Find the Roots (Zeros) x^3-2x^2+10x+136

$x^{3} - 2 x^{2} + 10 x + 136$

Step 1

Set $x^{3} - 2 x^{2} + 10 x + 136$ equal to $0$ .

$x^{3} - 2 x^{2} + 10 x + 136 = 0$

Step 2

Solve for $x$ .

Factor $x^{3} - 2 x^{2} + 10 x + 136$ 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.

$p = \pm 1, \pm 136, \pm 2, \pm 68, \pm 4, \pm 34, \pm 8, \pm 17 q = \pm 1$

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

$\pm 1, \pm 136, \pm 2, \pm 68, \pm 4, \pm 34, \pm 8, \pm 17$

Substitute $- 4$ and simplify the expression. In this case, the expression is equal to $0$ so $- 4$ is a root of the polynomial.

Substitute $- 4$ into the polynomial.

${(- 4)}^{3} - 2 {(- 4)}^{2} + 10 \cdot - 4 + 136$

Raise $- 4$ to the power of $3$ .

$- 64 - 2 {(- 4)}^{2} + 10 \cdot - 4 + 136$

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

$- 64 - 2 \cdot 16 + 10 \cdot - 4 + 136$

Multiply $- 2$ by $16$ .

$- 64 - 32 + 10 \cdot - 4 + 136$

Subtract $32$ from $- 64$ .

$- 96 + 10 \cdot - 4 + 136$

Multiply $10$ by $- 4$ .

$- 96 - 40 + 136$

Subtract $40$ from $- 96$ .

$- 136 + 136$

Add $- 136$ and $136$ .

$0$

Since $- 4$ is a known root, divide the polynomial by $x + 4$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{3} - 2 x^{2} + 10 x + 136}{x + 4}$

Divide $x^{3} - 2 x^{2} + 10 x + 136$ by $x + 4$ .

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

$x$

$4$

$x^{3}$

$2 x^{2}$

$10 x$

$136$

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

					$x^{2}$
	$x$	+	$4$		$x^{3}$	-	$2 x^{2}$	+	$10 x$	+	$136$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	+	$4$		$x^{3}$	-	$2 x^{2}$	+	$10 x$	+	$136$
			+	$x^{3}$	+	$4 x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $x^{3} + 4 x^{2}$

				$x^{2}$
$x$	+	$4$		$x^{3}$	-	$2 x^{2}$	+	$10 x$	+	$136$
			-	$x^{3}$	-	$4 x^{2}$

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

				$x^{2}$
$x$	+	$4$		$x^{3}$	-	$2 x^{2}$	+	$10 x$	+	$136$
			-	$x^{3}$	-	$4 x^{2}$
					-	$6 x^{2}$

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

				$x^{2}$
$x$	+	$4$		$x^{3}$	-	$2 x^{2}$	+	$10 x$	+	$136$
			-	$x^{3}$	-	$4 x^{2}$
					-	$6 x^{2}$	+	$10 x$

Divide the highest order term in the dividend $- 6 x^{2}$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$6 x$
$x$	+	$4$		$x^{3}$	-	$2 x^{2}$	+	$10 x$	+	$136$
			-	$x^{3}$	-	$4 x^{2}$
					-	$6 x^{2}$	+	$10 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$6 x$
$x$	+	$4$		$x^{3}$	-	$2 x^{2}$	+	$10 x$	+	$136$
			-	$x^{3}$	-	$4 x^{2}$
					-	$6 x^{2}$	+	$10 x$
					-	$6 x^{2}$	-	$24 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 6 x^{2} - 24 x$

				$x^{2}$	-	$6 x$
$x$	+	$4$		$x^{3}$	-	$2 x^{2}$	+	$10 x$	+	$136$
			-	$x^{3}$	-	$4 x^{2}$
					-	$6 x^{2}$	+	$10 x$
					+	$6 x^{2}$	+	$24 x$

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

				$x^{2}$	-	$6 x$
$x$	+	$4$		$x^{3}$	-	$2 x^{2}$	+	$10 x$	+	$136$
			-	$x^{3}$	-	$4 x^{2}$
					-	$6 x^{2}$	+	$10 x$
					+	$6 x^{2}$	+	$24 x$
							+	$34 x$

