Precalculus Examples

$2 x^{4} - 13 x^{3} + 6 x^{2} + 64 x - 32$

Step 1

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.

$p = \pm 1, \pm 2, \pm 4, \pm 8, \pm 16, \pm 32$

$q = \pm 1, \pm 2$

Step 2

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

$\pm 1, \pm \frac{1}{2}, \pm 2, \pm 4, \pm 8, \pm 16, \pm 32$

Step 3

Substitute the possible roots one by one into the polynomial to find the actual roots. Simplify to check if the value is $0$ , which means it is a root.

$2 {(\frac{1}{2})}^{4} - 13 {(\frac{1}{2})}^{3} + 6 {(\frac{1}{2})}^{2} + 64 (\frac{1}{2}) - 32$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = \frac{1}{2}$ is a root of the polynomial.

Tap for more steps...

Step 4.1

Simplify each term.

Tap for more steps...

Step 4.1.1

Apply the product rule to $\frac{1}{2}$ .

$2 \frac{1^{4}}{2^{4}} - 13 {(\frac{1}{2})}^{3} + 6 {(\frac{1}{2})}^{2} + 64 (\frac{1}{2}) - 32$

Step 4.1.2

One to any power is one.

$2 \frac{1}{2^{4}} - 13 {(\frac{1}{2})}^{3} + 6 {(\frac{1}{2})}^{2} + 64 (\frac{1}{2}) - 32$

Step 4.1.3

Raise $2$ to the power of $4$ .

$2 (\frac{1}{16}) - 13 {(\frac{1}{2})}^{3} + 6 {(\frac{1}{2})}^{2} + 64 (\frac{1}{2}) - 32$

Step 4.1.4

Cancel the common factor of $2$ .

Tap for more steps...

Step 4.1.4.1

Factor $2$ out of $16$ .

$2 \frac{1}{2 (8)} - 13 {(\frac{1}{2})}^{3} + 6 {(\frac{1}{2})}^{2} + 64 (\frac{1}{2}) - 32$

Step 4.1.4.2

Cancel the common factor.

$2 \frac{1}{2 \cdot 8} - 13 {(\frac{1}{2})}^{3} + 6 {(\frac{1}{2})}^{2} + 64 (\frac{1}{2}) - 32$

Step 4.1.4.3

Rewrite the expression.

$\frac{1}{8} - 13 {(\frac{1}{2})}^{3} + 6 {(\frac{1}{2})}^{2} + 64 (\frac{1}{2}) - 32$

Step 4.1.5

Apply the product rule to $\frac{1}{2}$ .

$\frac{1}{8} - 13 \frac{1^{3}}{2^{3}} + 6 {(\frac{1}{2})}^{2} + 64 (\frac{1}{2}) - 32$

Step 4.1.6

One to any power is one.

$\frac{1}{8} - 13 \frac{1}{2^{3}} + 6 {(\frac{1}{2})}^{2} + 64 (\frac{1}{2}) - 32$

Step 4.1.7

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

$\frac{1}{8} - 13 (\frac{1}{8}) + 6 {(\frac{1}{2})}^{2} + 64 (\frac{1}{2}) - 32$

Step 4.1.8

Combine $- 13$ and $\frac{1}{8}$ .

$\frac{1}{8} + \frac{- 13}{8} + 6 {(\frac{1}{2})}^{2} + 64 (\frac{1}{2}) - 32$

Step 4.1.9

Move the negative in front of the fraction.

$\frac{1}{8} - \frac{13}{8} + 6 {(\frac{1}{2})}^{2} + 64 (\frac{1}{2}) - 32$

Step 4.1.10

Apply the product rule to $\frac{1}{2}$ .

$\frac{1}{8} - \frac{13}{8} + 6 \frac{1^{2}}{2^{2}} + 64 (\frac{1}{2}) - 32$

Step 4.1.11

One to any power is one.

$\frac{1}{8} - \frac{13}{8} + 6 \frac{1}{2^{2}} + 64 (\frac{1}{2}) - 32$

Step 4.1.12

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

$\frac{1}{8} - \frac{13}{8} + 6 (\frac{1}{4}) + 64 (\frac{1}{2}) - 32$

Step 4.1.13

Cancel the common factor of $2$ .

