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 .
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:
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 is no remainder. We hope this is building some trust between us. Now if only we could persuade you to fall backward and let us catch you.
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:
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 as
5x2 – 9 = 5x2 + 0x – 9
without really changing it.
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.