Find the Factors Using the Factor Theorem 100x^3-200x^2-81x+162 ; x-2

$100 x^{3} - 200 x^{2} - 81 x + 162$ ; $x - 2$

Divide $\frac{100 x^{3} - 200 x^{2} - 81 x + 162}{x - 2}$ 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 $100 x^{3} - 200 x^{2} - 81 x + 162$ . If the remainder is not equal to $0$ , it means that $x - 2$ is not a factor for $100 x^{3} - 200 x^{2} - 81 x + 162$ .

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

$2$	$100$	$- 200$	$- 81$	$162$
	\xa0			\xa0

The first number in the dividend $(100)$ is put into the first position of the result area (below the horizontal line).

$2$	$100$	$- 200$	$- 81$	$162$
	\xa0			\xa0
	$100$

Multiply the newest entry in the result $(100)$ by the divisor $(2)$ and place the result of $(200)$ under the next term in the dividend $(- 200)$ .

$2$	$100$	$- 200$	$- 81$	$162$
	\xa0	$200$		\xa0
	$100$

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

$2$	$100$	$- 200$	$- 81$	$162$
	\xa0	$200$		\xa0
	$100$	$0$

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 $(- 81)$ .

$2$	$100$	$- 200$	$- 81$	$162$
	\xa0	$200$	$0$	\xa0
	$100$	$0$

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

$2$	$100$	$- 200$	$- 81$	$162$
	\xa0	$200$	$0$	\xa0
	$100$	$0$	$- 81$

Multiply the newest entry in the result $(- 81)$ by the divisor $(2)$ and place the result of $(- 162)$ under the next term in the dividend $(162)$ .

$2$	$100$	$- 200$	$- 81$	$162$
	\xa0	$200$	$0$	$- 162$
	$100$	$0$	$- 81$

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

$2$	$100$	$- 200$	$- 81$	$162$
	\xa0	$200$	$0$	$- 162$
	$100$	$0$	$- 81$	$0$

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

$100 x^{2} + 0 x - 81$

Simplify the quotient polynomial.

$100 x^{2} - 81$

The remainder from dividing $\frac{100 x^{3} - 200 x^{2} - 81 x + 162}{x - 2}$ is $0$ , which means that $x - 2$ is a factor for $100 x^{3} - 200 x^{2} - 81 x + 162$ .

$x - 2$ is a factor for $100 x^{3} - 200 x^{2} - 81 x + 162$

Find all the possible roots for $100 x^{2} - 81$ .

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.

$p = \pm 1, \pm 3, \pm 9, \pm 27, \pm 81$

$q = \pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20, \pm 25, \pm 50, \pm 100$

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

$\pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm \frac{1}{5}, \pm \frac{1}{10}, \pm \frac{1}{20}, \pm \frac{1}{25}, \pm \frac{1}{50}, \pm \frac{1}{100}, \pm 3, \pm \frac{3}{2}, \pm \frac{3}{4}, \pm \frac{3}{5}, \pm \frac{3}{10}, \pm \frac{3}{20}, \pm \frac{3}{25}, \pm \frac{3}{50}, \pm \frac{3}{100}, \pm 9, \pm \frac{9}{2}, \pm \frac{9}{4}, \pm \frac{9}{5}, \pm \frac{9}{10}, \pm \frac{9}{20}, \pm \frac{9}{25}, \pm \frac{9}{50}, \pm \frac{9}{100}, \pm 27, \pm \frac{27}{2}, \pm \frac{27}{4}, \pm \frac{27}{5}, \pm \frac{27}{10}, \pm \frac{27}{20}, \pm \frac{27}{25}, \pm \frac{27}{50}, \pm \frac{27}{100}, \pm 81, \pm \frac{81}{2}, \pm \frac{81}{4}, \pm \frac{81}{5}, \pm \frac{81}{10}, \pm \frac{81}{20}, \pm \frac{81}{25}, \pm \frac{81}{50}, \pm \frac{81}{100}$

Set up the next division to determine if $x - \frac{9}{10}$ is a factor of the polynomial $100 x^{2} - 81$ .

$\frac{100 x^{2} - 81}{x - \frac{9}{10}}$

Divide the expression using synthetic division to determine if it is a factor of the polynomial. Since $x - \frac{9}{10}$ divides evenly into $100 x^{2} - 81$ , $x - \frac{9}{10}$ is a factor of the polynomial and there is a remaining polynomial of $100 x + 90$ .

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

$\frac{9}{10}$	$100$	$0$	$- 81$
	\xa0		\xa0

The first number in the dividend $(100)$ is put into the first position of the result area (below the horizontal line).

$\frac{9}{10}$	$100$	$0$	$- 81$
	\xa0		\xa0
	$100$

Multiply the newest entry in the result $(100)$ by the divisor $(\frac{9}{10})$ and place the result of $(90)$ under the next term in the dividend $(0)$ .

$\frac{9}{10}$	$100$	$0$	$- 81$
	\xa0	$90$	\xa0
	$100$

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

$\frac{9}{10}$	$100$	$0$	$- 81$
	\xa0	$90$	\xa0
	$100$	$90$

Multiply the newest entry in the result $(90)$ by the divisor $(\frac{9}{10})$ and place the result of $(81)$ under the next term in the dividend $(- 81)$ .

$\frac{9}{10}$	$100$	$0$	$- 81$
	\xa0	$90$	$81$
	$100$	$90$

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

$\frac{9}{10}$	$100$	$0$	$- 81$
	\xa0	$90$	$81$
	$100$	$90$	$0$

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

$(100) x + 90$

Simplify the quotient polynomial.

$100 x + 90$

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

$100 x + 90$

The factored polynomial is $(x - 2) (x - \frac{9}{10}) (100 x + 90)$ .

$(x - 2) (x - \frac{9}{10}) (100 x + 90)$

Find the Factors Using the Factor Theorem 100x^3-200x^2-81x+162 ; x-2

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