Tap for more steps...

Step 4.1.13.1

Factor $2$ out of $6$ .

$\frac{1}{8} - \frac{13}{8} + 2 (3) \frac{1}{4} + 64 (\frac{1}{2}) - 32$

Step 4.1.13.2

Factor $2$ out of $4$ .

$\frac{1}{8} - \frac{13}{8} + 2 \cdot 3 \frac{1}{2 \cdot 2} + 64 (\frac{1}{2}) - 32$

Step 4.1.13.3

Cancel the common factor.

$\frac{1}{8} - \frac{13}{8} + 2 \cdot 3 \frac{1}{2 \cdot 2} + 64 (\frac{1}{2}) - 32$

Step 4.1.13.4

Rewrite the expression.

$\frac{1}{8} - \frac{13}{8} + 3 (\frac{1}{2}) + 64 (\frac{1}{2}) - 32$

Step 4.1.14

Combine $3$ and $\frac{1}{2}$ .

$\frac{1}{8} - \frac{13}{8} + \frac{3}{2} + 64 (\frac{1}{2}) - 32$

Step 4.1.15

Cancel the common factor of $2$ .

Tap for more steps...

Step 4.1.15.1

Factor $2$ out of $64$ .

$\frac{1}{8} - \frac{13}{8} + \frac{3}{2} + 2 (32) \frac{1}{2} - 32$

Step 4.1.15.2

Cancel the common factor.

$\frac{1}{8} - \frac{13}{8} + \frac{3}{2} + 2 \cdot 32 \frac{1}{2} - 32$

Step 4.1.15.3

Rewrite the expression.

$\frac{1}{8} - \frac{13}{8} + \frac{3}{2} + 32 - 32$

Step 4.2

Combine fractions.

Tap for more steps...

Step 4.2.1

Combine the numerators over the common denominator.

$32 - 32 + \frac{1 - 13}{8} + \frac{3}{2}$

Step 4.2.2

Subtract $13$ from $1$ .

$32 - 32 + \frac{- 12}{8} + \frac{3}{2}$

Step 4.3

Find the common denominator.

Tap for more steps...

Step 4.3.1

Write $32$ as a fraction with denominator $1$ .

$\frac{32}{1} - 32 + \frac{- 12}{8} + \frac{3}{2}$

Step 4.3.2

Multiply $\frac{32}{1}$ by $\frac{8}{8}$ .

$\frac{32}{1} \cdot \frac{8}{8} - 32 + \frac{- 12}{8} + \frac{3}{2}$

Step 4.3.3

Multiply $\frac{32}{1}$ by $\frac{8}{8}$ .

$\frac{32 \cdot 8}{8} - 32 + \frac{- 12}{8} + \frac{3}{2}$

Step 4.3.4

Write $- 32$ as a fraction with denominator $1$ .

$\frac{32 \cdot 8}{8} + \frac{- 32}{1} + \frac{- 12}{8} + \frac{3}{2}$

Step 4.3.5

Multiply $\frac{- 32}{1}$ by $\frac{8}{8}$ .

$\frac{32 \cdot 8}{8} + \frac{- 32}{1} \cdot \frac{8}{8} + \frac{- 12}{8} + \frac{3}{2}$

Step 4.3.6

Multiply $\frac{- 32}{1}$ by $\frac{8}{8}$ .

$\frac{32 \cdot 8}{8} + \frac{- 32 \cdot 8}{8} + \frac{- 12}{8} + \frac{3}{2}$

Step 4.3.7

Multiply $\frac{3}{2}$ by $\frac{4}{4}$ .

$\frac{32 \cdot 8}{8} + \frac{- 32 \cdot 8}{8} + \frac{- 12}{8} + \frac{3}{2} \cdot \frac{4}{4}$

Step 4.3.8

Multiply $\frac{3}{2}$ by $\frac{4}{4}$ .

