Dividing Polynomial Functions and the Remainder Theorem

6.3.4 Dividing Polynomial Functions and the Remainder Theorem

6.3.4 • Dividing Polynomial Functions and the Remainder Theorem

Activity

Divide the polynomial functions. Find the quotient value.

1. $f (x) = x^{2} - 5 x - 24$

$g (x) = x + 3$

$(\frac{f}{g}) (x) =$ ___________

$f (- 3) =$ ___________

Compare your answers:

$(\frac{f}{g}) (x) = x - 8$

$f (- 3) = 0$

2. $f (x) = x^{2} - 15 x + 54$

$g (x) = x - 9$

$(\frac{f}{g}) (x) =$

___________ $f (9) =$ ___________

Compare your answers:

$(\frac{f}{g}) (x) = x - 6$

$f (9) = 0$

3. $f (x) = x^{2} + 2 x - 3$

$g (x) = x + 3$

$(\frac{f}{g}) (x) =$ ___________

$f (- 3) =$ ___________

Compare your answers:

$(\frac{f}{g}) (x) = x - 1$

$f (- 3) = 0$

The Remainder Theorem states that if a polynomial function $f (x)$ is divided by $x - c$ , then the remainder is $f (c)$ . This means we can always compare the remainder by finding $f (c)$ when the divisor is written in the form $x - c$ .

4. Use the Remainder Theorem to find the remainder when $f (x) = x^{3} - 7 x + 12$ is divided by $x + 3$ .

The remainder is 6

5. Use the Remainder Theorem to find the remainder when $f (x) = 2 x^{3} - 6 x - 24$ is divided by $x - 3$ .

The remainder is 12

Self Check

For functions $f(x)=x^3+2x^2-19x+12$ and $g(x)=x-3$ :

Find $(\frac{f}{g})(x)$ and $(\frac{f}{g})(0)$ .

$(\frac{f}{g})(x)=x^2-5x+4$

$(\frac{f}{g})(0)=-4$
$(\frac{f}{g})(x)=x^2+5x+4$

$(\frac{f}{g})(0)=4$
$(\frac{f}{g})(x)=x^2+5x-4$

$(\frac{f}{g})(0)=12$
$(\frac{f}{g})(x)=x^2+5x-4$

$(\frac{f}{g})(0)=-4$

Additional Resources

Dividing Polynomial Functions

Just as polynomials can be divided, polynomial functions can also be divided.

DIVISION OF POLYNOMIAL FUNCTIONS

For functions $f (x)$ and $g (x)$ , where $g (x) \neq 0$ ,

$(\frac{f}{g}) (x) = \frac{f (x)}{g (x)}$

Example 1

For functions $f (x) = x^{2} - 5 x - 14$ and $g (x) = x + 2$ :

a. Find $(\frac{f}{g}) (x)$ .

Step 1 - $(\frac{f}{g}) (x) = \frac{f (x)}{g (x)}$ Substitute for $f (x)$ and $g (x)$ .

$(\frac{f}{g}) (x) = \frac{f (x) = x^{2} - 5 x - 14}{x + 2}$

Step 2 - Divide the polynomials using long division.

$\begin{array}{l} x - 7 \\ x + 2) \bar{x^{2} - 5 x - 14} \\ \underline{- (x^{2} + 2 x)} \\ - 7 x - 14 \\ \underline{- (- 7 x - 14)} \\ 0 \end{array}$

$(\frac{f}{g}) (x) = x - 7$

b. Find $(\frac{f}{g}) (- 4)$ .

Step 1 - In part (1), we found $(\frac{f}{g}) (x)$ .

$(\frac{f}{g}) (x) = x - 7$

Step 2 - To find $(\frac{f}{g}) (- 4)$ , substitute $x = - 4$ .

$(\frac{f}{g}) (- 4) = - 4 - 7$

$(\frac{f}{g}) (- 4) = - 11$

Try it

Dividing Polynomial Functions

For functions $f (x) = x^{2} - 5 x - 36$ and $g (x) = x + 4$ :

Find $(\frac{f}{g}) (x)$ .
Find $(\frac{f}{g}) (- 5)$ .

Here is how to solve these polynomial function division problems:

$(\frac{f}{g}) (x) = \frac{x^{2} - 5 x - 36}{x + 4} = x - 9$
$(\frac{f}{g}) (- 5) = x - 9 = (- 5) - 9 = - 14$

Using the Remainder Theorem

Let’s look at some division problems and their remainders. They are summarized in the chart below. If we take the dividend from each division problem and use it to define a function, we get the functions shown in the chart. When the divisor is written as $x - c$ , the value of the function at $c$ , $f (c)$ , is the same as the remainder from the division problem.

Dividend	Divisor	Remainder	Function	$f (c)$
$x^{4} - 7 x^{2} + 7 x + 6$	$x - (- 3)$	$3$	$f (x) = x^{4} - 7 x^{2} + 7 x + 6$	$f (- 3) = 3$
$f (x) = 3 x^{3} - 2 x^{2} - 10 x + 8$	$x - (- 3)$	$- 61$	$f (x) = 3 x^{3} - 2 x^{2} - 10 x + 8$	$f (- 3) = - 61$
$x^{4} - 16 x^{2} + 3 x + 15$	$x - (- 4)$	$3$	$f (x) = x^{4} - 16 x^{2} + 3 x + 15$	$f (- 4) = 3$

To see this more generally, we realize we can check a division problem by multiplying the quotient times the divisor and adding the remainder. In function notation, we could say: To get the dividend $f (x)$ , we multiply the quotient, $q (x)$ , times the divisor, $x - c$ , and add the remainder, $r$ .

$f (x) = q (x) (x - c) + r$

If we evaluate this at $c$ , we get:

$f (c) = q (c) (c - c) + r$

$f (c) = q (c) (0) + r$

$f (c) = r$

This leads us to the Remainder Theorem.

REMAINDER THEOREM

If the polynomial function $f (x)$ is divided by $x - c$ , then the remainder is $f (c)$ .

Example 2

Use the Remainder Theorem to find the remainder when $f (x) = x^{3} + 3 x + 19$ is divided by $x + 2$ .

To use the Remainder Theorem, we must use the divisor in the $x - c$ form. We can write the divisor $x + 2$ as $x - (- 2)$ . So, our $c$ is $- 2$ .

To find the remainder, we evaluate $f (c)$ , which is $f (- 2)$ .

$f (x) = x^{3} + 3 x + 19$

Step 1 - To evaluate $f (- 2)$ , substitute $x = - 2$ .

$f (- 2) = (- 2)^{3} + 3 (- 2) + 19$

Step 2 - Simplify.

$f (- 2) = (- 8) - 6 + 19$

$f (- 2) = 5$

The remainder is 5 when $f (x) = x^{3} + 3 x + 19$ is divided by $x + 2$ .

Step 3 - Check using synthetic division.

Synthetic division setup: The divisor is -2. The dividend’s coefficients are 1, 0, 3, and 19. The first row below shows -2, 4, and -14. The final row is 1, -2, 7, remainder 5.

The remainder is 5.

Try it

Using the Remainder Theorem

Use the Remainder Theorem to find the remainder when $f (x) = x^{3} + 4 x + 15$ is divided by $x + 2$ .

Here is how to solve a problem using the Remainder Theorem:

$f (x) = x^{3} + 4 x + 15$

$f (2) = (- 2)^{3} + 4 (- 2) + 15$

$f (2) = - 8 - 8 + 15 = - 1$

The remainder is $- 1$ .