Find the Roots/Zeros Using the Rational Roots Test x^3-2x^2+10x+136=0

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

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.

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

$q = \pm 1$

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

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

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.

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

Simplify the expression. In this case, the expression is equal to $0$ so $x = - 4$ is a root of the polynomial.

Simplify each term.

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

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

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

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

Multiply $- 2$ by $16$ .

$- 64 - 32 + 10 (- 4) + 136$

Multiply $10$ by $- 4$ .

$- 64 - 32 - 40 + 136$

Simplify by adding and subtracting.

Subtract $32$ from $- 64$ .

$- 96 - 40 + 136$

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.

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

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

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

$- 4$	$1$	$- 2$	$10$	$136$
	\xa0			\xa0

The first number in the dividend $(1)$ is put into the first position of the result area (below the horizontal line).

$- 4$	$1$	$- 2$	$10$	$136$
	\xa0			\xa0
	$1$

Multiply the newest entry in the result $(1)$ by the divisor $(- 4)$ and place the result of $(- 4)$ under the next term in the dividend $(- 2)$ .

$- 4$	$1$	$- 2$	$10$	$136$
	\xa0	$- 4$		\xa0
	$1$

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

$- 4$	$1$	$- 2$	$10$	$136$
	\xa0	$- 4$		\xa0
	$1$	$- 6$

Multiply the newest entry in the result $(- 6)$ by the divisor $(- 4)$ and place the result of $(24)$ under the next term in the dividend $(10)$ .

$- 4$	$1$	$- 2$	$10$	$136$
	\xa0	$- 4$	$24$	\xa0
	$1$	$- 6$

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

$- 4$	$1$	$- 2$	$10$	$136$
	\xa0	$- 4$	$24$	\xa0
	$1$	$- 6$	$34$

Multiply the newest entry in the result $(34)$ by the divisor $(- 4)$ and place the result of $(- 136)$ under the next term in the dividend $(136)$ .

$- 4$	$1$	$- 2$	$10$	$136$
	\xa0	$- 4$	$24$	$- 136$
	$1$	$- 6$	$34$

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

$- 4$	$1$	$- 2$	$10$	$136$
	\xa0	$- 4$	$24$	$- 136$
	$1$	$- 6$	$34$	$0$

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

$1 x^{2} + - 6 x + 34$

Simplify the quotient polynomial.

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

Solve the equation to find any remaining roots.

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 $- 4$ 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 $\frac{6 \pm 10 i}{2}$ .

$x = 3 \pm 5 i$

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

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 $- 4$ 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 $\frac{6 \pm 10 i}{2}$ .

$x = 3 \pm 5 i$

Change the $\pm$ to $+$ .

$x = 3 + 5 i$

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

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 $- 4$ 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 $\frac{6 \pm 10 i}{2}$ .

$x = 3 \pm 5 i$

Change the $\pm$ to $-$ .

$x = 3 - 5 i$

The final answer is the combination of both solutions.

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

The polynomial can be written as a set of linear factors.

$(x + 4) (x - 3 - 5 i) (x - 3 + 5 i)$

These are the roots (zeros) of the polynomial $x^{3} - 2 x^{2} + 10 x + 136$ .

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

Find the Roots/Zeros Using the Rational Roots Test x^3-2x^2+10x+136=0

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