Find the Factors Using the Factor Theorem x+4 , x^3+6x^2+5x-12

$x + 4$ , $x^{3} + 6 x^{2} + 5 x - 12$

Step 1

Divide $\frac{x^{3} + 6 x^{2} + 5 x - 12}{x + 4}$ using synthetic division and check if the remainder is equal to $0$ . If the remainder is equal to $0$ , it means that $x + 4$ is a factor for $x^{3} + 6 x^{2} + 5 x - 12$ . If the remainder is not equal to $0$ , it means that $x + 4$ is not a factor for $x^{3} + 6 x^{2} + 5 x - 12$ .

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

$- 4$	$1$	$6$	$5$	$- 12$
	\xa0			\xa0

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

$- 4$	$1$	$6$	$5$	$- 12$
	\xa0			\xa0
	$1$

Multiply the newest entry in the result $(1)$ by the divisor $(- 4)$ and place the result of $(- 4)$ under the next term in the dividend $(6)$ .

$- 4$	$1$	$6$	$5$	$- 12$
	\xa0	$- 4$		\xa0
	$1$

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

$- 4$	$1$	$6$	$5$	$- 12$
	\xa0	$- 4$		\xa0
	$1$	$2$

Multiply the newest entry in the result $(2)$ by the divisor $(- 4)$ and place the result of $(- 8)$ under the next term in the dividend $(5)$ .

$- 4$	$1$	$6$	$5$	$- 12$
	\xa0	$- 4$	$- 8$	\xa0
	$1$	$2$

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

$- 4$	$1$	$6$	$5$	$- 12$
	\xa0	$- 4$	$- 8$	\xa0
	$1$	$2$	$- 3$

Multiply the newest entry in the result $(- 3)$ by the divisor $(- 4)$ and place the result of $(12)$ under the next term in the dividend $(- 12)$ .

$- 4$	$1$	$6$	$5$	$- 12$
	\xa0	$- 4$	$- 8$	$12$
	$1$	$2$	$- 3$

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

$- 4$	$1$	$6$	$5$	$- 12$
	\xa0	$- 4$	$- 8$	$12$
	$1$	$2$	$- 3$	$0$

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

$1 x^{2} + 2 x - 3$

Simplify the quotient polynomial.

$x^{2} + 2 x - 3$

Step 2

The remainder from dividing $\frac{x^{3} + 6 x^{2} + 5 x - 12}{x + 4}$ is $0$ , which means that $x + 4$ is a factor for $x^{3} + 6 x^{2} + 5 x - 12$ .

$x + 4$ is a factor for $x^{3} + 6 x^{2} + 5 x - 12$

Step 3

Find all the possible roots for $x^{2} + 2 x - 3$ .

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.

$p = \pm 1, \pm 3$

$q = \pm 1$

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

$\pm 1, \pm 3$

Step 4

Set up the next division to determine if $x - 1$ is a factor of the polynomial $x^{2} + 2 x - 3$ .

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

Step 5

Divide the expression using synthetic division to determine if it is a factor of the polynomial. Since $x - 1$ divides evenly into $x^{2} + 2 x - 3$ , $x - 1$ is a factor of the polynomial and there is a remaining polynomial of $x + 3$ .

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

$1$	$1$	$2$	$- 3$
	\xa0		\xa0

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

$1$	$1$	$2$	$- 3$
	\xa0		\xa0
	$1$

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

$1$	$1$	$2$	$- 3$
	\xa0	$1$	\xa0
	$1$

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

$1$	$1$	$2$	$- 3$
	\xa0	$1$	\xa0
	$1$	$3$

Multiply the newest entry in the result $(3)$ by the divisor $(1)$ and place the result of $(3)$ under the next term in the dividend $(- 3)$ .

$1$	$1$	$2$	$- 3$
	\xa0	$1$	$3$
	$1$	$3$

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

$1$	$1$	$2$	$- 3$
	\xa0	$1$	$3$
	$1$	$3$	$0$

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

$(1) x + 3$

Simplify the quotient polynomial.

$x + 3$

Step 6

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

$x + 3$

Step 7

The factored polynomial is $(x + 4) (x - 1) (x + 3)$ .

$(x + 4) (x - 1) (x + 3)$

Find the Factors Using the Factor Theorem x+4 , x^3+6x^2+5x-12

Related Questions

Support LCMGCF.com by disabling your adblocker.