Polynomials have tons of similarities to integers, and throughout this section we'll be coming back a lot to the integer analogy to make our points. We'll look at polynomial division in a few different situations, each of which can be related to integer division. See? We did it already.
A lot of ideas in this section come from the section on fractions, so it's a good idea to make sure you're comfortable with that stuff before diving into this section. Slipping into your Snuggie might help you feel more comfortable as well. If anyone's willing, you may also want to think about requesting a plate of apple wedges and a Capri Sun.
We'll be using fraction notation to express polynomial division. For example:
(3x2 + 9x) divided by (5x – 2) is written as .
We'll treat these expressions like fractions. The number on top is still called the numerator, the number on the bottom is still called the denominator, and we still aren't allowed to divide by zero. Bummer.
To start with, we'll look at polynomial division where the divisor goes into the dividend with no remainder. As a matter of fact, we can guarantee that all the division problems in this particular unit will work out evenly. Absolutely no leftovers. Our apologies to your dog.
For polynomial division problems, there are three main things to do:
The easiest case of polynomial division is when the divisor is a single term that happens to be a common factor of the dividend. You'll love when this happens. For the sake of the integer analogy, we'll do a couple of examples with integers first. You'll probably notice that we're not doing the integer problems in the most efficient way. That's not because we're being lazy or unsure of what we're doing. By approaching them in a certain manner, we'll be doing them in a way that will translate nicely to polynomials. Ah, there's a method to our madness...
Write as a sum of two integers.
The fraction is the sum of the two fractions and . Since and , we have:
Write as a sum of integers.
This is like the last problem, except we'll be splitting the fraction into a sum of four other fractions. Once you're done, try splitting an atom. Or your pants. Anything, really, as long as you're getting in some good splitting practice.
Performing division on each term, we find:
Now we'll do something similar with polynomials. Take five, integers. If the divisor is a common factor of the dividend, then the divisor divides (goes evenly into, with nothing left over) every term of the dividend, meaning we can break up the polynomial like we broke up the fractions above. Contrary to what you may have heard, breaking up isn't that hard to do.
We can split up the terms of the dividend and rewrite this as .
Simplifying each term gives us the final answer: 4x + 8. Just cancel out those x's in the denominators.
Another way to look at this sort of problem is to factor. We'll let the integers be the opening act again. They did such a great job when they opened for Coldplay last weekend that they deserve to be given another opportunity.
Write as a sum of two integers.
Pull out the common factor of 5 from the numerator and the denominator.
Then we can write the fraction as and cancel the 5 from the top and bottom to find .
Good job canceling the 5, but don't assume you'll never hear from it again. You're still on its email list.
After a bit of practice, you can skip the step of physically writing down where you pull the common factor out of each term...as long as you can do it mentally and still wind up with the correct answer, that is. If you can master this skill, who knows? You may be able to start doing all kinds of things mentally. Don't go all Chronicle on us, though.
When asked to divide a polynomial by a single term, you can use whichever method you like. You'll find the same answer whether you break up the polynomial first and then simplify the resulting terms, or pull out the common factor first. Don't do both at the same time, or you may disrupt the space-time continuum.
Factoring isn't only for when the divisor is a single term. We can use factoring to divide one polynomial by another polynomial, as long as the polynomial in the denominator is a factor of the polynomial in the numerator. We're sorry. We know you're keeping track of a lot of nomials.
For example, we can simplify by factoring 10 as 2 × 5 and canceling 2 from the top and bottom, and we can do similar things with polynomials.
When factoring, keep an eye out for common factors that are one term. No sense in making more work for yourself if you can avoid it. We know what a workaholic you are, but tone it down a bit.
Find the quotient: .
The polynomial on top factors as (6x + 5)(6x – 5), so we can rewrite the expression as:
Those all match, so now we can cancel the factor of (6x + 5) from the top and bottom to get our final answer:
(6x – 5)
What is 4x4 + 4x2 + 1 divided by 2x2 + 1?
We want to find:
The numerator factors as (2x2 + 1)(2x2 + 1), so we can rewrite the problem like this:
Then we cancel the factor of (2x2 + 1) from the top and bottom:
2x2 + 1
Find the quotient: .
The numerator and denominator look like something should cancel, right? Remember, we have that guarantee that the problems in this section will work out evenly, so we can take that to the bank. Actually, don't try taking that to the bank. They'll give you a weird look.
