Solve by Factoring x^3+8x^2+11x-20=(x+5)

$x^{3} + 8 x^{2} + 11 x - 20 = (x + 5)$

Step 1

Subtract $(x + 5)$ from both sides of the equation.

$x^{3} + 8 x^{2} + 11 x - 20 - (x + 5) = 0$

Step 2

Simplify $x^{3} + 8 x^{2} + 11 x - 20 - (x + 5)$ .

Simplify each term.

Apply the distributive property.

$x^{3} + 8 x^{2} + 11 x - 20 - x - 1 \cdot 5 = 0$

Multiply $- 1$ by $5$ .

$x^{3} + 8 x^{2} + 11 x - 20 - x - 5 = 0$

Subtract $x$ from $11 x$ .

$x^{3} + 8 x^{2} + 10 x - 20 - 5 = 0$

Subtract $5$ from $- 20$ .

$x^{3} + 8 x^{2} + 10 x - 25 = 0$

Step 3

Factor $x^{3} + 8 x^{2} + 10 x - 25$ 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.

$p = \pm 1, \pm 25, \pm 5 q = \pm 1$

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

$\pm 1, \pm 25, \pm 5$

Substitute $- 5$ and simplify the expression. In this case, the expression is equal to $0$ so $- 5$ is a root of the polynomial.

Substitute $- 5$ into the polynomial.

${(- 5)}^{3} + 8 {(- 5)}^{2} + 10 \cdot - 5 - 25$

Raise $- 5$ to the power of $3$ .

$- 125 + 8 {(- 5)}^{2} + 10 \cdot - 5 - 25$

Raise $- 5$ to the power of $2$ .

$- 125 + 8 \cdot 25 + 10 \cdot - 5 - 25$

Multiply $8$ by $25$ .

$- 125 + 200 + 10 \cdot - 5 - 25$

Add $- 125$ and $200$ .

$75 + 10 \cdot - 5 - 25$

Multiply $10$ by $- 5$ .

$75 - 50 - 25$

Subtract $50$ from $75$ .

$25 - 25$

Subtract $25$ from $25$ .

$0$

Since $- 5$ is a known root, divide the polynomial by $x + 5$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{3} + 8 x^{2} + 10 x - 25}{x + 5}$

Divide $x^{3} + 8 x^{2} + 10 x - 25$ by $x + 5$ .

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

$x$

$5$

$x^{3}$

$8 x^{2}$

$10 x$

$25$

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

					$x^{2}$
	$x$	+	$5$		$x^{3}$	+	$8 x^{2}$	+	$10 x$	-	$25$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	+	$5$		$x^{3}$	+	$8 x^{2}$	+	$10 x$	-	$25$
			+	$x^{3}$	+	$5 x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $x^{3} + 5 x^{2}$

				$x^{2}$
$x$	+	$5$		$x^{3}$	+	$8 x^{2}$	+	$10 x$	-	$25$
			-	$x^{3}$	-	$5 x^{2}$

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

				$x^{2}$
$x$	+	$5$		$x^{3}$	+	$8 x^{2}$	+	$10 x$	-	$25$
			-	$x^{3}$	-	$5 x^{2}$
					+	$3 x^{2}$

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

				$x^{2}$
$x$	+	$5$		$x^{3}$	+	$8 x^{2}$	+	$10 x$	-	$25$
			-	$x^{3}$	-	$5 x^{2}$
					+	$3 x^{2}$	+	$10 x$

Divide the highest order term in the dividend $3 x^{2}$ by the highest order term in divisor $x$ .

