# 12.5: Division of Polynomials

**At Grade**Created by: CK-12

**Practice**Division of Polynomials

What if you had a polynomial like

### Try This

To check your answers to long division problems involving polynomials, try the solver at http://calc101.com/webMathematica/long-divide.jsp. It shows the long division steps so you can tell where you may have made a mistake.

### Watch This

CK-12 Foundation: 1205S Division of Polynomials

### Guidance

A **rational expression** is formed by taking the quotient of two polynomials.

Some examples of rational expressions are

Just as with rational numbers, the expression on the top is called the **numerator** and the expression on the bottom is called the **denominator**. In special cases we can simplify a rational expression by dividing the numerator by the denominator.

**Divide a Polynomial by a Monomial**

We’ll start by dividing a polynomial by a monomial. To do this, we divide each term of the polynomial by the monomial. When the numerator has more than one term, the monomial on the bottom of the fraction serves as the **common** denominator to all the terms in the numerator.

#### Example A

*Divide.*

a)

b)

c)

**Solution**

a)

b)

c)

A common error is to cancel the denominator with just one term in the numerator.

Consider the quotient

Remember that the denominator of 4 is common to both the terms in the numerator. In other words we are dividing both of the terms in the numerator by the number 4.

The correct way to simplify is:

A common mistake is to cross out the number 4 from the numerator and the denominator, leaving just

#### Example B

*Divide 5x3−10x2+x−25−5x2.*

**Solution**

The negative sign in the denominator changes all the signs of the fractions:

**Divide a Polynomial by a Binomial**

We divide polynomials using a method that’s a lot like long division with numbers. We’ll explain the method by doing an example.

#### Example C

*Divide x2+4x+5x+3.*

**Solution**

When we perform division, the expression in the numerator is called the **dividend** and the expression in the denominator is called the **divisor**.

To start the division we rewrite the problem in the following form:

We start by dividing the first term in the dividend by the first term in the divisor:

We place the answer on the line above the

Next, we multiply the

Now we subtract

Now, we bring down the 5, the next term in the dividend.

And now we go through that procedure once more. First we divide the first term of

Multiply 1 by the divisor,

Subtract

Since there are no more terms from the dividend to bring down, we are done. The quotient is

Remember that for a division with a remainder the answer is

**Check**

To check the answer to a long division problem we use the fact that

For the problem above, here’s how we apply that fact to check our solution:

**The answer checks out.**

Watch this video for help with the Examples above.

CK-12 Foundation: Division of Polynomials

### Guided Practice

*Divide \begin{align*}\frac{x^2+8x+17}{x+4}\end{align*} x2+8x+17x+4.*

**Solution**

When we perform division, the expression in the numerator is called the **dividend** and the expression in the denominator is called the **divisor**.

To start the division we rewrite the problem in the following form:

\begin{align*}& {x+4 \overline{ ) x^2+8x+17 }}\end{align*}

We start by dividing the first term in the dividend by the first term in the divisor: \begin{align*}\frac{x^2}{x}=x\end{align*}

We place the answer on the line above the \begin{align*}x\end{align*}

\begin{align*}& \overset{\qquad x}{x+4 \overline{ ) x^2+8x+17 \;}}\end{align*}

Next, we multiply the \begin{align*}x\end{align*}

\begin{align*}& \overset{\qquad x}{x+4 \overline{ ) x^2+8x+17\;}}\\ & \qquad \ \ x^2 + 4x\end{align*}

Now we subtract \begin{align*}x^2+4x\end{align*}

\begin{align*}& \overset{\qquad x}{x+4 \overline{ ) x^2+8x+17 \;}}\\ & \qquad \underline{-x^2 - 4x}\\ & \qquad \qquad \quad \ 4x\end{align*}

Now, we bring down the 17, the next term in the dividend.

\begin{align*}& \overset{\qquad x}{x+4 \overline{ ) x^2+8x+17 \;}}\\ & \qquad \underline{-x^2 - 4x\;\;\;\;\;\;}\\ & \qquad \qquad \quad \ 4x + 17\end{align*}

And now we go through that procedure once more. First we divide the first term of \begin{align*}4x + 17\end{align*} by the first term of the divisor. \begin{align*}4x\end{align*} divided by \begin{align*}x\end{align*} is 4, so we place this answer on the line above the constant term of the dividend:

\begin{align*}& \overset{\qquad \qquad \quad x \ + \ 4}{x+4 \overline{ ) x^2+8x+17 \;}}\\ & \qquad \underline{-x^2 - 4x\;\;\;\;\;\;}\\ & \qquad \qquad \quad \ x + 17\end{align*}

Multiply 4 by the divisor, \begin{align*}x + 4\end{align*}, and write the answer below \begin{align*}4x + 16\end{align*}, matching like terms.