$\frac{32 \cdot 8}{8} + \frac{- 32 \cdot 8}{8} + \frac{- 12}{8} + \frac{3 \cdot 4}{2 \cdot 4}$

Step 4.3.9

Multiply $2$ by $4$ .

$\frac{32 \cdot 8}{8} + \frac{- 32 \cdot 8}{8} + \frac{- 12}{8} + \frac{3 \cdot 4}{8}$

Step 4.4

Combine the numerators over the common denominator.

$\frac{32 \cdot 8 - 32 \cdot 8 - 12 + 3 \cdot 4}{8}$

Step 4.5

Simplify each term.

Tap for more steps...

Step 4.5.1

Multiply $32$ by $8$ .

$\frac{256 - 32 \cdot 8 - 12 + 3 \cdot 4}{8}$

Step 4.5.2

Multiply $- 32$ by $8$ .

$\frac{256 - 256 - 12 + 3 \cdot 4}{8}$

Step 4.5.3

Multiply $3$ by $4$ .

$\frac{256 - 256 - 12 + 12}{8}$

Step 4.6

Simplify the expression.

Tap for more steps...

Step 4.6.1

Subtract $256$ from $256$ .

$\frac{0 - 12 + 12}{8}$

Step 4.6.2

Subtract $12$ from $0$ .

$\frac{- 12 + 12}{8}$

Step 4.6.3

Add $- 12$ and $12$ .

$\frac{0}{8}$

Step 4.6.4

Divide $0$ by $8$ .

$0$

Step 5

Since $\frac{1}{2}$ is a known root, divide the polynomial by $x - \frac{1}{2}$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{2 x^{4} - 13 x^{3} + 6 x^{2} + 64 x - 32}{x - \frac{1}{2}}$

Step 6

Next, find the roots of the remaining polynomial. The order of the polynomial has been reduced by $1$ .

Tap for more steps...

Step 6.1

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

$\frac{1}{2}$	$2$	$- 13$	$6$	$64$	$- 32$

Step 6.2

The first number in the dividend $(2)$ is put into the first position of the result area (below the horizontal line).

$\frac{1}{2}$	$2$	$- 13$	$6$	$64$	$- 32$

	$2$

Step 6.3

Multiply the newest entry in the result $(2)$ by the divisor $(\frac{1}{2})$ and place the result of $(1)$ under the next term in the dividend $(- 13)$ .

$\frac{1}{2}$	$2$	$- 13$	$6$	$64$	$- 32$
		$1$
	$2$

Step 6.4

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

$\frac{1}{2}$	$2$	$- 13$	$6$	$64$	$- 32$
		$1$
	$2$	$- 12$

Step 6.5

Multiply the newest entry in the result $(- 12)$ by the divisor $(\frac{1}{2})$ and place the result of $(- 6)$ under the next term in the dividend $(6)$ .

$\frac{1}{2}$	$2$	$- 13$	$6$	$64$	$- 32$
		$1$	$- 6$
	$2$	$- 12$

Step 6.6

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

$\frac{1}{2}$	$2$	$- 13$	$6$	$64$	$- 32$
		$1$	$- 6$
	$2$	$- 12$	$0$

Step 6.7

Multiply the newest entry in the result $(0)$ by the divisor $(\frac{1}{2})$ and place the result of $(0)$ under the next term in the dividend $(64)$ .

$\frac{1}{2}$	$2$	$- 13$	$6$	$64$	$- 32$
		$1$	$- 6$	$0$
	$2$	$- 12$	$0$

Step 6.8

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

$\frac{1}{2}$	$2$	$- 13$	$6$	$64$	$- 32$
		$1$	$- 6$	$0$
	$2$	$- 12$	$0$	$64$

Step 6.9

Multiply the newest entry in the result $(64)$ by the divisor $(\frac{1}{2})$ and place the result of $(32)$ under the next term in the dividend $(- 32)$ .

$\frac{1}{2}$	$2$	$- 13$	$6$	$64$	$- 32$
		$1$	$- 6$	$0$	$32$
	$2$	$- 12$	$0$	$64$

Step 6.10

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

$\frac{1}{2}$	$2$	$- 13$	$6$	$64$	$- 32$
		$1$	$- 6$	$0$	$32$
	$2$	$- 12$	$0$	$64$	$0$

Step 6.11

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

$2 x^{3} + - 12 x^{2} + (0) x + 64$

Step 6.12

Simplify the quotient polynomial.