				$x^{2}$	+	$3 x$
$x$	+	$5$		$x^{3}$	+	$8 x^{2}$	+	$10 x$	-	$25$
			-	$x^{3}$	-	$5 x^{2}$
					+	$3 x^{2}$	+	$10 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$3 x$
$x$	+	$5$		$x^{3}$	+	$8 x^{2}$	+	$10 x$	-	$25$
			-	$x^{3}$	-	$5 x^{2}$
					+	$3 x^{2}$	+	$10 x$
					+	$3 x^{2}$	+	$15 x$

The expression needs to be subtracted from the dividend, so change all the signs in $3 x^{2} + 15 x$

				$x^{2}$	+	$3 x$
$x$	+	$5$		$x^{3}$	+	$8 x^{2}$	+	$10 x$	-	$25$
			-	$x^{3}$	-	$5 x^{2}$
					+	$3 x^{2}$	+	$10 x$
					-	$3 x^{2}$	-	$15 x$

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

				$x^{2}$	+	$3 x$
$x$	+	$5$		$x^{3}$	+	$8 x^{2}$	+	$10 x$	-	$25$
			-	$x^{3}$	-	$5 x^{2}$
					+	$3 x^{2}$	+	$10 x$
					-	$3 x^{2}$	-	$15 x$
							-	$5 x$

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

				$x^{2}$	+	$3 x$
$x$	+	$5$		$x^{3}$	+	$8 x^{2}$	+	$10 x$	-	$25$
			-	$x^{3}$	-	$5 x^{2}$
					+	$3 x^{2}$	+	$10 x$
					-	$3 x^{2}$	-	$15 x$
							-	$5 x$	-	$25$

Divide the highest order term in the dividend $- 5 x$ by the highest order term in divisor $x$ .

				$x^{2}$	+	$3 x$	-	$5$
$x$	+	$5$		$x^{3}$	+	$8 x^{2}$	+	$10 x$	-	$25$
			-	$x^{3}$	-	$5 x^{2}$
					+	$3 x^{2}$	+	$10 x$
					-	$3 x^{2}$	-	$15 x$
							-	$5 x$	-	$25$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$3 x$	-	$5$
$x$	+	$5$		$x^{3}$	+	$8 x^{2}$	+	$10 x$	-	$25$
			-	$x^{3}$	-	$5 x^{2}$
					+	$3 x^{2}$	+	$10 x$
					-	$3 x^{2}$	-	$15 x$
							-	$5 x$	-	$25$
							-	$5 x$	-	$25$

The expression needs to be subtracted from the dividend, so change all the signs in $- 5 x - 25$

				$x^{2}$	+	$3 x$	-	$5$
$x$	+	$5$		$x^{3}$	+	$8 x^{2}$	+	$10 x$	-	$25$
			-	$x^{3}$	-	$5 x^{2}$
					+	$3 x^{2}$	+	$10 x$
					-	$3 x^{2}$	-	$15 x$
							-	$5 x$	-	$25$
							+	$5 x$	+	$25$

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

				$x^{2}$	+	$3 x$	-	$5$
$x$	+	$5$		$x^{3}$	+	$8 x^{2}$	+	$10 x$	-	$25$
			-	$x^{3}$	-	$5 x^{2}$
					+	$3 x^{2}$	+	$10 x$
					-	$3 x^{2}$	-	$15 x$
							-	$5 x$	-	$25$
							+	$5 x$	+	$25$
										$0$

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

$x^{2} + 3 x - 5$

Write $x^{3} + 8 x^{2} + 10 x - 25$ as a set of factors.

$(x + 5) (x^{2} + 3 x - 5) = 0$

Step 4

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

$x + 5 = 0$

$x^{2} + 3 x - 5 = 0$

Step 5

Set $x + 5$ equal to $0$ and solve for $x$ .

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 6

Set $x^{2} + 3 x - 5$ equal to $0$ and solve for $x$ .

Set $x^{2} + 3 x - 5$ equal to $0$ .

$x^{2} + 3 x - 5 = 0$

Solve $x^{2} + 3 x - 5 = 0$ for $x$ .

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = 3$ , and $c = - 5$ into the quadratic formula and solve for $x$ .