The first thing we're supposed to do is look for common factors, but there are no common factors shared by both the numerator and the denominator. They could probably learn a lot about sharing if they watched The Wiggles once in a while. Anyway, what we can do is pull out a common factor of 2 in the numerator to simplify that guy at least. Hopefully, doing this will shed some light on where we go next. We can now rewrite our original problem as:
The expressions 5x – 1 and 1 – 5x are negatives of each other. We're getting closer, we can feel it already:
(1 – 5x) = (-1)(5x – 1)
If we factor out -1 from the denominator, we have:
Now we can cancel 5x – 1 from the top and bottom:
When asked to divide polynomials, look for cases like this where expressions in the numerator and denominator are negatives of each other. Then factor out -1 and force them to get along.
Factoring works on multivariable polynomials too. Oh, happy day. We were starting to miss our old friends y and z.
After looking for single-term factors and finishing all our factoring, the next thing we do to divide polynomials is use good old-fashioned long division. You know, old-fashioned like the donut: delicious, time-tested, and kind of weird to think about in too much detail.
First, here's a reminder of how long division works with integers.
Find 611 ÷ 13.
Using long division notation, we set up the division problem as .
First off, 13 goes into 6 a grand total of zero times, no matter how forcefully you may try to squeeze it in there. We write 0 above 6, subtract 13 × 0 from 6, and bring down the next digit in the dividend:
Moving briskly along, 13 goes into 61 four times, so we write the digit 4 on top of the digit 1, subtract 13 × 4 = 52 from 61, and bring down the final 1:
Finally, 13 goes into 91 seven times (13 × 7 = 91):
We conclude that 611 ÷ 13 = 47. We can check this answer by multiplying 13 and 47 and making sure we find 611. Try it. It works! As advertised, there's no remainder. We hope this is building some trust between us. If only we could persuade you to fall backward and let us catch you.
Now we'll do some long division with polynomials.
Find using long division.
First we set up the long division, making sure the terms of each polynomial are written in decreasing order by exponent, like so:
Then we look at the first term of the divisor and see how many times it goes into the first term of the dividend:
Hmm...x2 goes into x5 a total of x3 times (that is, x2 × x3 = x5). We're dealing with exponents, so it's not the same as figuring out how many times the number 2 goes into the number 5. Here, we're subtracting exponents rather than dividing one number by another. That's why exponents look like they're floating away, because we "take them away."
We write x3 on top of x5:
We subtract (x3)(x2 + 4) = x5 + 4x3 from the dividend, the same as with normal long division, and then we "bring down" all of the remaining terms of the dividend:
Now we have a new polynomial. It even still has that new polynomial smell.
Next, we see how many times the first term of the divisor goes into the first term of this new polynomial:
Since is 3x, x2 goes into 3x3 a total of 3x times.
We subtract (3x)(x2 + 4) = 3x3 + 12x, and drop down the remaining terms. Whoa, careful. Don't drop them down so fast. These things are fragile.
Now we see that x2 goes into 7x2 a total of 7 times, and subtract 7(x2 + 4). We have 0 left over, so we're finished. You didn't think we'd finish all of that without a remainder, did you? Oh ye of little faith.
To make it official, the final answer is:
x3 + 3x + 7
Check the answer to the above example by multiplying (x2 + 4) and (x3 + 3x + 7). You should get x5 + 7x3 + 7x2 + 12x + 28.
In the example above, there were a few places where we wrote 0x2 or 0x. We did this so that like terms would line up nicely in the division problem. Obviously, 0x2 is the same as 0, but see how much easier it is to keep track of everything when you do it our way? Come to the Shmoop side...
Sometimes problems involve polynomials that "skip terms." Sort of like how some people—not you, obviously—skip classes, but without the negative repercussions. The polynomial 5x2 – 9, for example, "skips" the x term. A more mathematical and impressive way to say this is "the coefficient on the x term is 0." Aren't you impressed? We thought so. An even more impressive way to say this is "the coefficient of the first-degree term is 0." The most impressive way to say this is backwards and in Russian while doing a handstand. However, we don't want to put any undue pressure on you, so saying it the first way will be fine.
We can rewrite the polynomial like this without really changing it:
5x2 – 9 = 5x2 + 0x – 9
If the dividend skips terms when you're setting up polynomial long division, write them down anyway to help yourself keep things straight. If the dividend is 5x2 – 9, write 5x2 + 0x – 9 instead. We might need that column for subtraction somewhere in the middle of the long division. You'll be glad you stuck it in there.