Find the Roots (Zeros) x^5+3x^4-9x^3-31x^2+36

$x^{5} + 3 x^{4} - 9 x^{3} - 31 x^{2} + 36$

Step 1

Set $x^{5} + 3 x^{4} - 9 x^{3} - 31 x^{2} + 36$ equal to $0$ .

$x^{5} + 3 x^{4} - 9 x^{3} - 31 x^{2} + 36 = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Regroup terms.

$x^{5} + 3 x^{4} - 9 x^{3} - 31 x^{2} + 36 = 0$

Factor $x^{4}$ out of $x^{5} + 3 x^{4}$ .

Factor $x^{4}$ out of $x^{5}$ .

$x^{4} x + 3 x^{4} - 9 x^{3} - 31 x^{2} + 36 = 0$

Factor $x^{4}$ out of $3 x^{4}$ .

$x^{4} x + x^{4} \cdot 3 - 9 x^{3} - 31 x^{2} + 36 = 0$

Factor $x^{4}$ out of $x^{4} x + x^{4} \cdot 3$ .

$x^{4} (x + 3) - 9 x^{3} - 31 x^{2} + 36 = 0$

Factor $- 9 x^{3} - 31 x^{2} + 36$ using the rational roots test.

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.

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

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

$\pm 1, \pm 0.\overline{1}, \pm 0.\overline{3}, \pm 36, \pm 4, \pm 12, \pm 2, \pm 0.\overline{2}, \pm 0.\overline{6}, \pm 18, \pm 6, \pm 3, \pm 1.\overline{3}, \pm 0.\overline{4}, \pm 9$

Substitute $- 3$ and simplify the expression. In this case, the expression is equal to $0$ so $- 3$ is a root of the polynomial.

Substitute $- 3$ into the polynomial.

$- 9 {(- 3)}^{3} - 31 {(- 3)}^{2} + 36$

Raise $- 3$ to the power of $3$ .

$- 9 \cdot - 27 - 31 {(- 3)}^{2} + 36$

Multiply $- 9$ by $- 27$ .

$243 - 31 {(- 3)}^{2} + 36$

Raise $- 3$ to the power of $2$ .

$243 - 31 \cdot 9 + 36$

Multiply $- 31$ by $9$ .

$243 - 279 + 36$

Subtract $279$ from $243$ .

$- 36 + 36$

Add $- 36$ and $36$ .

$0$

Since $- 3$ is a known root, divide the polynomial by $x + 3$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{- 9 x^{3} - 31 x^{2} + 36}{x + 3}$

Divide $- 9 x^{3} - 31 x^{2} + 36$ by $x + 3$ .

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$x$

$3$

$9 x^{3}$

$31 x^{2}$

$0 x$

$36$

Divide the highest order term in the dividend $- 9 x^{3}$ by the highest order term in divisor $x$ .

				-	$9 x^{2}$
	$x$	+	$3$	-	$9 x^{3}$	-	$31 x^{2}$	+	$0 x$	+	$36$

Multiply the new quotient term by the divisor.

			-	$9 x^{2}$
$x$	+	$3$	-	$9 x^{3}$	-	$31 x^{2}$	+	$0 x$	+	$36$
			-	$9 x^{3}$	-	$27 x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $- 9 x^{3} - 27 x^{2}$

			-	$9 x^{2}$
$x$	+	$3$	-	$9 x^{3}$	-	$31 x^{2}$	+	$0 x$	+	$36$
			+	$9 x^{3}$	+	$27 x^{2}$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

			-	$9 x^{2}$
$x$	+	$3$	-	$9 x^{3}$	-	$31 x^{2}$	+	$0 x$	+	$36$
			+	$9 x^{3}$	+	$27 x^{2}$
					-	$4 x^{2}$

Pull the next terms from the original dividend down into the current dividend.

			-	$9 x^{2}$
$x$	+	$3$	-	$9 x^{3}$	-	$31 x^{2}$	+	$0 x$	+	$36$
			+	$9 x^{3}$	+	$27 x^{2}$
					-	$4 x^{2}$	+	$0 x$

Divide the highest order term in the dividend $- 4 x^{2}$ by the highest order term in divisor $x$ .