$2 x^{3} - 12 x^{2} + 64$

Step 7

Factor $2$ out of $2 x^{3} - 12 x^{2} + 64$ .

Tap for more steps...

Step 7.1

Factor $2$ out of $2 x^{3}$ .

$(x - \frac{1}{2}) (2 (x^{3}) - 12 x^{2} + 64)$

Step 7.2

Factor $2$ out of $- 12 x^{2}$ .

$(x - \frac{1}{2}) (2 (x^{3}) + 2 (- 6 x^{2}) + 64)$

Step 7.3

Factor $2$ out of $64$ .

$(x - \frac{1}{2}) (2 x^{3} + 2 (- 6 x^{2}) + 2 \cdot 32)$

Step 7.4

Factor $2$ out of $2 x^{3} + 2 (- 6 x^{2})$ .

$(x - \frac{1}{2}) (2 (x^{3} - 6 x^{2}) + 2 \cdot 32)$

Step 7.5

Factor $2$ out of $2 (x^{3} - 6 x^{2}) + 2 \cdot 32$ .

$(x - \frac{1}{2}) (2 (x^{3} - 6 x^{2} + 32))$

Step 8

Factor the left side of the equation.

Tap for more steps...

Step 8.1

Regroup terms.

$64 x - 32 + 2 x^{4} - 13 x^{3} + 6 x^{2} = 0$

Step 8.2

Factor $32$ out of $64 x - 32$ .

Tap for more steps...

Step 8.2.1

Factor $32$ out of $64 x$ .

$32 (2 x) - 32 + 2 x^{4} - 13 x^{3} + 6 x^{2} = 0$

Step 8.2.2

Factor $32$ out of $- 32$ .

$32 (2 x) + 32 (- 1) + 2 x^{4} - 13 x^{3} + 6 x^{2} = 0$

Step 8.2.3

Factor $32$ out of $32 (2 x) + 32 (- 1)$ .

$32 (2 x - 1) + 2 x^{4} - 13 x^{3} + 6 x^{2} = 0$

Step 8.3

Factor $x^{2}$ out of $2 x^{4} - 13 x^{3} + 6 x^{2}$ .

Tap for more steps...

Step 8.3.1

Factor $x^{2}$ out of $2 x^{4}$ .

$32 (2 x - 1) + x^{2} (2 x^{2}) - 13 x^{3} + 6 x^{2} = 0$

Step 8.3.2

Factor $x^{2}$ out of $- 13 x^{3}$ .

$32 (2 x - 1) + x^{2} (2 x^{2}) + x^{2} (- 13 x) + 6 x^{2} = 0$

Step 8.3.3

Factor $x^{2}$ out of $6 x^{2}$ .

$32 (2 x - 1) + x^{2} (2 x^{2}) + x^{2} (- 13 x) + x^{2} \cdot 6 = 0$

Step 8.3.4

Factor $x^{2}$ out of $x^{2} (2 x^{2}) + x^{2} (- 13 x)$ .

$32 (2 x - 1) + x^{2} (2 x^{2} - 13 x) + x^{2} \cdot 6 = 0$

Step 8.3.5

Factor $x^{2}$ out of $x^{2} (2 x^{2} - 13 x) + x^{2} \cdot 6$ .

$32 (2 x - 1) + x^{2} (2 x^{2} - 13 x + 6) = 0$

Step 8.4

Factor.

Tap for more steps...

Step 8.4.1

Factor by grouping.

Tap for more steps...

Step 8.4.1.1

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 $a \cdot c = 2 \cdot 6 = 12$ and whose sum is $b = - 13$ .

Tap for more steps...

Step 8.4.1.1.1

Factor $- 13$ out of $- 13 x$ .

