point 3 was on the ceiling. You could connect all 3 points with a wall - which is a real-life example of a plane. Yes, I know walls have thickness and planes technically don't, but with real-life examples you do have to be a bit fanciful, no?)

I can try to explain the rest of that list as well, but I figured that starting off with those is a good base, as that's where you seem to be having the most trouble at present.

If you need me to go more in-depth, or if you have more specific examples that you'd like me to look at instead, I will. I don't remember too much about geometry, but I remember enough to visualise and conceptualise it for others. Do let me know what else you require.

::::

@[member="Princess Victoria"]
I'm always glad to have a bit more help, and it helps to have a second mind on-hand to assist. I'm definitely interested, though I would like to know what you would like to cover - or, rather, how you want to cover it. My preliminary idea would be to see what guides you have (whenever you write them), and then I would put them up where needed, crediting you as the author.

I would need to know a bit more about how you like to explain things, though - essentially, I'd like to know your teaching plan and style. I don't know how you do things and I'd like to know how, but I'm certainly not adverse to the idea. Please, tell me more.