Precalculus Examples

$x^{4} + 2 x^{3} - 8 x - 16$ , $x + 2$

Step 1

Divide $\frac{x^{4} + 2 x^{3} - 8 x - 16}{x + 2}$ using synthetic division and check if the remainder is equal to $0$ . If the remainder is equal to $0$ , it means that $x + 2$ is a factor for $x^{4} + 2 x^{3} - 8 x - 16$ . If the remainder is not equal to $0$ , it means that $x + 2$ is not a factor for $x^{4} + 2 x^{3} - 8 x - 16$ .

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Step 1.1

Place the numbers representing the divisor and the dividend into a division-like configuration.

$- 2$	$1$	$2$	$0$	$- 8$	$- 16$

Step 1.2

The first number in the dividend $(1)$ is put into the first position of the result area (below the horizontal line).

$- 2$	$1$	$2$	$0$	$- 8$	$- 16$

	$1$

Step 1.3

Multiply the newest entry in the result $(1)$ by the divisor $(- 2)$ and place the result of $(- 2)$ under the next term in the dividend $(2)$ .

$- 2$	$1$	$2$	$0$	$- 8$	$- 16$
		$- 2$
	$1$

Step 1.4

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$- 2$	$1$	$2$	$0$	$- 8$	$- 16$
		$- 2$
	$1$	$0$

Step 1.5

Multiply the newest entry in the result $(0)$ by the divisor $(- 2)$ and place the result of $(0)$ under the next term in the dividend $(0)$ .

$- 2$	$1$	$2$	$0$	$- 8$	$- 16$
		$- 2$	$0$
	$1$	$0$

Step 1.6

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$- 2$	$1$	$2$	$0$	$- 8$	$- 16$
		$- 2$	$0$
	$1$	$0$	$0$

Step 1.7

Multiply the newest entry in the result $(0)$ by the divisor $(- 2)$ and place the result of $(0)$ under the next term in the dividend $(- 8)$ .

$- 2$	$1$	$2$	$0$	$- 8$	$- 16$
		$- 2$	$0$	$0$
	$1$	$0$	$0$

Step 1.8

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$- 2$	$1$	$2$	$0$	$- 8$	$- 16$
		$- 2$	$0$	$0$
	$1$	$0$	$0$	$- 8$

Step 1.9

Multiply the newest entry in the result $(- 8)$ by the divisor $(- 2)$ and place the result of $(16)$ under the next term in the dividend $(- 16)$ .

$- 2$	$1$	$2$	$0$	$- 8$	$- 16$
		$- 2$	$0$	$0$	$16$
	$1$	$0$	$0$	$- 8$

Step 1.10

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$- 2$	$1$	$2$	$0$	$- 8$	$- 16$
		$- 2$	$0$	$0$	$16$
	$1$	$0$	$0$	$- 8$	$0$

Step 1.11

All numbers except the last become the coefficients of the quotient polynomial. The last value in the result line is the remainder.

$1 x^{3} + 0 x^{2} + (0) x - 8$

Step 1.12

Simplify the quotient polynomial.

$x^{3} - 8$

Step 2

The remainder from dividing $\frac{x^{4} + 2 x^{3} - 8 x - 16}{x + 2}$ is $0$ , which means that $x + 2$ is a factor for $x^{4} + 2 x^{3} - 8 x - 16$ .

$x + 2$ is a factor for $x^{4} + 2 x^{3} - 8 x - 16$

Step 3

Find all the possible roots for $x^{3} - 8$ .

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Step 3.1

If a polynomial function has integer coefficients, then every rational zero will have the form $\frac{p}{q}$ where $p$ is a factor of the constant and $q$ is a factor of the leading coefficient.

$p = \pm 1, \pm 2, \pm 4, \pm 8$

$q = \pm 1$

Step 3.2

Find every combination of $\pm \frac{p}{q}$ . These are the possible roots of the polynomial function.

$\pm 1, \pm 2, \pm 4, \pm 8$

Step 4

Set up the next division to determine if $x - 2$ is a factor of the polynomial $x^{3} - 8$ .

$\frac{x^{3} - 8}{x - 2}$

Step 5

Divide the expression using synthetic division to determine if it is a factor of the polynomial. Since $x - 2$ divides evenly into $x^{3} - 8$ , $x - 2$ is a factor of the polynomial and there is a remaining polynomial of $x^{2} + 2 x + 4$ .

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Step 5.1

Place the numbers representing the divisor and the dividend into a division-like configuration.

$2$	$1$	$0$	$0$	$- 8$

Step 5.2

The first number in the dividend $(1)$ is put into the first position of the result area (below the horizontal line).

$2$	$1$	$0$	$0$	$- 8$

	$1$

Step 5.3

Multiply the newest entry in the result $(1)$ by the divisor $(2)$ and place the result of $(2)$ under the next term in the dividend $(0)$ .

$2$	$1$	$0$	$0$	$- 8$
		$2$
	$1$

Step 5.4

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$2$	$1$	$0$	$0$	$- 8$
		$2$
	$1$	$2$

Step 5.5

Multiply the newest entry in the result $(2)$ by the divisor $(2)$ and place the result of $(4)$ under the next term in the dividend $(0)$ .

$2$	$1$	$0$	$0$	$- 8$
		$2$	$4$
	$1$	$2$

Step 5.6

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$2$	$1$	$0$	$0$	$- 8$
		$2$	$4$
	$1$	$2$	$4$

Step 5.7

Multiply the newest entry in the result $(4)$ by the divisor $(2)$ and place the result of $(8)$ under the next term in the dividend $(- 8)$ .

$2$	$1$	$0$	$0$	$- 8$
		$2$	$4$	$8$
	$1$	$2$	$4$

Step 5.8

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$2$	$1$	$0$	$0$	$- 8$
		$2$	$4$	$8$
	$1$	$2$	$4$	$0$

Step 5.9

All numbers except the last become the coefficients of the quotient polynomial. The last value in the result line is the remainder.

$1 x^{2} + 2 x + 4$

Step 5.10

Simplify the quotient polynomial.

$x^{2} + 2 x + 4$

Step 6

Find all the possible roots for $x^{2} + 2 x + 4$ .

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Step 6.1

If a polynomial function has integer coefficients, then every rational zero will have the form $\frac{p}{q}$ where $p$ is a factor of the constant and $q$ is a factor of the leading coefficient.

$p = \pm 1, \pm 2, \pm 4$

$q = \pm 1$

Step 6.2

Find every combination of $\pm \frac{p}{q}$ . These are the possible roots of the polynomial function.

$\pm 1, \pm 2, \pm 4$

Step 7

The final factor is the only factor left over from the synthetic division.

$x^{2} + 2 x + 4$

Step 8

The factored polynomial is $(x + 2) (x - 2) (x^{2} + 2 x + 4)$ .

$(x + 2) (x - 2) (x^{2} + 2 x + 4)$