			-	$9 x^{2}$	-	$4 x$
$x$	+	$3$	-	$9 x^{3}$	-	$31 x^{2}$	+	$0 x$	+	$36$
			+	$9 x^{3}$	+	$27 x^{2}$
					-	$4 x^{2}$	+	$0 x$

Multiply the new quotient term by the divisor.

			-	$9 x^{2}$	-	$4 x$
$x$	+	$3$	-	$9 x^{3}$	-	$31 x^{2}$	+	$0 x$	+	$36$
			+	$9 x^{3}$	+	$27 x^{2}$
					-	$4 x^{2}$	+	$0 x$
					-	$4 x^{2}$	-	$12 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 4 x^{2} - 12 x$

			-	$9 x^{2}$	-	$4 x$
$x$	+	$3$	-	$9 x^{3}$	-	$31 x^{2}$	+	$0 x$	+	$36$
			+	$9 x^{3}$	+	$27 x^{2}$
					-	$4 x^{2}$	+	$0 x$
					+	$4 x^{2}$	+	$12 x$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

			-	$9 x^{2}$	-	$4 x$
$x$	+	$3$	-	$9 x^{3}$	-	$31 x^{2}$	+	$0 x$	+	$36$
			+	$9 x^{3}$	+	$27 x^{2}$
					-	$4 x^{2}$	+	$0 x$
					+	$4 x^{2}$	+	$12 x$
							+	$12 x$

Pull the next terms from the original dividend down into the current dividend.

			-	$9 x^{2}$	-	$4 x$
$x$	+	$3$	-	$9 x^{3}$	-	$31 x^{2}$	+	$0 x$	+	$36$
			+	$9 x^{3}$	+	$27 x^{2}$
					-	$4 x^{2}$	+	$0 x$
					+	$4 x^{2}$	+	$12 x$
							+	$12 x$	+	$36$

Divide the highest order term in the dividend $12 x$ by the highest order term in divisor $x$ .

			-	$9 x^{2}$	-	$4 x$	+	$12$
$x$	+	$3$	-	$9 x^{3}$	-	$31 x^{2}$	+	$0 x$	+	$36$
			+	$9 x^{3}$	+	$27 x^{2}$
					-	$4 x^{2}$	+	$0 x$
					+	$4 x^{2}$	+	$12 x$
							+	$12 x$	+	$36$

Multiply the new quotient term by the divisor.

			-	$9 x^{2}$	-	$4 x$	+	$12$
$x$	+	$3$	-	$9 x^{3}$	-	$31 x^{2}$	+	$0 x$	+	$36$
			+	$9 x^{3}$	+	$27 x^{2}$
					-	$4 x^{2}$	+	$0 x$
					+	$4 x^{2}$	+	$12 x$
							+	$12 x$	+	$36$
							+	$12 x$	+	$36$

The expression needs to be subtracted from the dividend, so change all the signs in $12 x + 36$

			-	$9 x^{2}$	-	$4 x$	+	$12$
$x$	+	$3$	-	$9 x^{3}$	-	$31 x^{2}$	+	$0 x$	+	$36$
			+	$9 x^{3}$	+	$27 x^{2}$
					-	$4 x^{2}$	+	$0 x$
					+	$4 x^{2}$	+	$12 x$
							+	$12 x$	+	$36$
							-	$12 x$	-	$36$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

			-	$9 x^{2}$	-	$4 x$	+	$12$
$x$	+	$3$	-	$9 x^{3}$	-	$31 x^{2}$	+	$0 x$	+	$36$
			+	$9 x^{3}$	+	$27 x^{2}$
					-	$4 x^{2}$	+	$0 x$
					+	$4 x^{2}$	+	$12 x$
							+	$12 x$	+	$36$
							-	$12 x$	-	$36$
										$0$

Since the remander is $0$ , the final answer is the quotient.

$- 9 x^{2} - 4 x + 12$

Write $- 9 x^{3} - 31 x^{2} + 36$ as a set of factors.

