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# 12.3: Division of Polynomials

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Divide a polynomials by a monomial.
• Divide a polynomial by a binomial.
• Rewrite and graph rational functions.

## Introduction

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

Some examples of rational expressions are

a) $\frac {2x}{x^2-1}$

b) $\frac {4x^2-3x+4}{2x}$

c) $\frac {9x^2+4x-5}{x^2+5x-1}$

d) $\frac {2x^3}{2x+3}$

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 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 different terms, the term on the bottom of the fraction serves as common denominator to all the terms in the numerator.

Example 1

Divide.

a) $\frac {8x^2-4x+16}{2}$

b) $\frac {3x^3-6x-1}{x}$

c) $\frac {-3x^2-18x+6}{9x}$

Solution

$\frac {8x^2-4x+16}{2} & =\frac {8x^2}{2}-\frac {4x}{2}+\frac {16}{2}={4x^2-2x+8} \\\frac {3x^3-6x-1}{x} & =\frac {3x^3}{x}+\frac {6x}{x}-\frac {1}{x}=3x^2+6 -\frac {1}{x} \\\frac {-3x^2-18x+6}{9x} & =\frac {3x^2}{9x}-\frac {18x}{9x}+\frac {6}{9x}=-\frac {x}{3}-2+\frac {2}{3x}$

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

Consider the quotient $\frac {3x+4}{4}$

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

$\frac {3x+4}{4}=\frac {3x}{4}+\frac {4}{4}=\frac {3x}{4}+1$

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

$\xcancel{\frac{3x+4}{4}}=3x$

This is incorrect because the term $3x$ does not get divided by 4 as it should be.

Example 2

Divide $\frac {5x^3-10x^2+x-25}{-5x^2}$.

Solution

$\frac {5x^3-10x^2+x-25}{-5x^2}=\frac {5x^3}{-5x^2}-\frac {10x^2}{-5x^2}+\frac {x}{-5x^2}-\frac {25}{-5x^2}$

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

$-\frac {5x^3}{5x^2}+\frac {10x^2}{5x^2}-\frac {x}{5x^2}+\frac {25}{5x^2}=-x+2-\frac {1}{5x}+\frac {5}{x^2}$

## Divide a Polynomial by a Binomial

We divide polynomials in a similar way that we perform long division with numbers. We will explain the method by doing an example.

Example 3

Divide $\frac {x^2+4x+5}{x+3}$.

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 $\frac {x^2}{x}=x$.

We place the answer on the line above the $x$ term.

Next, we multiply the $x$ term in the answer by each of the $x + 3$ in the divisor and place the result under the divided matching like terms.

Now subtract $x^2 + 3x$ from $x^2+4x+ 5$. It is useful to change the signs of the terms of $x^2 + 3x$ to $-x^2 - 3x$ and add like terms vertically.

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

We repeat the procedure.

First divide the first term of $x + 5$ by the first term of the divisor $\left ( \frac{x}{x} \right ) = 1$.

Place this answer on the line above the constant term of the dividend,

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

Subtract $x + 3$ from $x + 5$ by changing the signs of $x + 3$ to $-x-3$ and adding like terms.

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

The answer is $x + 1$ with a remainder of 2.

Remember that for a division with a remainder the answer is $\text{quotient} + \frac{\text{remainder}}{\text{divisor}}$

$\frac {x^2+4x+5}{x+3}=x+1+\frac {2}{x+3}$

Check

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

$\text{divisor} \cdot \text{qiotient} + \text{remainder} = \text{divisor}$

For the problem above here is the check of our solution.

$(x+3)(x+1)+2 & = x^2+4x+3+2 \\& = x^2+4x+5$

Example 4

Divide $\frac {4x^2-25x-21}{x-7}$.

Solution

Answer $\frac {4x^2-25x-21}{x-7}=4x+3$

Check $(4x + 3)(x - 7) + 0 = 4x^2-25x-21$. The answer checks out.

## Rewrite and Graph Rational Functions

In the last section we saw how to find vertical and horizontal asymptotes. Remember that the horizontal asymptote shows the value of $y$ that the function approaches for large values of $x$. Let’s review the method for finding horizontal asymptotes and see how it is related to polynomial division.

We can look at different types of rational functions.

Case 1 The polynomial in the numerator has a lower degree than the polynomial in the denominator. Take for example, $y=\frac {2}{x-1}$

We see that we cannot divide 2 by $x - 1$ and $y$ approaches zero because the number in the denominator is bigger than the number in the numerator for large values of $x$.

The horizontal asymptote is $y = 0$.

Case 2 The polynomial in the numerator has the same degree as the polynomial in the denominator. Take for example, $y=\frac {3x+2}{x-1}$

In this case, we can divide the two polynomials and obtain.

The quotient is $y=3+\frac {5}{x-1}$.

