5.5 Rational expression concepts -- dividing polynomials

Simplifying, multiplying, dividing, adding, and subtracting rational expressions are all based on the basic skills of working with fractions. Dividing polynomials is based on an even earlier skill, one that pretty much everyone remembers with horror: long division.

To refresh your memory, try dividing $\frac{\text{745}}{3}$ by hand. You should end up with something that looks something like this:

So we conclude that $\frac{\text{745}}{3}$ is 248 with a remainder of 1; or, to put it another way, $\frac{\text{745}}{3}=\text{248}\frac{1}{3}$ .

You may have decided years ago that you could forget this skill, since calculators will do it for you. But now it comes roaring back, because here is a problem that your calculator will not solve for you: $\frac{{\mathrm{6x}}^3-{\mathrm{8x}}^2+\mathrm{4x}-2}{\mathrm{2x}-4}$ . You can solve this problem in much the same way as the previous problem.

So we conclude that $\frac{{\mathrm{6x}}^3-{\mathrm{8x}}^2+\mathrm{4x}-2}{\mathrm{2x}-4}$ is ${\mathrm{3x}}^2+\mathrm{2x}+6$ with a remainder of 22, or, to put it another way, ${\mathrm{3x}}^2+\mathrm{2x}+6+\frac{\text{22}}{\mathrm{2x}-4}$ .

Checking your answers

As always, checking your answers is not just a matter of catching careless errors: it is a way of making sure that you know what you have come up with . There are two different ways to check the answer to a division problem, and both provide valuable insight

The first is by plugging in numbers. We have created an algebraic generalization:

In order to be valid, this generalization must hold for $x=3$ , $x=-4$ , $x=0$ , $x=\varpi$ ,or any other value except $x=2$ (which is outside the domain). Let’s try $x=3$ .

Checking the answer by plugging in $x=3$

The second method is by multiplying back. Remember what division is: it is the opposite of multiplication! If $\frac{\text{745}}{3}$ is 248 with a remainder of 1, that means that $\text{248}\cdot 3$ will be 745, with 1 left over. Similarly, if our long division was correct, then $\left({\mathrm{3x}}^2+\mathrm{2x}+6\right)\left(\mathrm{2x}-4\right)+\text{22}$ should be ${\mathrm{6x}}^3-{\mathrm{8x}}^2+\mathrm{4x}-2$ .

Checking the answer by multiplying back