Identify the Zeros and Their Multiplicities (x^3-15x^2+68x-96)/(x-3)

$\frac{(x^{3} - 15 x^{2} + 68 x - 96)}{(x - 3)}$

Step 1

Set $\frac{x^{3} - 15 x^{2} + 68 x - 96}{x - 3}$ equal to $0$ .

$\frac{x^{3} - 15 x^{2} + 68 x - 96}{x - 3} = 0$

Step 2

Solve for $x$ .

Set the numerator equal to zero.

$x^{3} - 15 x^{2} + 68 x - 96 = 0$

Solve the equation for $x$ .

Factor the left side of the equation.

Factor $x^{3} - 15 x^{2} + 68 x - 96$ 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.

$p = \pm 1, \pm 96, \pm 2, \pm 48, \pm 3, \pm 32, \pm 4, \pm 24, \pm 6, \pm 16, \pm 8, \pm 12 q = \pm 1$

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

$\pm 1, \pm 96, \pm 2, \pm 48, \pm 3, \pm 32, \pm 4, \pm 24, \pm 6, \pm 16, \pm 8, \pm 12$

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.

$3^{3} - 15 \cdot 3^{2} + 68 \cdot 3 - 96$

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

$27 - 15 \cdot 3^{2} + 68 \cdot 3 - 96$

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

$27 - 15 \cdot 9 + 68 \cdot 3 - 96$

Multiply $- 15$ by $9$ .

$27 - 135 + 68 \cdot 3 - 96$

Subtract $135$ from $27$ .

$- 108 + 68 \cdot 3 - 96$

Multiply $68$ by $3$ .

$- 108 + 204 - 96$

Add $- 108$ and $204$ .

$96 - 96$

Subtract $96$ from $96$ .

$0$

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.

$\frac{x^{3} - 15 x^{2} + 68 x - 96}{x - 3}$

Divide $x^{3} - 15 x^{2} + 68 x - 96$ 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 $0$ .

$x$

$3$

$x^{3}$

$15 x^{2}$

$68 x$

$96$

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

					$x^{2}$
	$x$	-	$3$		$x^{3}$	-	$15 x^{2}$	+	$68 x$	-	$96$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	-	$3$		$x^{3}$	-	$15 x^{2}$	+	$68 x$	-	$96$
			+	$x^{3}$	-	$3 x^{2}$

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

				$x^{2}$
$x$	-	$3$		$x^{3}$	-	$15 x^{2}$	+	$68 x$	-	$96$
			-	$x^{3}$	+	$3 x^{2}$

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

				$x^{2}$
$x$	-	$3$		$x^{3}$	-	$15 x^{2}$	+	$68 x$	-	$96$
			-	$x^{3}$	+	$3 x^{2}$
					-	$12 x^{2}$

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

				$x^{2}$
$x$	-	$3$		$x^{3}$	-	$15 x^{2}$	+	$68 x$	-	$96$
			-	$x^{3}$	+	$3 x^{2}$
					-	$12 x^{2}$	+	$68 x$

Divide the highest order term in the dividend $- 12 x^{2}$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$12 x$
$x$	-	$3$		$x^{3}$	-	$15 x^{2}$	+	$68 x$	-	$96$
			-	$x^{3}$	+	$3 x^{2}$
					-	$12 x^{2}$	+	$68 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$12 x$
$x$	-	$3$		$x^{3}$	-	$15 x^{2}$	+	$68 x$	-	$96$
			-	$x^{3}$	+	$3 x^{2}$
					-	$12 x^{2}$	+	$68 x$
					-	$12 x^{2}$	+	$36 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 12 x^{2} + 36 x$

				$x^{2}$	-	$12 x$
$x$	-	$3$		$x^{3}$	-	$15 x^{2}$	+	$68 x$	-	$96$
			-	$x^{3}$	+	$3 x^{2}$
					-	$12 x^{2}$	+	$68 x$
					+	$12 x^{2}$	-	$36 x$

