Find the Factors Using the Factor Theorem 16x^3-144x^2-81x+729 ; x-9

$16 x^{3} - 144 x^{2} - 81 x + 729$ ; $x - 9$

Step 1

Divide $\frac{16 x^{3} - 144 x^{2} - 81 x + 729}{x - 9}$ using synthetic division and check if the remainder is equal to $0$ . If the remainder is equal to $0$ , it means that $x - 9$ is a factor for $16 x^{3} - 144 x^{2} - 81 x + 729$ . If the remainder is not equal to $0$ , it means that $x - 9$ is not a factor for $16 x^{3} - 144 x^{2} - 81 x + 729$ .

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

$9$	$16$	$- 144$	$- 81$	$729$
	\xa0			\xa0

The first number in the dividend $(16)$ is put into the first position of the result area (below the horizontal line).

$9$	$16$	$- 144$	$- 81$	$729$
	\xa0			\xa0
	$16$

Multiply the newest entry in the result $(16)$ by the divisor $(9)$ and place the result of $(144)$ under the next term in the dividend $(- 144)$ .

$9$	$16$	$- 144$	$- 81$	$729$
	\xa0	$144$		\xa0
	$16$

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

$9$	$16$	$- 144$	$- 81$	$729$
	\xa0	$144$		\xa0
	$16$	$0$

Multiply the newest entry in the result $(0)$ by the divisor $(9)$ and place the result of $(0)$ under the next term in the dividend $(- 81)$ .

$9$	$16$	$- 144$	$- 81$	$729$
	\xa0	$144$	$0$	\xa0
	$16$	$0$

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

$9$	$16$	$- 144$	$- 81$	$729$
	\xa0	$144$	$0$	\xa0
	$16$	$0$	$- 81$

Multiply the newest entry in the result $(- 81)$ by the divisor $(9)$ and place the result of $(- 729)$ under the next term in the dividend $(729)$ .

$9$	$16$	$- 144$	$- 81$	$729$
	\xa0	$144$	$0$	$- 729$
	$16$	$0$	$- 81$

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

$9$	$16$	$- 144$	$- 81$	$729$
	\xa0	$144$	$0$	$- 729$
	$16$	$0$	$- 81$	$0$

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

$16 x^{2} + 0 x - 81$

Simplify the quotient polynomial.

$16 x^{2} - 81$

Step 2

The remainder from dividing $\frac{16 x^{3} - 144 x^{2} - 81 x + 729}{x - 9}$ is $0$ , which means that $x - 9$ is a factor for $16 x^{3} - 144 x^{2} - 81 x + 729$ .

$x - 9$ is a factor for $16 x^{3} - 144 x^{2} - 81 x + 729$

Step 3

Find all the possible roots for $16 x^{2} - 81$ .

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.

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

$q = \pm 1, \pm 2, \pm 4, \pm 8, \pm 16$

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

$\pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm \frac{1}{8}, \pm \frac{1}{16}, \pm 3, \pm \frac{3}{2}, \pm \frac{3}{4}, \pm \frac{3}{8}, \pm \frac{3}{16}, \pm 9, \pm \frac{9}{2}, \pm \frac{9}{4}, \pm \frac{9}{8}, \pm \frac{9}{16}, \pm 27, \pm \frac{27}{2}, \pm \frac{27}{4}, \pm \frac{27}{8}, \pm \frac{27}{16}, \pm 81, \pm \frac{81}{2}, \pm \frac{81}{4}, \pm \frac{81}{8}, \pm \frac{81}{16}$

Step 4

Set up the next division to determine if $x - \frac{9}{4}$ is a factor of the polynomial $16 x^{2} - 81$ .

$\frac{16 x^{2} - 81}{x - \frac{9}{4}}$

Step 5

Divide the expression using synthetic division to determine if it is a factor of the polynomial. Since $x - \frac{9}{4}$ divides evenly into $16 x^{2} - 81$ , $x - \frac{9}{4}$ is a factor of the polynomial and there is a remaining polynomial of $16 x + 36$ .

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

$\frac{9}{4}$	$16$	$0$	$- 81$
	\xa0		\xa0

The first number in the dividend $(16)$ is put into the first position of the result area (below the horizontal line).

$\frac{9}{4}$	$16$	$0$	$- 81$
	\xa0		\xa0
	$16$

Multiply the newest entry in the result $(16)$ by the divisor $(\frac{9}{4})$ and place the result of $(36)$ under the next term in the dividend $(0)$ .

$\frac{9}{4}$	$16$	$0$	$- 81$
	\xa0	$36$	\xa0
	$16$

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

$\frac{9}{4}$	$16$	$0$	$- 81$
	\xa0	$36$	\xa0
	$16$	$36$

Multiply the newest entry in the result $(36)$ by the divisor $(\frac{9}{4})$ and place the result of $(81)$ under the next term in the dividend $(- 81)$ .

$\frac{9}{4}$	$16$	$0$	$- 81$
	\xa0	$36$	$81$
	$16$	$36$

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

$\frac{9}{4}$	$16$	$0$	$- 81$
	\xa0	$36$	$81$
	$16$	$36$	$0$

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

$(16) x + 36$

Simplify the quotient polynomial.

$16 x + 36$

Step 6

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

$16 x + 36$

Step 7

The factored polynomial is $(x - 9) (x - \frac{9}{4}) (16 x + 36)$ .

$(x - 9) (x - \frac{9}{4}) (16 x + 36)$

Find the Factors Using the Factor Theorem 16x^3-144x^2-81x+729 ; x-9

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