Because the number in the denominator of the remainder is bigger than the number in the numerator of the remainder, the reminder will approach zero for large values of $x$ leaving only the 3, thus $y$ will approach the value of 3 for large values of $x$.

The horizontal asymptote is $y = 3$.

Case 3 The polynomial in the numerator has a degree that is one more than the polynomial in the denominator. Take for example, $y=\frac {4x^2+3x+2}{x-1}$.

The quotient is: $y=4x+7+\frac {9}{x-1}$.

The remainder approaches the value of zero for large values of $x$ and the function $y$ approaches the straight line $y = 4x + 7$.When the rational function approaches a straight line for large values of $x$, we say that the rational function has an oblique asymptote. (Sometimes oblique asymptotes are also called slant asymptotes). The oblique asymptote is $y = 4x + 7$.

Case 4 The polynomial in the numerator has a degree that in two or more than the degree in the denominator. For example, $y=\frac {x^3}{x-1}$.

In this case the polynomial has no horizontal or oblique asymptotes.

Example 5

Find the horizontal or oblique asymptotes of the following rational functions.

a) $y=\frac {3x^2}{x^2+4}$

b) $y=\frac {x-1}{3x^2-6}$

c) $y=\frac {x^4+1}{x-5}$

d) $y=\frac {x^3-3x^2+4x-1}{x^2-2}$

Solution

a) We can perform the division

The answer to the division is $y=3-\frac {12}{x^2+4}$

There is a horizontal asymptote at $y = 3$.

b) We cannot divide the two polynomials.

There is a horizontal asymptote at $y = 0$.

c) The power of the numerator is 3 more than the power of the denominator. There are no horizontal or oblique asymptotes.

d) We can perform the division

The answer to the division is $y=x-3+\frac {6x-7}{x^2-2}$

There is an oblique asymptote at $y = x - 3$.

Notice that a rational function will either have a horizontal asymptote, an oblique asymptote or neither kind. In other words horizontal or oblique asymptotes cannot exist together for the same rational function. As $x$ gets large, $y$ values can approach a horizontal line or an oblique line but not both. On the other hand, a rational function can have any number of vertical asymptotes at the same time that it has horizontal or oblique asymptotes.

## Review Questions

Divide the following polynomials:

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

Find all asymptotes of the following rational functions:

1. $\frac{x^2}{x - 2}$
2. $\frac{1}{x + 4}$
3. $\frac{x^2 - 1}{x^2 + 1}$
4. $\frac{x - 4}{x^2 - 9}$
5. $\frac{x^2 + 2x + 1}{3x + 4}$
6. $\frac{x^3 + 1}{4x - 1}$
7. $\frac{x - x^3}{x^2 - 6x - 7}$
8. $\frac{x^4 - 2x}{8x + 24}$

Graph the following rational functions. Indicate all asymptotes on the graph:

1. $\frac{x^2}{x + 2}$
2. $\frac{x^3 - 1}{x^2 - 4}$
3. $\frac{x^2 + 1}{2x - 4}$
4. $\frac{x - x^2}{3x + 2}$

1. $x + 2$
2. $1 - \frac{4}{x}$
3. $1 - \frac{7}{x}$
4. $x + 2 - \frac{5}{x}$
5. $-x-3 + \frac{9}{x}$
6. $1 + \frac{5}{x} + \frac{7}{2x^2}$
7. $-\frac{x}{2} + \frac{1}{2x}$
8. $\frac{5x^3}{3} - \frac{3}{x}$
9. $\frac{x}{12} - 1 + \frac{1}{4x} - \frac{1}{3x^2}$
10. $-\frac{1}{3x^3} + \frac{2}{3x^2} - \frac{1}{9}$
11. $x + 2 + \frac{4}{x + 1}$
12. $x - 8 - \frac{2}{x - 1}$
13. $x + 1$
14. $x - 5$
15. $x - 17 - \frac{39}{x - 3}$
16. $3x + 5 + \frac{15}{x - 2}$
17. $9x - 34 + \frac{128}{x + 4}$
18. $x - \frac{1}{3} - \frac{11}{3(3x + 1)}$
19. $\frac{5}{2}x + \frac{9}{4} - \frac{27}{4(2x - 1)}$
20. $\frac{1}{5}x - \frac{34}{25} - \frac{164}{25(5x + 4)}$
21. vertical: $x = 2$, oblique: $y = x$
22. vertical: $x = -4$, horizontal: $y = 0$
23. horizontal: $y = 1$
24. vertical: $x = 3, x = -3,$ horizontal: $y = 0$
25. vertical: $x = \frac{-4}{3}$, oblique: $y = \frac{x}{3} + \frac{2}{9}$
26. vertical: $x = \frac{1}{4}$
27. vertical: $x = 7, x = -1,$ oblique: $y = -x - 6$
28. vertical: $x = -3$

Feb 22, 2012

Sep 17, 2014