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

				$x^{2}$	-	$12 x$
$x$	-	$3$		$x^{3}$	-	$15 x^{2}$	+	$68 x$	-	$96$
			-	$x^{3}$	+	$3 x^{2}$
					-	$12 x^{2}$	+	$68 x$
					+	$12 x^{2}$	-	$36 x$
							+	$32 x$

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

				$x^{2}$	-	$12 x$
$x$	-	$3$		$x^{3}$	-	$15 x^{2}$	+	$68 x$	-	$96$
			-	$x^{3}$	+	$3 x^{2}$
					-	$12 x^{2}$	+	$68 x$
					+	$12 x^{2}$	-	$36 x$
							+	$32 x$	-	$96$

Divide the highest order term in the dividend $32 x$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$12 x$	+	$32$
$x$	-	$3$		$x^{3}$	-	$15 x^{2}$	+	$68 x$	-	$96$
			-	$x^{3}$	+	$3 x^{2}$
					-	$12 x^{2}$	+	$68 x$
					+	$12 x^{2}$	-	$36 x$
							+	$32 x$	-	$96$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$12 x$	+	$32$
$x$	-	$3$		$x^{3}$	-	$15 x^{2}$	+	$68 x$	-	$96$
			-	$x^{3}$	+	$3 x^{2}$
					-	$12 x^{2}$	+	$68 x$
					+	$12 x^{2}$	-	$36 x$
							+	$32 x$	-	$96$
							+	$32 x$	-	$96$

The expression needs to be subtracted from the dividend, so change all the signs in $32 x - 96$

				$x^{2}$	-	$12 x$	+	$32$
$x$	-	$3$		$x^{3}$	-	$15 x^{2}$	+	$68 x$	-	$96$
			-	$x^{3}$	+	$3 x^{2}$
					-	$12 x^{2}$	+	$68 x$
					+	$12 x^{2}$	-	$36 x$
							+	$32 x$	-	$96$
							-	$32 x$	+	$96$

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

				$x^{2}$	-	$12 x$	+	$32$
$x$	-	$3$		$x^{3}$	-	$15 x^{2}$	+	$68 x$	-	$96$
			-	$x^{3}$	+	$3 x^{2}$
					-	$12 x^{2}$	+	$68 x$
					+	$12 x^{2}$	-	$36 x$
							+	$32 x$	-	$96$
							-	$32 x$	+	$96$
										$0$

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

$x^{2} - 12 x + 32$

Write $x^{3} - 15 x^{2} + 68 x - 96$ as a set of factors.

$(x - 3) (x^{2} - 12 x + 32) = 0$

Factor $x^{2} - 12 x + 32$ 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 $32$ and whose sum is $- 12$ .

$- 8, - 4$

Write the factored form using these integers.

$(x - 3) ((x - 8) (x - 4)) = 0$

Remove unnecessary parentheses.

$(x - 3) (x - 8) (x - 4) = 0$

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

$x - 3 = 0$

$x - 8 = 0$

$x - 4 = 0$

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.

$x = 3$

Set $x - 8$ equal to $0$ and solve for $x$ .

Set $x - 8$ equal to $0$ .

$x - 8 = 0$

Add $8$ to both sides of the equation.

$x = 8$

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 $(x - 3) (x - 8) (x - 4) = 0$ true. The multiplicity of a root is the number of times the root appears.

$x = 3$ (Multiplicity of $1$ )

$x = 8$ (Multiplicity of $1$ )

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

$x = 3$ (Multiplicity of $1$ )

$x = 8$ (Multiplicity of $1$ )

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

Exclude the solutions that do not make $\frac{x^{3} - 15 x^{2} + 68 x - 96}{x - 3} = 0$ true.

$x = 8, 4$

Step 3

Identify the Zeros and Their Multiplicities (x^3-15x^2+68x-96)/(x-3)

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