$32 (2 x - 1) + x^{2} (2 x^{2} - 13 x + 6) = 0$

Step 8.4.1.1.2

Rewrite $- 13$ as $- 1$ plus $- 12$

$32 (2 x - 1) + x^{2} (2 x^{2} + (- 1 - 12) x + 6) = 0$

Step 8.4.1.1.3

Apply the distributive property.

$32 (2 x - 1) + x^{2} (2 x^{2} - 1 x - 12 x + 6) = 0$

Step 8.4.1.2

Factor out the greatest common factor from each group.

Tap for more steps...

Step 8.4.1.2.1

Group the first two terms and the last two terms.

$32 (2 x - 1) + x^{2} ((2 x^{2} - 1 x) - 12 x + 6) = 0$

Step 8.4.1.2.2

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

$32 (2 x - 1) + x^{2} (x (2 x - 1) - 6 (2 x - 1)) = 0$

Step 8.4.1.3

Factor the polynomial by factoring out the greatest common factor, $2 x - 1$ .

$32 (2 x - 1) + x^{2} ((2 x - 1) (x - 6)) = 0$

Step 8.4.2

Remove unnecessary parentheses.

$32 (2 x - 1) + x^{2} (2 x - 1) (x - 6) = 0$

Step 8.5

Factor $2 x - 1$ out of $32 (2 x - 1) + x^{2} (2 x - 1) (x - 6)$ .

Tap for more steps...

Step 8.5.1

Factor $2 x - 1$ out of $32 (2 x - 1)$ .

$(2 x - 1) \cdot 32 + x^{2} (2 x - 1) (x - 6) = 0$

Step 8.5.2

Factor $2 x - 1$ out of $x^{2} (2 x - 1) (x - 6)$ .

$(2 x - 1) \cdot 32 + (2 x - 1) (x^{2} (x - 6)) = 0$

Step 8.5.3

Factor $2 x - 1$ out of $(2 x - 1) \cdot 32 + (2 x - 1) (x^{2} (x - 6))$ .

$(2 x - 1) (32 + x^{2} (x - 6)) = 0$

Step 8.6

Apply the distributive property.

$(2 x - 1) (32 + x^{2} x + x^{2} \cdot - 6) = 0$

Step 8.7

Multiply $x^{2}$ by $x$ by adding the exponents.

Tap for more steps...

Step 8.7.1

Multiply $x^{2}$ by $x$ .

Tap for more steps...

Step 8.7.1.1

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

$(2 x - 1) (32 + x^{2} x + x^{2} \cdot - 6) = 0$

Step 8.7.1.2

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

$(2 x - 1) (32 + x^{2 + 1} + x^{2} \cdot - 6) = 0$

Step 8.7.2

Add $2$ and $1$ .

$(2 x - 1) (32 + x^{3} + x^{2} \cdot - 6) = 0$

Step 8.8

Move $- 6$ to the left of $x^{2}$ .

$(2 x - 1) (32 + x^{3} - 6 x^{2}) = 0$

Step 8.9

Reorder terms.

$(2 x - 1) (x^{3} - 6 x^{2} + 32) = 0$

Step 8.10

Factor.

Tap for more steps...

Step 8.10.1

Rewrite $x^{3} - 6 x^{2} + 32$ in a factored form.

Tap for more steps...

Step 8.10.1.1

Factor $x^{3} - 6 x^{2} + 32$ using the rational roots test.

Tap for more steps...

Step 8.10.1.1.1

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.

$p = \pm 1, \pm 32, \pm 2, \pm 16, \pm 4, \pm 8$

$q = \pm 1$

Step 8.10.1.1.2

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

$\pm 1, \pm 32, \pm 2, \pm 16, \pm 4, \pm 8$

Step 8.10.1.1.3

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

Tap for more steps...

Step 8.10.1.1.3.1

Substitute $- 2$ into the polynomial.

${(- 2)}^{3} - 6 {(- 2)}^{2} + 32$

Step 8.10.1.1.3.2

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

$- 8 - 6 {(- 2)}^{2} + 32$

Step 8.10.1.1.3.3

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

$- 8 - 6 \cdot 4 + 32$

Step 8.10.1.1.3.4

Multiply $- 6$ by $4$ .

