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**Long Division**: At a Glance

- Topics At a Glance
**Polynomial Division**- Common Factors
- Factoring
**Long Division**- More Polynomial Division
- Common Factors
- Factoring
- Long Division
- A Clever Trick
- Rational Expressions
- Evaluating Rational Expressions
- Simplifying Rational Expressions
- Multiplying and Dividing Rational Expressions
- Adding and Subtracting Rational Expressions
- Simplifying Complex Rational Expressions
- Equations Involving Rational Expressions
- Word Problems
- In the Real World
- I Like Abstract Stuff; Why Should I Care?
- How to Solve a Math Problem

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:

*x*^{2} goes into *x*^{5} a total of *x*^{3} times (that is, *x*^{2} × *x*^{3} = *x*^{5}). 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 *x*^{3} on top of *x*^{5}:

We subtract (*x*^{3})(*x*^{2} + 4) = *x*^{5} + 4*x*^{3} 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 3*x*, *x*^{2} goes into 3*x*^{3} a total of 3*x* times.

We subtract (3*x*)(*x*^{2} + 4) = 3*x*^{3} + 12*x*, and drop down the remaining terms. Whoa, careful. Don't drop them down so fast. These things are fragile.

Now we see that *x*^{2} goes into 7*x*^{2} a total of 7 times, and subtract 7(*x*^{2} + 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

*x*^{3} + 3*x* + 7

Check the answer to the above example by multiplying (*x*^{2} + 4) and (*x*^{3} + 3*x* + 7). You should get *x*^{5} + 7*x*^{3} + 7*x*^{2} + 12*x* + 28.

In the example above, there were a few places where we wrote 0*x*^{2} or 0*x*. We did this so that like terms would line up nicely in the division problem. Obviously, 0*x*^{2} 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 5*x*^{2} – 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

5*x*^{2} – 9 = 5*x*^{2} + 0*x* – 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 5*x*^{2} – 9, write 5*x*^{2} + 0*x* – 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.

Example 1

Use polynomial long division to find . |

Example 2

Find using polynomial long division. |

Exercise 1

Find the following quotient using polynomial long division and check your answer by multiplication *before* you peek at the solution: .

Exercise 2

Find the following quotient using polynomial long division and check your answer by multiplication *before* you peek at the solution: .

Exercise 3

Find the following quotient using polynomial long division and check your answer by multiplication *before* you peek at the solution: .

Exercise 4

*before* you peek at the solution: .

Exercise 5

*before* you peek at the solution: .