$x^{4} (x + 3) + (x + 3) (- 9 x^{2} - 4 x + 12) = 0$

Factor $x + 3$ out of $x^{4} (x + 3) + (x + 3) (- 9 x^{2} - 4 x + 12)$ .

Factor $x + 3$ out of $x^{4} (x + 3)$ .

$(x + 3) x^{4} + (x + 3) (- 9 x^{2} - 4 x + 12) = 0$

Factor $x + 3$ out of $(x + 3) x^{4} + (x + 3) (- 9 x^{2} - 4 x + 12)$ .

$(x + 3) (x^{4} - 9 x^{2} - 4 x + 12) = 0$

Factor.

Rewrite $x^{4} - 9 x^{2} - 4 x + 12$ in a factored form.

Factor $x^{4} - 9 x^{2} - 4 x + 12$ using the rational roots test.

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.

$p = \pm 1, \pm 12, \pm 2, \pm 6, \pm 3, \pm 4 q = \pm 1$

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

$\pm 1, \pm 12, \pm 2, \pm 6, \pm 3, \pm 4$

Substitute $1$ and simplify the expression. In this case, the expression is equal to $0$ so $1$ is a root of the polynomial.

Substitute $1$ into the polynomial.

$1^{4} - 9 \cdot 1^{2} - 4 \cdot 1 + 12$

Raise $1$ to the power of $4$ .

$1 - 9 \cdot 1^{2} - 4 \cdot 1 + 12$

Raise $1$ to the power of $2$ .

$1 - 9 \cdot 1 - 4 \cdot 1 + 12$

Multiply $- 9$ by $1$ .

$1 - 9 - 4 \cdot 1 + 12$

Subtract $9$ from $1$ .

$- 8 - 4 \cdot 1 + 12$

Multiply $- 4$ by $1$ .

$- 8 - 4 + 12$

Subtract $4$ from $- 8$ .

$- 12 + 12$

Add $- 12$ and $12$ .

$0$

Since $1$ is a known root, divide the polynomial by $x - 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{4} - 9 x^{2} - 4 x + 12}{x - 1}$

Divide $x^{4} - 9 x^{2} - 4 x + 12$ by $x - 1$ .

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$x$

$1$

$x^{4}$

$0 x^{3}$

$9 x^{2}$

$4 x$

$12$

Divide the highest order term in the dividend $x^{4}$ by the highest order term in divisor $x$ .

$x^{3}$

$x$

$1$

$x^{4}$

$0 x^{3}$

$9 x^{2}$

$4 x$

$12$

Multiply the new quotient term by the divisor.

$x^{3}$

$x$

$1$

$x^{4}$

$0 x^{3}$

$9 x^{2}$

$4 x$

$12$

$x^{4}$

$x^{3}$

The expression needs to be subtracted from the dividend, so change all the signs in $x^{4} - x^{3}$

$x^{3}$

$x$

$1$

$x^{4}$

$0 x^{3}$

$9 x^{2}$

$4 x$

$12$

$x^{4}$

$x^{3}$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

$x^{3}$

$x$

$1$

$x^{4}$

$0 x^{3}$

$9 x^{2}$

$4 x$

$12$

$x^{4}$

$x^{3}$

Pull the next terms from the original dividend down into the current dividend.

$x^{3}$

$x$

$1$

$x^{4}$

$0 x^{3}$

$9 x^{2}$

$4 x$

$12$

$x^{4}$

$x^{3}$

$9 x^{2}$

Divide the highest order term in the dividend $x^{3}$ by the highest order term in divisor $x$ .

$x^{3}$

$x^{2}$

$x$

$1$

$x^{4}$

$0 x^{3}$

$9 x^{2}$

$4 x$

$12$

$x^{4}$

$x^{3}$

$9 x^{2}$

Multiply the new quotient term by the divisor.

