Mathematics » Polynomials II » Divide Polynomials

# Dividing a Polynomial By a Binomial

## Dividing a Polynomial By a Binomial

To divide a polynomial by a binomial, we follow a procedure very similar to long division of numbers. So let’s look carefully the steps we take when we divide a 3-digit number, 875, by a 2-digit number, 25.

 We write the long division  We divide the first two digits, 87, by 25.  We multiply 3 times 25 and write the product under the 87.  Now we subtract 75 from 87.  Then we bring down the third digit of the dividend, 5.  Repeat the process, dividing 25 into 125.  We check division by multiplying the quotient by the divisor.

If we did the division correctly, the product should equal the dividend.

$$\begin{array}{c}\hfill 35·25\hfill \\ \hfill 875\phantom{\rule{0.2em}{0ex}}\text{✓}\hfill \end{array}$$

Now we will divide a trinomial by a binomial. As you read through the example, notice how similar the steps are to the numerical example above.

## Example

Find the quotient: $$\left({x}^{2}+9x+20\right)÷\left(x+5\right).$$

### Solution  Write it as a long division problem. Be sure the dividend is in standard form.  Divide x2 by x. It may help to ask yourself, “What do I need to multiply x by to get x2?” Put the answer, x, in the quotient over the x term.  Multiply x times x + 5. Line up the like terms under the dividend.  Subtract x2 + 5x from x2 + 9x.  Then bring down the last term, 20.  Divide 4x by x. It may help to ask yourself, “What do I need to multiply x by to get 4x?” Put the answer, 4, in the quotient over the constant term.  Multiply 4 times x + 5.  Subtract 4x + 20 from 4x + 20.  Check: Multiply the quotient by the divisor. (x + 4)(x + 5) You should get the dividend. x2 + 9x + 20✓

When the divisor has subtraction sign, we must be extra careful when we multiply the partial quotient and then subtract. It may be safer to show that we change the signs and then add.

## Example

Find the quotient: $$\left(2{x}^{2}-5x-3\right)÷\left(x-3\right).$$

### Solution  Write it as a long division problem. Be sure the dividend is in standard form.  Divide 2x2 by x. Put the answer, 2x, in the quotient over the x term.  Multiply 2x times x − 3. Line up the like terms under the dividend.  Subtract 2x2 − 6x from 2x2 − 5x. Change the signs and then add. Then bring down the last term.  Divide x by x. Put the answer, 1, in the quotient over the constant term.  Multiply 1 times x − 3.  Subtract x − 3 from x − 3 by changing the signs and adding.  To check, multiply (x − 3)(2x + 1). The result should be 2x2 − 5x − 3.

When we divided 875 by 25, we had no remainder. But sometimes division of numbers does leave a remainder. The same is true when we divide polynomials. In the example below, we’ll have a division that leaves a remainder. We write the remainder as a fraction with the divisor as the denominator.

## Example

Find the quotient: $$\left({x}^{3}-{x}^{2}+x+4\right)÷\left(x+1\right).$$

### Solution  Write it as a long division problem. Be sure the dividend is in standard form.  Divide x3 by x. Put the answer, x2, in the quotient over the x2 term. Multiply x2 times x + 1. Line up the like terms under the dividend.  Subtract x3 + x2 from x3 − x2 by changing the signs and adding. Then bring down the next term.  Divide −2x2 by x. Put the answer, −2x, in the quotient over the x term. Multiply −2x times x + 1. Line up the like terms under the dividend.  Subtract −2x2 − 2x from −2x2 + x by changing the signs and adding. Then bring down the last term.  Divide 3x by x. Put the answer, 3, in the quotient over the constant term. Multiply 3 times x + 1. Line up the like terms under the dividend.  Subtract 3x + 3 from 3x + 4 by changing the signs and adding. Write the remainder as a fraction with the divisor as the denominator.  To check, multiply $$\left(x+1\right)\left({x}^{2}-2x+3+\frac{1}{x+1}\right).$$ The result should be $${x}^{3}-{x}^{2}+x+4$$.

Look back at the dividends in the examples above. The terms were written in descending order of degrees, and there were no missing degrees. The dividend in the example below will be $${x}^{4}-{x}^{2}+5x-2$$. It is missing an $${x}^{3}$$ term. We will add in $$0{x}^{3}$$ as a placeholder.

## Example

Find the quotient: $$\left({x}^{4}-{x}^{2}+5x-2\right)÷\left(x+2\right).$$

### Solution

Notice that there is no $${x}^{3}$$ term in the dividend. We will add $$0{x}^{3}$$ as a placeholder.  Write it as a long division problem. Be sure the dividend is in standard form with placeholders for missing terms.  Divide x4 by x. Put the answer, x3, in the quotient over the x3 term. Multiply x3 times x + 2. Line up the like terms. Subtract and then bring down the next term.  Divide −2x3 by x. Put the answer, −2x2, in the quotient over the x2 term. Multiply −2x2 times x + 1. Line up the like terms. Subtract and bring down the next term.  Divide 3x2 by x. Put the answer, 3x, in the quotient over the x term. Multiply 3x times x + 1. Line up the like terms. Subtract and bring down the next term.  Divide −x by x. Put the answer, −1, in the quotient over the constant term. Multiply −1 times x + 1. Line up the like terms. Change the signs, add.  To check, multiply $$\left(x+2\right)\left({x}^{3}-2{x}^{2}+3x-1\right)$$. The result should be $${x}^{4}-{x}^{2}+5x-2$$.

In the example below, we will divide by $$2a-3$$. As we divide we will have to consider the constants as well as the variables.

## Example

Find the quotient: $$\left(8{a}^{3}+27\right)÷\left(2a+3\right).$$

### Solution

This time we will show the division all in one step. We need to add two placeholders in order to divide. To check, multiply $$\left(2a+3\right)\left(4{a}^{2}-6a+9\right)$$.

The result should be $$8{a}^{3}+27$$.

## Optional Videos:

You can watch the following videos for additional instruction and practice with dividing polynomials:

## Key Concepts

• If $$a,b,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}c$$ are numbers where $$c\ne 0$$, then
$$\frac{a}{c}+\frac{b}{c}=\frac{a+b}{c}\phantom{\rule{0.5em}{0ex}}\text{and}\phantom{\rule{0.5em}{0ex}}\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}$$