\begin{align*}& \overset{\qquad \qquad \quad x \ + \ 4}{x+4 \overline{ ) x^2+8x+17 \;}}\\ & \qquad \underline{-x^2 - 4x\;\;\;\;\;\;}\\ & \qquad \qquad \quad \ 4x + 17\\ & \qquad \qquad \quad \ 4x + 16\end{align*}

Subtract \begin{align*}4x + 16\end{align*} from \begin{align*}4x + 17\end{align*} by changing the signs of \begin{align*}4x + 16\end{align*} to \begin{align*}-4x -16\end{align*} and adding like terms:

\begin{align*}& \overset{\qquad \qquad \quad x \ + \ 4}{x+4 \overline{ ) x^2+8x+17 \;}}\\ & \qquad \underline{-x^2 - 4x\;\;\;\;\;\;}\\ & \qquad \qquad \quad \ x + 17\\ & \qquad \qquad \ \ \underline{-4x - 16}\\ & \qquad \qquad \qquad \quad 1\end{align*}

Since there are no more terms from the dividend to bring down, we are done. The quotient is \begin{align*}x + 4\end{align*} and the remainder is 1.

Remember that for a division with a remainder the answer is \begin{align*}\text{quotient}+\frac{\text{remainder}}{\text{divisor}}\end{align*}. So the answer to this division problem is \begin{align*}\frac{x^2+8x+17}{x+4}=x+4+\frac{1}{x+4}\end{align*}.

**Check**

To check the answer to a long division problem we use the fact that

\begin{align*}(\text{divisor} \times \text{quotient}) + \text{remainder} = \text{dividend}\end{align*}

For the problem above, here’s how we apply that fact to check our solution:

\begin{align*}(x+4)(x+4)+1 & = x^2+8x+16+1\\ & = x^2+8x+17\end{align*}

**The answer checks out.**

### Explore More

Divide the following polynomials:

- \begin{align*}\frac{2x+4}{2}\end{align*}
- \begin{align*}\frac{x-4}{x}\end{align*}
- \begin{align*}\frac{5x-35}{5x}\end{align*}
- \begin{align*}\frac{x^2+2x-5}{x}\end{align*}
- \begin{align*}\frac{4x^2+12x-36}{-4x}\end{align*}
- \begin{align*}\frac{2x^2+10x+7}{2x^2}\end{align*}
- \begin{align*}\frac{x^3-x}{-2x^2}\end{align*}
- \begin{align*}\frac{5x^4-9}{3x}\end{align*}
- \begin{align*}\frac{x^3-12x^2+3x-4}{12x^2}\end{align*}
- \begin{align*}\frac{3-6x+x^3}{-9x^3}\end{align*}
- \begin{align*}\frac{x^2+3x+6}{x+1}\end{align*}
- \begin{align*}\frac{x^2-9x+6}{x-1}\end{align*}
- \begin{align*}\frac{x^2+5x+4}{x+4}\end{align*}
- \begin{align*}\frac{x^2-10x+25}{x-5}\end{align*}
- \begin{align*}\frac{x^2-20x+12}{x-3}\end{align*}
- \begin{align*}\frac{3x^2-x+5}{x-2}\end{align*}
- \begin{align*}\frac{9x^2+2x-8}{x+4}\end{align*}
- \begin{align*}\frac{3x^2-4}{3x+1}\end{align*}
- \begin{align*}\frac{5x^2+2x-9}{2x-1}\end{align*}
- \begin{align*}\frac{x^2-6x-12}{5x^4}\end{align*}

### Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 12.5.

Denominator

The denominator of a fraction (rational number) is the number on the bottom and indicates the total number of equal parts in the whole or the group. has denominator .Dividend

In a division problem, the dividend is the number or expression that is being divided.divisor

In a division problem, the divisor is the number or expression that is being divided into the dividend. For example: In the expression , 6 is the divisor and 152 is the dividend.Polynomial long division

Polynomial long division is the standard method of long division, applied to the division of polynomials.Rational Expression

A rational expression is a fraction with polynomials in the numerator and the denominator.Rational Root Theorem

The rational root theorem states that for a polynomial, , where are integers, the rational roots can be determined from the factors of and . More specifically, if is a factor of and is a factor of , then all the rational factors will have the form .Remainder Theorem

The remainder theorem states that if , then is the remainder when dividing by .Synthetic Division

Synthetic division is a shorthand version of polynomial long division where only the coefficients of the polynomial are used.### Image Attributions

## Description

## Learning Objectives

Here you'll learn how to divide polynomials by monomials and binomials.

## Related Materials

## Difficulty Level:

At Grade## Subjects:

## Search Keywords:

## Concept Nodes:

## Date Created:

Oct 01, 2012## Last Modified:

Jan 31, 2016## Vocabulary

**Save or share your relevant files like activites, homework and worksheet.**

To add resources, you must be the owner of the Modality. Click Customize to make your own copy.