$x^{3}$

$x^{2}$

$x$

$1$

$x^{4}$

$0 x^{3}$

$9 x^{2}$

$4 x$

$12$

$x^{4}$

$x^{3}$

$9 x^{2}$

$x^{3}$

$x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $x^{3} - x^{2}$

$x^{3}$

$x^{2}$

$x$

$1$

$x^{4}$

$0 x^{3}$

$9 x^{2}$

$4 x$

$12$

$x^{4}$

$x^{3}$

$9 x^{2}$

$x^{3}$

$x^{2}$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

$x^{3}$

$x^{2}$

$x$

$1$

$x^{4}$

$0 x^{3}$

$9 x^{2}$

$4 x$

$12$

$x^{4}$

$x^{3}$

$9 x^{2}$

$x^{3}$

$x^{2}$

$8 x^{2}$

Pull the next terms from the original dividend down into the current dividend.

$x^{3}$

$x^{2}$

$x$

$1$

$x^{4}$

$0 x^{3}$

$9 x^{2}$

$4 x$

$12$

$x^{4}$

$x^{3}$

$9 x^{2}$

$x^{3}$

$x^{2}$

$8 x^{2}$

$4 x$

Divide the highest order term in the dividend $- 8 x^{2}$ by the highest order term in divisor $x$ .

$x^{3}$

$x^{2}$

$8 x$

$x$

$1$

$x^{4}$

$0 x^{3}$

$9 x^{2}$

$4 x$

$12$

$x^{4}$

$x^{3}$

$9 x^{2}$

$x^{3}$

$x^{2}$

$8 x^{2}$

$4 x$

Multiply the new quotient term by the divisor.

$x^{3}$

$x^{2}$

$8 x$

$x$

$1$

$x^{4}$

$0 x^{3}$

$9 x^{2}$

$4 x$

$12$

$x^{4}$

$x^{3}$

$9 x^{2}$

$x^{3}$

$x^{2}$

$8 x^{2}$

$4 x$

$8 x^{2}$

$8 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 8 x^{2} + 8 x$

$x^{3}$

$x^{2}$

$8 x$

$x$

$1$

$x^{4}$

$0 x^{3}$

$9 x^{2}$

$4 x$

$12$

$x^{4}$

$x^{3}$

$9 x^{2}$

$x^{3}$

$x^{2}$

$8 x^{2}$

$4 x$

$8 x^{2}$

$8 x$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

$x^{3}$

$x^{2}$

$8 x$

$x$

$1$

$x^{4}$

$0 x^{3}$

$9 x^{2}$

$4 x$

$12$

$x^{4}$

$x^{3}$

$9 x^{2}$

$x^{3}$

$x^{2}$

$8 x^{2}$

$4 x$

$8 x^{2}$

$8 x$

$12 x$

Pull the next terms from the original dividend down into the current dividend.

$x^{3}$

$x^{2}$

$8 x$

$x$

$1$

$x^{4}$

$0 x^{3}$

$9 x^{2}$

$4 x$

$12$

$x^{4}$

$x^{3}$

$9 x^{2}$

$x^{3}$

$x^{2}$

$8 x^{2}$

$4 x$

$8 x^{2}$

$8 x$

$12 x$

$12$

Divide the highest order term in the dividend $- 12 x$ by the highest order term in divisor $x$ .

$x^{3}$

$x^{2}$

$8 x$

$12$

$x$

$1$

$x^{4}$

$0 x^{3}$

$9 x^{2}$

$4 x$

$12$

$x^{4}$

$x^{3}$

$9 x^{2}$

$x^{3}$

$x^{2}$

$8 x^{2}$

$4 x$

$8 x^{2}$

$8 x$

$12 x$

$12$

Multiply the new quotient term by the divisor.

$x^{3}$

$x^{2}$

$8 x$

$12$

$x$

$1$

$x^{4}$

$0 x^{3}$

$9 x^{2}$

$4 x$

$12$

$x^{4}$

$x^{3}$

$9 x^{2}$

$x^{3}$

$x^{2}$

$8 x^{2}$

$4 x$

$8 x^{2}$

$8 x$

$12 x$

$12$

$12 x$

$12$

The expression needs to be subtracted from the dividend, so change all the signs in $- 12 x + 12$

$x^{3}$

$x^{2}$

$8 x$

$12$

$x$

$1$

$x^{4}$

$0 x^{3}$

$9 x^{2}$

$4 x$

$12$

$x^{4}$

$x^{3}$

$9 x^{2}$

$x^{3}$

$x^{2}$

$8 x^{2}$

$4 x$

$8 x^{2}$

$8 x$

$12 x$

$12$

$12 x$

$12$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

$x^{3}$

$x^{2}$

$8 x$

$12$

$x$

$1$

$x^{4}$

$0 x^{3}$

$9 x^{2}$

$4 x$

$12$

$x^{4}$

$x^{3}$

$9 x^{2}$

$x^{3}$

$x^{2}$

$8 x^{2}$

$4 x$

$8 x^{2}$

$8 x$

$12 x$

$12$

$12 x$

$12$

$0$

Since the remander is $0$ , the final answer is the quotient.

