Find the Factors Using the Factor Theorem 36x^3+60x^2-143x-242 , x-2

$36 x^{3} + 60 x^{2} - 143 x - 242$ , $x - 2$

Step 1

Divide $\frac{36 x^{3} + 60 x^{2} - 143 x - 242}{x - 2}$ 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 $36 x^{3} + 60 x^{2} - 143 x - 242$ . If the remainder is not equal to $0$ , it means that $x - 2$ is not a factor for $36 x^{3} + 60 x^{2} - 143 x - 242$ .

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

$2$	$36$	$60$	$- 143$	$- 242$
	\xa0			\xa0

The first number in the dividend $(36)$ is put into the first position of the result area (below the horizontal line).

$2$	$36$	$60$	$- 143$	$- 242$
	\xa0			\xa0
	$36$

Multiply the newest entry in the result $(36)$ by the divisor $(2)$ and place the result of $(72)$ under the next term in the dividend $(60)$ .

$2$	$36$	$60$	$- 143$	$- 242$
	\xa0	$72$		\xa0
	$36$

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

$2$	$36$	$60$	$- 143$	$- 242$
	\xa0	$72$		\xa0
	$36$	$132$

Multiply the newest entry in the result $(132)$ by the divisor $(2)$ and place the result of $(264)$ under the next term in the dividend $(- 143)$ .

$2$	$36$	$60$	$- 143$	$- 242$
	\xa0	$72$	$264$	\xa0
	$36$	$132$

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

$2$	$36$	$60$	$- 143$	$- 242$
	\xa0	$72$	$264$	\xa0
	$36$	$132$	$121$

Multiply the newest entry in the result $(121)$ by the divisor $(2)$ and place the result of $(242)$ under the next term in the dividend $(- 242)$ .

$2$	$36$	$60$	$- 143$	$- 242$
	\xa0	$72$	$264$	$242$
	$36$	$132$	$121$

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

$2$	$36$	$60$	$- 143$	$- 242$
	\xa0	$72$	$264$	$242$
	$36$	$132$	$121$	$0$

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

$36 x^{2} + 132 x + 121$

Step 2

The remainder from dividing $\frac{36 x^{3} + 60 x^{2} - 143 x - 242}{x - 2}$ is $0$ , which means that $x - 2$ is a factor for $36 x^{3} + 60 x^{2} - 143 x - 242$ .

$x - 2$ is a factor for $36 x^{3} + 60 x^{2} - 143 x - 242$

Step 3

Find all the possible roots for $36 x^{2} + 132 x + 121$ .

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.

$p = \pm 1, \pm 11, \pm 121$

$q = \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 9, \pm 12, \pm 18, \pm 36$

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

$\pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{4}, \pm \frac{1}{6}, \pm \frac{1}{9}, \pm \frac{1}{12}, \pm \frac{1}{18}, \pm \frac{1}{36}, \pm 11, \pm \frac{11}{2}, \pm \frac{11}{3}, \pm \frac{11}{4}, \pm \frac{11}{6}, \pm \frac{11}{9}, \pm \frac{11}{12}, \pm \frac{11}{18}, \pm \frac{11}{36}, \pm 121, \pm \frac{121}{2}, \pm \frac{121}{3}, \pm \frac{121}{4}, \pm \frac{121}{6}, \pm \frac{121}{9}, \pm \frac{121}{12}, \pm \frac{121}{18}, \pm \frac{121}{36}$

Step 4

Set up the next division to determine if $x + \frac{11}{6}$ is a factor of the polynomial $36 x^{2} + 132 x + 121$ .

$\frac{36 x^{2} + 132 x + 121}{x + \frac{11}{6}}$

Step 5

Divide the expression using synthetic division to determine if it is a factor of the polynomial. Since $x + \frac{11}{6}$ divides evenly into $36 x^{2} + 132 x + 121$ , $x + \frac{11}{6}$ is a factor of the polynomial and there is a remaining polynomial of $36 x + 66$ .

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

$- \frac{11}{6}$	$36$	$132$	$121$
	\xa0		\xa0

The first number in the dividend $(36)$ is put into the first position of the result area (below the horizontal line).

$- \frac{11}{6}$	$36$	$132$	$121$
	\xa0		\xa0
	$36$

Multiply the newest entry in the result $(36)$ by the divisor $(- \frac{11}{6})$ and place the result of $(- 66)$ under the next term in the dividend $(132)$ .

$- \frac{11}{6}$	$36$	$132$	$121$
	\xa0	$- 66$	\xa0
	$36$

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

$- \frac{11}{6}$	$36$	$132$	$121$
	\xa0	$- 66$	\xa0
	$36$	$66$

Multiply the newest entry in the result $(66)$ by the divisor $(- \frac{11}{6})$ and place the result of $(- 121)$ under the next term in the dividend $(121)$ .

$- \frac{11}{6}$	$36$	$132$	$121$
	\xa0	$- 66$	$- 121$
	$36$	$66$

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

$- \frac{11}{6}$	$36$	$132$	$121$
	\xa0	$- 66$	$- 121$
	$36$	$66$	$0$

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

$(36) x + 66$

Simplify the quotient polynomial.

$36 x + 66$

Step 6

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

$36 x + 66$

Step 7

The factored polynomial is $(x - 2) (x + \frac{11}{6}) (36 x + 66)$ .

$(x - 2) (x + \frac{11}{6}) (36 x + 66)$

Find the Factors Using the Factor Theorem 36x^3+60x^2-143x-242 , x-2

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