$- 8 - 24 + 32$

Step 8.10.1.1.3.5

Subtract $24$ from $- 8$ .

$- 32 + 32$

Step 8.10.1.1.3.6

Add $- 32$ and $32$ .

$0$

Step 8.10.1.1.4

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.

$\frac{x^{3} - 6 x^{2} + 32}{x + 2}$

Step 8.10.1.1.5

Divide $x^{3} - 6 x^{2} + 32$ by $x + 2$ .

Tap for more steps...

Step 8.10.1.1.5.1

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

$x$

$2$

$x^{3}$

$6 x^{2}$

$0 x$

$32$

Step 8.10.1.1.5.2

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

					$x^{2}$
	$x$	+	$2$		$x^{3}$	-	$6 x^{2}$	+	$0 x$	+	$32$

Step 8.10.1.1.5.3

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	+	$2$		$x^{3}$	-	$6 x^{2}$	+	$0 x$	+	$32$
			+	$x^{3}$	+	$2 x^{2}$

Step 8.10.1.1.5.4

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

				$x^{2}$
$x$	+	$2$		$x^{3}$	-	$6 x^{2}$	+	$0 x$	+	$32$
			-	$x^{3}$	-	$2 x^{2}$

Step 8.10.1.1.5.5

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

				$x^{2}$
$x$	+	$2$		$x^{3}$	-	$6 x^{2}$	+	$0 x$	+	$32$
			-	$x^{3}$	-	$2 x^{2}$
					-	$8 x^{2}$

Step 8.10.1.1.5.6

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

				$x^{2}$
$x$	+	$2$		$x^{3}$	-	$6 x^{2}$	+	$0 x$	+	$32$
			-	$x^{3}$	-	$2 x^{2}$
					-	$8 x^{2}$	+	$0 x$

Step 8.10.1.1.5.7

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

				$x^{2}$	-	$8 x$
$x$	+	$2$		$x^{3}$	-	$6 x^{2}$	+	$0 x$	+	$32$
			-	$x^{3}$	-	$2 x^{2}$
					-	$8 x^{2}$	+	$0 x$

Step 8.10.1.1.5.8

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$8 x$
$x$	+	$2$		$x^{3}$	-	$6 x^{2}$	+	$0 x$	+	$32$
			-	$x^{3}$	-	$2 x^{2}$
					-	$8 x^{2}$	+	$0 x$
					-	$8 x^{2}$	-	$16 x$

Step 8.10.1.1.5.9

The expression needs to be subtracted from the dividend, so change all the signs in $- 8 x^{2} - 16 x$

				$x^{2}$	-	$8 x$
$x$	+	$2$		$x^{3}$	-	$6 x^{2}$	+	$0 x$	+	$32$
			-	$x^{3}$	-	$2 x^{2}$
					-	$8 x^{2}$	+	$0 x$
					+	$8 x^{2}$	+	$16 x$

Step 8.10.1.1.5.10

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

				$x^{2}$	-	$8 x$
$x$	+	$2$		$x^{3}$	-	$6 x^{2}$	+	$0 x$	+	$32$
			-	$x^{3}$	-	$2 x^{2}$
					-	$8 x^{2}$	+	$0 x$
					+	$8 x^{2}$	+	$16 x$
							+	$16 x$

Step 8.10.1.1.5.11

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

				$x^{2}$	-	$8 x$
$x$	+	$2$		$x^{3}$	-	$6 x^{2}$	+	$0 x$	+	$32$
			-	$x^{3}$	-	$2 x^{2}$
					-	$8 x^{2}$	+	$0 x$
					+	$8 x^{2}$	+	$16 x$
							+	$16 x$	+	$32$

Step 8.10.1.1.5.12

Divide the highest order term in the dividend $16 x$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$8 x$	+	$16$
$x$	+	$2$		$x^{3}$	-	$6 x^{2}$	+	$0 x$	+	$32$
			-	$x^{3}$	-	$2 x^{2}$
					-	$8 x^{2}$	+	$0 x$
					+	$8 x^{2}$	+	$16 x$
							+	$16 x$	+	$32$