$x^{3} + x^{2} - 8 x - 12$

Write $x^{4} - 9 x^{2} - 4 x + 12$ as a set of factors.

$(x + 3) ((x - 1) (x^{3} + x^{2} - 8 x - 12)) = 0$

Factor $x^{3} + x^{2} - 8 x - 12$ using the rational roots test.

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.

$p = \pm 1, \pm 12, \pm 2, \pm 6, \pm 3, \pm 4 q = \pm 1$

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

$\pm 1, \pm 12, \pm 2, \pm 6, \pm 3, \pm 4$

Substitute $- 2$ and simplify the expression. In this case, the expression is equal to $0$ so $- 2$ is a root of the polynomial.

Substitute $- 2$ into the polynomial.

${(- 2)}^{3} + {(- 2)}^{2} - 8 \cdot - 2 - 12$

Raise $- 2$ to the power of $3$ .

$- 8 + {(- 2)}^{2} - 8 \cdot - 2 - 12$

Raise $- 2$ to the power of $2$ .

$- 8 + 4 - 8 \cdot - 2 - 12$

Add $- 8$ and $4$ .

$- 4 - 8 \cdot - 2 - 12$

Multiply $- 8$ by $- 2$ .

$- 4 + 16 - 12$

Add $- 4$ and $16$ .

$12 - 12$

Subtract $12$ from $12$ .

$0$

Since $- 2$ is a known root, divide the polynomial by $x + 2$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{3} + x^{2} - 8 x - 12}{x + 2}$

Divide $x^{3} + x^{2} - 8 x - 12$ by $x + 2$ .

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$x$

$2$

$x^{3}$

$x^{2}$

$8 x$

$12$

Divide the highest order term in the dividend $x^{3}$ by the highest order term in divisor $x$ .

					$x^{2}$
	$x$	+	$2$		$x^{3}$	+	$x^{2}$	-	$8 x$	-	$12$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	+	$2$		$x^{3}$	+	$x^{2}$	-	$8 x$	-	$12$
			+	$x^{3}$	+	$2 x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $x^{3} + 2 x^{2}$

				$x^{2}$
$x$	+	$2$		$x^{3}$	+	$x^{2}$	-	$8 x$	-	$12$
			-	$x^{3}$	-	$2 x^{2}$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$x^{2}$
$x$	+	$2$		$x^{3}$	+	$x^{2}$	-	$8 x$	-	$12$
			-	$x^{3}$	-	$2 x^{2}$
					-	$x^{2}$

Pull the next terms from the original dividend down into the current dividend.

				$x^{2}$
$x$	+	$2$		$x^{3}$	+	$x^{2}$	-	$8 x$	-	$12$
			-	$x^{3}$	-	$2 x^{2}$
					-	$x^{2}$	-	$8 x$

Divide the highest order term in the dividend $- x^{2}$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$x$
$x$	+	$2$		$x^{3}$	+	$x^{2}$	-	$8 x$	-	$12$
			-	$x^{3}$	-	$2 x^{2}$
					-	$x^{2}$	-	$8 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$x$
$x$	+	$2$		$x^{3}$	+	$x^{2}$	-	$8 x$	-	$12$
			-	$x^{3}$	-	$2 x^{2}$
					-	$x^{2}$	-	$8 x$
					-	$x^{2}$	-	$2 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- x^{2} - 2 x$

				$x^{2}$	-	$x$
$x$	+	$2$		$x^{3}$	+	$x^{2}$	-	$8 x$	-	$12$
			-	$x^{3}$	-	$2 x^{2}$
					-	$x^{2}$	-	$8 x$
					+	$x^{2}$	+	$2 x$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$x^{2}$	-	$x$
$x$	+	$2$		$x^{3}$	+	$x^{2}$	-	$8 x$	-	$12$
			-	$x^{3}$	-	$2 x^{2}$
					-	$x^{2}$	-	$8 x$
					+	$x^{2}$	+	$2 x$
							-	$6 x$

Pull the next terms from the original dividend down into the current dividend.

