Identify the Zeros and Their Multiplicities -x^3-2x^2+15x+36

$- x^{3} - 2 x^{2} + 15 x + 36$

Step 1

Set $- x^{3} - 2 x^{2} + 15 x + 36$ equal to $0$ .

$- x^{3} - 2 x^{2} + 15 x + 36 = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Factor $- x^{3} - 2 x^{2} + 15 x + 36$ 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.

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

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

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

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.

$- {(- 3)}^{3} - 2 {(- 3)}^{2} + 15 \cdot - 3 + 36$

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

$- - 27 - 2 {(- 3)}^{2} + 15 \cdot - 3 + 36$

Multiply $- 1$ by $- 27$ .

$27 - 2 {(- 3)}^{2} + 15 \cdot - 3 + 36$

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

$27 - 2 \cdot 9 + 15 \cdot - 3 + 36$

Multiply $- 2$ by $9$ .

$27 - 18 + 15 \cdot - 3 + 36$

Subtract $18$ from $27$ .

$9 + 15 \cdot - 3 + 36$

Multiply $15$ by $- 3$ .

$9 - 45 + 36$

Subtract $45$ from $9$ .

$- 36 + 36$

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.

$\frac{- x^{3} - 2 x^{2} + 15 x + 36}{x + 3}$

Divide $- x^{3} - 2 x^{2} + 15 x + 36$ 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 $0$ .

$x$

$3$

$x^{3}$

$2 x^{2}$

$15 x$

$36$

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

				-	$x^{2}$
	$x$	+	$3$	-	$x^{3}$	-	$2 x^{2}$	+	$15 x$	+	$36$

Multiply the new quotient term by the divisor.

			-	$x^{2}$
$x$	+	$3$	-	$x^{3}$	-	$2 x^{2}$	+	$15 x$	+	$36$
			-	$x^{3}$	-	$3 x^{2}$

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

			-	$x^{2}$
$x$	+	$3$	-	$x^{3}$	-	$2 x^{2}$	+	$15 x$	+	$36$
			+	$x^{3}$	+	$3 x^{2}$

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

			-	$x^{2}$
$x$	+	$3$	-	$x^{3}$	-	$2 x^{2}$	+	$15 x$	+	$36$
			+	$x^{3}$	+	$3 x^{2}$
					+	$x^{2}$

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

			-	$x^{2}$
$x$	+	$3$	-	$x^{3}$	-	$2 x^{2}$	+	$15 x$	+	$36$
			+	$x^{3}$	+	$3 x^{2}$
					+	$x^{2}$	+	$15 x$

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

			-	$x^{2}$	+	$x$
$x$	+	$3$	-	$x^{3}$	-	$2 x^{2}$	+	$15 x$	+	$36$
			+	$x^{3}$	+	$3 x^{2}$
					+	$x^{2}$	+	$15 x$

Multiply the new quotient term by the divisor.

			-	$x^{2}$	+	$x$
$x$	+	$3$	-	$x^{3}$	-	$2 x^{2}$	+	$15 x$	+	$36$
			+	$x^{3}$	+	$3 x^{2}$
					+	$x^{2}$	+	$15 x$
					+	$x^{2}$	+	$3 x$

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

			-	$x^{2}$	+	$x$
$x$	+	$3$	-	$x^{3}$	-	$2 x^{2}$	+	$15 x$	+	$36$
			+	$x^{3}$	+	$3 x^{2}$
					+	$x^{2}$	+	$15 x$
					-	$x^{2}$	-	$3 x$

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

			-	$x^{2}$	+	$x$
$x$	+	$3$	-	$x^{3}$	-	$2 x^{2}$	+	$15 x$	+	$36$
			+	$x^{3}$	+	$3 x^{2}$
					+	$x^{2}$	+	$15 x$
					-	$x^{2}$	-	$3 x$
							+	$12 x$

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

			-	$x^{2}$	+	$x$
$x$	+	$3$	-	$x^{3}$	-	$2 x^{2}$	+	$15 x$	+	$36$
			+	$x^{3}$	+	$3 x^{2}$
					+	$x^{2}$	+	$15 x$
					-	$x^{2}$	-	$3 x$
							+	$12 x$	+	$36$

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

			-	$x^{2}$	+	$x$	+	$12$
$x$	+	$3$	-	$x^{3}$	-	$2 x^{2}$	+	$15 x$	+	$36$
			+	$x^{3}$	+	$3 x^{2}$
					+	$x^{2}$	+	$15 x$
					-	$x^{2}$	-	$3 x$
							+	$12 x$	+	$36$

Multiply the new quotient term by the divisor.

			-	$x^{2}$	+	$x$	+	$12$
$x$	+	$3$	-	$x^{3}$	-	$2 x^{2}$	+	$15 x$	+	$36$
			+	$x^{3}$	+	$3 x^{2}$
					+	$x^{2}$	+	$15 x$
					-	$x^{2}$	-	$3 x$
							+	$12 x$	+	$36$
							+	$12 x$	+	$36$

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

			-	$x^{2}$	+	$x$	+	$12$
$x$	+	$3$	-	$x^{3}$	-	$2 x^{2}$	+	$15 x$	+	$36$
			+	$x^{3}$	+	$3 x^{2}$
					+	$x^{2}$	+	$15 x$
					-	$x^{2}$	-	$3 x$
							+	$12 x$	+	$36$
							-	$12 x$	-	$36$

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

			-	$x^{2}$	+	$x$	+	$12$
$x$	+	$3$	-	$x^{3}$	-	$2 x^{2}$	+	$15 x$	+	$36$
			+	$x^{3}$	+	$3 x^{2}$
					+	$x^{2}$	+	$15 x$
					-	$x^{2}$	-	$3 x$
							+	$12 x$	+	$36$
							-	$12 x$	-	$36$
										$0$

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

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

Write $- x^{3} - 2 x^{2} + 15 x + 36$ as a set of factors.

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

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 1 \cdot 12 = - 12$ and whose sum is $b = 1$ .

Multiply by $1$ .

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

Rewrite $1$ as $- 3$ plus $4$

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

Factor out the greatest common factor (GCF) from each group.

$(x + 3) (x (- x - 3) - 4 (- x - 3)) = 0$

Factor the polynomial by factoring out the greatest common factor, $- x - 3$ .

$(x + 3) ((- x - 3) (x - 4)) = 0$

Remove unnecessary parentheses.

$(x + 3) (- x - 3) (x - 4) = 0$

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

$x + 3 = 0$

$- x - 3 = 0$

$x - 4 = 0$

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 $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

The final solution is all the values that make $(x + 3) (- x - 3) (x - 4) = 0$ true. The multiplicity of a root is the number of times the root appears.

$x = - 3$ (Multiplicity of $2$ )

$x = 4$ (Multiplicity of $1$ )

$x = - 3$ (Multiplicity of $2$ )

$x = 4$ (Multiplicity of $1$ )

Step 3

Identify the Zeros and Their Multiplicities -x^3-2x^2+15x+36

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