$\frac{- 3 \pm \sqrt{3^{2} - 4 \cdot (1 \cdot - 5)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

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

$x = \frac{- 3 \pm \sqrt{9 - 4 \cdot 1 \cdot - 5}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 5$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 3 \pm \sqrt{9 - 4 \cdot - 5}}{2 \cdot 1}$

Multiply $- 4$ by $- 5$ .

$x = \frac{- 3 \pm \sqrt{9 + 20}}{2 \cdot 1}$

Add $9$ and $20$ .

$x = \frac{- 3 \pm \sqrt{29}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 3 \pm \sqrt{29}}{2}$

Simplify the expression to solve for the $+$ portion of the $\pm$ .

Simplify the numerator.

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

$x = \frac{- 3 \pm \sqrt{9 - 4 \cdot 1 \cdot - 5}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 5$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 3 \pm \sqrt{9 - 4 \cdot - 5}}{2 \cdot 1}$

Multiply $- 4$ by $- 5$ .

$x = \frac{- 3 \pm \sqrt{9 + 20}}{2 \cdot 1}$

Add $9$ and $20$ .

$x = \frac{- 3 \pm \sqrt{29}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 3 \pm \sqrt{29}}{2}$

Change the $\pm$ to $+$ .

$x = \frac{- 3 + \sqrt{29}}{2}$

Rewrite $- 3$ as $- 1 (3)$ .

$x = \frac{- 1 \cdot 3 + \sqrt{29}}{2}$

Factor $- 1$ out of $\sqrt{29}$ .

$x = \frac{- 1 \cdot 3 - 1 (- \sqrt{29})}{2}$

Factor $- 1$ out of $- 1 (3) - 1 (- \sqrt{29})$ .

$x = \frac{- 1 (3 - \sqrt{29})}{2}$

Move the negative in front of the fraction.

$x = - \frac{3 - \sqrt{29}}{2}$

Simplify the expression to solve for the $-$ portion of the $\pm$ .

Simplify the numerator.

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

$x = \frac{- 3 \pm \sqrt{9 - 4 \cdot 1 \cdot - 5}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 5$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 3 \pm \sqrt{9 - 4 \cdot - 5}}{2 \cdot 1}$

Multiply $- 4$ by $- 5$ .

$x = \frac{- 3 \pm \sqrt{9 + 20}}{2 \cdot 1}$

Add $9$ and $20$ .

$x = \frac{- 3 \pm \sqrt{29}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 3 \pm \sqrt{29}}{2}$

Change the $\pm$ to $-$ .

$x = \frac{- 3 - \sqrt{29}}{2}$

Rewrite $- 3$ as $- 1 (3)$ .

$x = \frac{- 1 \cdot 3 - \sqrt{29}}{2}$

Factor $- 1$ out of $- \sqrt{29}$ .

$x = \frac{- 1 \cdot 3 - (\sqrt{29})}{2}$

Factor $- 1$ out of $- 1 (3) - (\sqrt{29})$ .

$x = \frac{- 1 (3 + \sqrt{29})}{2}$

Move the negative in front of the fraction.

$x = - \frac{3 + \sqrt{29}}{2}$

The final answer is the combination of both solutions.

$x = - \frac{3 - \sqrt{29}}{2}, - \frac{3 + \sqrt{29}}{2}$

Step 7

The final solution is all the values that make $(x + 5) (x^{2} + 3 x - 5) = 0$ true.

$x = - 5, - \frac{3 - \sqrt{29}}{2}, - \frac{3 + \sqrt{29}}{2}$

Step 8

The result can be shown in multiple forms.

Exact Form:

$x = - 5, - \frac{3 - \sqrt{29}}{2}, - \frac{3 + \sqrt{29}}{2}$

Decimal Form:

$x = - 5, 1.19258240 \dots, - 4.19258240 \dots$

Solve by Factoring x^3+8x^2+11x-20=(x+5)

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