				$x^{2}$	-	$x$
$x$	+	$2$		$x^{3}$	+	$x^{2}$	-	$8 x$	-	$12$
			-	$x^{3}$	-	$2 x^{2}$
					-	$x^{2}$	-	$8 x$
					+	$x^{2}$	+	$2 x$
							-	$6 x$	-	$12$

Divide the highest order term in the dividend $- 6 x$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$x$	-	$6$
$x$	+	$2$		$x^{3}$	+	$x^{2}$	-	$8 x$	-	$12$
			-	$x^{3}$	-	$2 x^{2}$
					-	$x^{2}$	-	$8 x$
					+	$x^{2}$	+	$2 x$
							-	$6 x$	-	$12$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$x$	-	$6$
$x$	+	$2$		$x^{3}$	+	$x^{2}$	-	$8 x$	-	$12$
			-	$x^{3}$	-	$2 x^{2}$
					-	$x^{2}$	-	$8 x$
					+	$x^{2}$	+	$2 x$
							-	$6 x$	-	$12$
							-	$6 x$	-	$12$

The expression needs to be subtracted from the dividend, so change all the signs in $- 6 x - 12$

				$x^{2}$	-	$x$	-	$6$
$x$	+	$2$		$x^{3}$	+	$x^{2}$	-	$8 x$	-	$12$
			-	$x^{3}$	-	$2 x^{2}$
					-	$x^{2}$	-	$8 x$
					+	$x^{2}$	+	$2 x$
							-	$6 x$	-	$12$
							+	$6 x$	+	$12$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$x^{2}$	-	$x$	-	$6$
$x$	+	$2$		$x^{3}$	+	$x^{2}$	-	$8 x$	-	$12$
			-	$x^{3}$	-	$2 x^{2}$
					-	$x^{2}$	-	$8 x$
					+	$x^{2}$	+	$2 x$
							-	$6 x$	-	$12$
							+	$6 x$	+	$12$
										$0$

Since the remander is $0$ , the final answer is the quotient.

$x^{2} - x - 6$

Write $x^{3} + x^{2} - 8 x - 12$ as a set of factors.

$(x + 3) ((x - 1) ((x + 2) (x^{2} - x - 6))) = 0$

Factor $x^{2} - x - 6$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 6$ and whose sum is $- 1$ .

$- 3, 2$

Write the factored form using these integers.

$(x + 3) ((x - 1) ((x + 2) ((x - 3) (x + 2)))) = 0$

Remove unnecessary parentheses.

$(x + 3) ((x - 1) ((x + 2) (x - 3) (x + 2))) = 0$

Combine exponents.

Raise $x + 2$ to the power of $1$ .

$(x + 3) ((x - 1) ((x + 2) (x + 2) (x - 3))) = 0$

Raise $x + 2$ to the power of $1$ .

$(x + 3) ((x - 1) ((x + 2) (x + 2) (x - 3))) = 0$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Add $1$ and $1$ .

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

Remove unnecessary parentheses.

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

Remove unnecessary parentheses.

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

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x + 3 = 0$

$x - 1 = 0$

${(x + 2)}^{2} = 0$

$x - 3 = 0$

Set $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Set ${(x + 2)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(x + 2)}^{2}$ equal to $0$ .

${(x + 2)}^{2} = 0$

Solve ${(x + 2)}^{2} = 0$ for $x$ .

Set the $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

The final solution is all the values that make $(x + 3) (x - 1) {(x + 2)}^{2} (x - 3) = 0$ true.

$x = - 3, 1, - 2, 3$

Step 3

Find the Roots (Zeros) x^5+3x^4-9x^3-31x^2+36

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