## Dividing Polynomials by Binomials

Read the section on dividing a polynomial by a binomial. Pay attention to the review of long division, since it will help you understand the technique. Review the solution to example 6.84 to see how to divide a trinomial by a binomial. Then, review the solution to example 6.85 to see how we handle dividing by a subtraction binomial. Be careful, and make sure you keep track of the negative sign.

After you study these examples, complete questions 6.167 through 6.170 in the Try It section.

### Divide 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.

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 6.84

#### Solution

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 6.85

#### Solution

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 Example 6.86, we'll have a division that leaves a remainder. We write the remainder as a fraction with the divisor as the denominator.

#### EXAMPLE 6.86

#### Solution

Look back at the dividends in Example 6.84, Example 6.85, and Example 6.86. The terms were written in descending order of degrees, and there were no missing degrees. The dividend in Example 6.87 will be x4−x2+5x−2. It is missing an x3 term. We will add in 0x3 as a placeholder.

#### EXAMPLE 6.87

#### Solution

Notice that there is no x3 term in the dividend. We will add 0x3 as a placeholder.

##### TRY IT 6.173

Find the quotient: \left(x^{3}+3 x+14\right) \div(x+2).

In Example 6.88, we will divide by . As we divide we will have to consider the constants as well as the variables.

#### EXAMPLE 6.88

Source: OpenStax, https://openstax.org/books/elementary-algebra/pages/6-6-divide-polynomials

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