Step 8.10.1.1.5.13

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$8 x$	+	$16$
$x$	+	$2$		$x^{3}$	-	$6 x^{2}$	+	$0 x$	+	$32$
			-	$x^{3}$	-	$2 x^{2}$
					-	$8 x^{2}$	+	$0 x$
					+	$8 x^{2}$	+	$16 x$
							+	$16 x$	+	$32$
							+	$16 x$	+	$32$

Step 8.10.1.1.5.14

The expression needs to be subtracted from the dividend, so change all the signs in $16 x + 32$

				$x^{2}$	-	$8 x$	+	$16$
$x$	+	$2$		$x^{3}$	-	$6 x^{2}$	+	$0 x$	+	$32$
			-	$x^{3}$	-	$2 x^{2}$
					-	$8 x^{2}$	+	$0 x$
					+	$8 x^{2}$	+	$16 x$
							+	$16 x$	+	$32$
							-	$16 x$	-	$32$

Step 8.10.1.1.5.15

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

				$x^{2}$	-	$8 x$	+	$16$
$x$	+	$2$		$x^{3}$	-	$6 x^{2}$	+	$0 x$	+	$32$
			-	$x^{3}$	-	$2 x^{2}$
					-	$8 x^{2}$	+	$0 x$
					+	$8 x^{2}$	+	$16 x$
							+	$16 x$	+	$32$
							-	$16 x$	-	$32$
										$0$

Step 8.10.1.1.5.16

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

$x^{2} - 8 x + 16$

Step 8.10.1.1.6

Write $x^{3} - 6 x^{2} + 32$ as a set of factors.

$(2 x - 1) ((x + 2) (x^{2} - 8 x + 16)) = 0$

Step 8.10.1.2

Factor using the perfect square rule.

Tap for more steps...

Step 8.10.1.2.1

Rewrite $16$ as $4^{2}$ .

$(2 x - 1) ((x + 2) (x^{2} - 8 x + 4^{2})) = 0$

Step 8.10.1.2.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$8 x = 2 \cdot x \cdot 4$

Step 8.10.1.2.3

Rewrite the polynomial.

$(2 x - 1) ((x + 2) (x^{2} - 2 \cdot x \cdot 4 + 4^{2})) = 0$

Step 8.10.1.2.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 4$ .

$(2 x - 1) ((x + 2) {(x - 4)}^{2}) = 0$

Step 8.10.2

Remove unnecessary parentheses.

$(2 x - 1) (x + 2) {(x - 4)}^{2} = 0$

Step 9

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

$2 x - 1 = 0$

$x + 2 = 0$

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

Step 10

Set $2 x - 1$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 10.1

Set $2 x - 1$ equal to $0$ .

$2 x - 1 = 0$

Step 10.2

Solve $2 x - 1 = 0$ for $x$ .

Tap for more steps...

Step 10.2.1

Add $1$ to both sides of the equation.

$2 x = 1$

Step 10.2.2

Divide each term in $2 x = 1$ by $2$ and simplify.

Tap for more steps...

Step 10.2.2.1

Divide each term in $2 x = 1$ by $2$ .

$\frac{2 x}{2} = \frac{1}{2}$

Step 10.2.2.2

Simplify the left side.

Tap for more steps...

Step 10.2.2.2.1

Cancel the common factor of $2$ .

Tap for more steps...

Step 10.2.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{1}{2}$

Step 10.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{1}{2}$

Step 11

Set $x + 2$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 11.1

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

$x + 2 = 0$

Step 11.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 12

Set ${(x - 4)}^{2}$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 12.1

Set ${(x - 4)}^{2}$ equal to $0$ .

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

Step 12.2

Solve ${(x - 4)}^{2} = 0$ for $x$ .

Tap for more steps...

Step 12.2.1

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

$x - 4 = 0$

Step 12.2.2

Add $4$ to both sides of the equation.

$x = 4$

Step 13

The final solution is all the values that make $(2 x - 1) (x + 2) {(x - 4)}^{2} = 0$ true.

$x = \frac{1}{2}, - 2, 4$

Step 14