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12.5: Division of Polynomials

Difficulty Level: At Grade Created by: CK-12
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What if you had a polynomial like 2x2+5x3 and you wanted to divide it by a monomial like x or a binomial like x+1? How would you do so? After completing this Concept, you'll be able to divide polynomials like this one by monomials and binomials.

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

2xx214x23x+42x9x2+4x5x2+5x12x32x+3

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) 8x24x+162

b) 3x2+6x1x

c) 3x218x+69x

Solution

a) 8x24x+162=8x224x2+162=4x22x+8

b) 3x3+6x1x=3x3x+6xx1x=3x2+61x

c) 3x218x+69x=3x29x18x9x+69x=x32+23x

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

Consider the quotient 3x+44.

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:

3x+44=3x4+44=3x4+1

A common mistake is to cross out the number 4 from the numerator and the denominator, leaving just 3x. This is incorrect, because the entire numerator needs to be divided by 4.

Example B

Divide 5x310x2+x255x2.

Solution

5x310x2+x255x2=5x35x210x25x2+x5x2255x2

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

5x35x2+10x25x2x5x2+255x2=x+215x+5x2

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:

x+3)x2+4x+5¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

We start by dividing the first term in the dividend by the first term in the divisor: x2x=x.

We place the answer on the line above the x term:

x+3)x2+4x+5¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯x

Next, we multiply the x term in the answer by the divisor, x+3, and place the result under the dividend, matching like terms. x times (x+3) is x2+3x, so we put that under the divisor:

x+3)x2+4x+5¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯x  x2+3x

Now we subtract x2+3x from x2+4x+5. It is useful to change the signs of the terms of x2+3x to x23x and add like terms vertically:

x+3)x2+4x+5¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯xx23x x

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

x+3)x2+4x+5¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯xx23x x+5

And now we go through that procedure once more. First we divide the first term of x+5 by the first term of the divisor. x divided by x is 1, so we place this answer on the line above the constant term of the dividend:

x+3)x2+4x+5¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯x + 1x23x x+5

Multiply 1 by the divisor, x+3, and write the answer below x+5, matching like terms.

x+3)x2+4x+5¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯x + 1x23x x+5 x+3

Subtract x+3 from x+5 by changing the signs of x+3 to x3 and adding like terms:

x+3)x2+4x+5¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯x + 1x23x x+5  x32

Since there are no more terms from the dividend to bring down, we are done. The quotient is x+1 and the remainder is 2.

Remember that for a division with a remainder the answer is quotient+remainderdivisor. So the answer to this division problem is x2+4x+5x+3=x+1+2x+3.

Check

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

(divisor×quotient)+remainder=dividend

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

(x+3)(x+1)+2=x2+4x+3+2=x2+4x+5

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*}

x+4)x2+8x+17¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

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*}x2x=x.

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

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

x+4)x2+8x+17¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯x

Next, we multiply the \begin{align*}x\end{align*}x term in the answer by the divisor, \begin{align*}x + 4\end{align*}x+4, and place the result under the dividend, matching like terms. \begin{align*}x\end{align*}x times \begin{align*}(x + 4)\end{align*}(x+4) is \begin{align*}x^2+4x\end{align*}x2+4x, so we put that under the divisor:

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

x+4)x2+8x+17¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯x  x2+4x

Now we subtract \begin{align*}x^2+4x\end{align*}x2+4x from \begin{align*}x^2+8x+17\end{align*}x2+8x+17. It is useful to change the signs of the terms of \begin{align*}x^2+4x\end{align*}x2+4x to \begin{align*}-x^2-4x\end{align*}x24x and add like terms vertically:

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

x+4)x2+8x+17¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯xx24x 4x

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:

  1. \begin{align*}\frac{2x+4}{2}\end{align*}
  2. \begin{align*}\frac{x-4}{x}\end{align*}
  3. \begin{align*}\frac{5x-35}{5x}\end{align*}
  4. \begin{align*}\frac{x^2+2x-5}{x}\end{align*}
  5. \begin{align*}\frac{4x^2+12x-36}{-4x}\end{align*}
  6. \begin{align*}\frac{2x^2+10x+7}{2x^2}\end{align*}
  7. \begin{align*}\frac{x^3-x}{-2x^2}\end{align*}
  8. \begin{align*}\frac{5x^4-9}{3x}\end{align*}
  9. \begin{align*}\frac{x^3-12x^2+3x-4}{12x^2}\end{align*}
  10. \begin{align*}\frac{3-6x+x^3}{-9x^3}\end{align*}
  11. \begin{align*}\frac{x^2+3x+6}{x+1}\end{align*}
  12. \begin{align*}\frac{x^2-9x+6}{x-1}\end{align*}
  13. \begin{align*}\frac{x^2+5x+4}{x+4}\end{align*}
  14. \begin{align*}\frac{x^2-10x+25}{x-5}\end{align*}
  15. \begin{align*}\frac{x^2-20x+12}{x-3}\end{align*}
  16. \begin{align*}\frac{3x^2-x+5}{x-2}\end{align*}
  17. \begin{align*}\frac{9x^2+2x-8}{x+4}\end{align*}
  18. \begin{align*}\frac{3x^2-4}{3x+1}\end{align*}
  19. \begin{align*}\frac{5x^2+2x-9}{2x-1}\end{align*}
  20. \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. 

Vocabulary

Denominator

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. \frac{5}{8} has denominator 8.
Dividend

Dividend

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

divisor

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

Polynomial long division

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

Rational Expression

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

Rational Root Theorem

The rational root theorem states that for a polynomial, f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0, where a_n, a_{n-1}, \cdots a_0 are integers, the rational roots can be determined from the factors of a_n and a_0. More specifically, if p is a factor of a_0 and q is a factor of a_n, then all the rational factors will have the form \pm \frac{p}{q}.
Remainder Theorem

Remainder Theorem

The remainder theorem states that if f(k) = r, then r is the remainder when dividing f(x) by (x - k).
Synthetic Division

Synthetic Division

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

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Difficulty Level:

At Grade

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Date Created:

Oct 01, 2012

Last Modified:

Jan 31, 2016
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