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

				$x^{2}$	-	$6 x$
$x$	+	$4$		$x^{3}$	-	$2 x^{2}$	+	$10 x$	+	$136$
			-	$x^{3}$	-	$4 x^{2}$
					-	$6 x^{2}$	+	$10 x$
					+	$6 x^{2}$	+	$24 x$
							+	$34 x$	+	$136$

Divide the highest order term in the dividend $34 x$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$6 x$	+	$34$
$x$	+	$4$		$x^{3}$	-	$2 x^{2}$	+	$10 x$	+	$136$
			-	$x^{3}$	-	$4 x^{2}$
					-	$6 x^{2}$	+	$10 x$
					+	$6 x^{2}$	+	$24 x$
							+	$34 x$	+	$136$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$6 x$	+	$34$
$x$	+	$4$		$x^{3}$	-	$2 x^{2}$	+	$10 x$	+	$136$
			-	$x^{3}$	-	$4 x^{2}$
					-	$6 x^{2}$	+	$10 x$
					+	$6 x^{2}$	+	$24 x$
							+	$34 x$	+	$136$
							+	$34 x$	+	$136$

The expression needs to be subtracted from the dividend, so change all the signs in $34 x + 136$

				$x^{2}$	-	$6 x$	+	$34$
$x$	+	$4$		$x^{3}$	-	$2 x^{2}$	+	$10 x$	+	$136$
			-	$x^{3}$	-	$4 x^{2}$
					-	$6 x^{2}$	+	$10 x$
					+	$6 x^{2}$	+	$24 x$
							+	$34 x$	+	$136$
							-	$34 x$	-	$136$

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

				$x^{2}$	-	$6 x$	+	$34$
$x$	+	$4$		$x^{3}$	-	$2 x^{2}$	+	$10 x$	+	$136$
			-	$x^{3}$	-	$4 x^{2}$
					-	$6 x^{2}$	+	$10 x$
					+	$6 x^{2}$	+	$24 x$
							+	$34 x$	+	$136$
							-	$34 x$	-	$136$
										$0$

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

$x^{2} - 6 x + 34$

Write $x^{3} - 2 x^{2} + 10 x + 136$ as a set of factors.

$(x + 4) (x^{2} - 6 x + 34) = 0$

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

$x + 4 = 0$

$x^{2} - 6 x + 34 = 0$

Set $x + 4$ equal to $0$ and solve for $x$ .

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

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

Set $x^{2} - 6 x + 34$ equal to $0$ and solve for $x$ .

Set $x^{2} - 6 x + 34$ equal to $0$ .

$x^{2} - 6 x + 34 = 0$

Solve $x^{2} - 6 x + 34 = 0$ for $x$ .

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = - 6$ , and $c = 34$ into the quadratic formula and solve for $x$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (1 \cdot 34)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $- 6$ to the power of $2$ .

$x = \frac{6 \pm \sqrt{36 - 4 \cdot 1 \cdot 34}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 34$ .

Multiply $- 4$ by $1$ .

$x = \frac{6 \pm \sqrt{36 - 4 \cdot 34}}{2 \cdot 1}$

Multiply $- 4$ by $34$ .

$x = \frac{6 \pm \sqrt{36 - 136}}{2 \cdot 1}$

Subtract $136$ from $36$ .

$x = \frac{6 \pm \sqrt{- 100}}{2 \cdot 1}$

Rewrite $- 100$ as $- 1 (100)$ .

$x = \frac{6 \pm \sqrt{- 1 \cdot 100}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (100)}$ as $\sqrt{- 1} \cdot \sqrt{100}$ .

$x = \frac{6 \pm \sqrt{- 1} \cdot \sqrt{100}}{2 \cdot 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{6 \pm i \cdot \sqrt{100}}{2 \cdot 1}$

Rewrite $100$ as $10^{2}$ .

$x = \frac{6 \pm i \cdot \sqrt{10^{2}}}{2 \cdot 1}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{6 \pm i \cdot 10}{2 \cdot 1}$

Move $10$ to the left of $i$ .

$x = \frac{6 \pm 10 i}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{6 \pm 10 i}{2}$

Simplify $\frac{6 \pm 10 i}{2}$ .

$x = 3 \pm 5 i$

The final answer is the combination of both solutions.

$x = 3 + 5 i, 3 - 5 i$

The final solution is all the values that make $(x + 4) (x^{2} - 6 x + 34) = 0$ true.

$x = - 4, 3 + 5 i, 3 - 5 i$

Step 3

Find the Roots (Zeros) x^3-2x^2+10x+136

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