Find the Roots/Zeros Using the Rational Roots Test h(x)=x^3-5x^2+2x+12

$h (x) = x^{3} - 5 x^{2} + 2 x + 12$

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.

$p = \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$

$q = \pm 1$

Step 2

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

$\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$

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.

${(3)}^{3} - 5 {(3)}^{2} + 2 (3) + 12$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = 3$ is a root of the polynomial.

Simplify each term.

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

$27 - 5 {(3)}^{2} + 2 (3) + 12$

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

$27 - 5 \cdot 9 + 2 (3) + 12$

Multiply $- 5$ by $9$ .

$27 - 45 + 2 (3) + 12$

Multiply $2$ by $3$ .

$27 - 45 + 6 + 12$

Simplify by adding and subtracting.

Subtract $45$ from $27$ .

$- 18 + 6 + 12$

Add $- 18$ and $6$ .

$- 12 + 12$

Add $- 12$ and $12$ .

$0$

Step 5

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.

$\frac{x^{3} - 5 x^{2} + 2 x + 12}{x - 3}$

Step 6

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.

$3$	$1$	$- 5$	$2$	$12$
	\xa0			\xa0

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

$3$	$1$	$- 5$	$2$	$12$
	\xa0			\xa0
	$1$

Multiply the newest entry in the result $(1)$ by the divisor $(3)$ and place the result of $(3)$ under the next term in the dividend $(- 5)$ .

$3$	$1$	$- 5$	$2$	$12$
	\xa0	$3$		\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.

$3$	$1$	$- 5$	$2$	$12$
	\xa0	$3$		\xa0
	$1$	$- 2$

Multiply the newest entry in the result $(- 2)$ by the divisor $(3)$ and place the result of $(- 6)$ under the next term in the dividend $(2)$ .

$3$	$1$	$- 5$	$2$	$12$
	\xa0	$3$	$- 6$	\xa0
	$1$	$- 2$

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

$3$	$1$	$- 5$	$2$	$12$
	\xa0	$3$	$- 6$	\xa0
	$1$	$- 2$	$- 4$

Multiply the newest entry in the result $(- 4)$ by the divisor $(3)$ and place the result of $(- 12)$ under the next term in the dividend $(12)$ .

$3$	$1$	$- 5$	$2$	$12$
	\xa0	$3$	$- 6$	$- 12$
	$1$	$- 2$	$- 4$

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

$3$	$1$	$- 5$	$2$	$12$
	\xa0	$3$	$- 6$	$- 12$
	$1$	$- 2$	$- 4$	$0$

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

$1 x^{2} + - 2 x - 4$

Simplify the quotient polynomial.

$x^{2} - 2 x - 4$

Step 7

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 $a = 1$ , $b = - 2$ , and $c = - 4$ into the quadratic formula and solve for $x$ .

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

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $- 4$ .

$x = \frac{2 \pm \sqrt{4 + 16}}{2 \cdot 1}$

Add $4$ and $16$ .

$x = \frac{2 \pm \sqrt{20}}{2 \cdot 1}$

Rewrite $20$ as $2^{2} \cdot 5$ .

Factor $4$ out of $20$ .

$x = \frac{2 \pm \sqrt{4 (5)}}{2 \cdot 1}$

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

$x = \frac{2 \pm \sqrt{2^{2} \cdot 5}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{2 \pm 2 \sqrt{5}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{2 \pm 2 \sqrt{5}}{2}$

Simplify $\frac{2 \pm 2 \sqrt{5}}{2}$ .

$x = 1 \pm \sqrt{5}$

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

Simplify the numerator.

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

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $- 4$ .

$x = \frac{2 \pm \sqrt{4 + 16}}{2 \cdot 1}$

Add $4$ and $16$ .

$x = \frac{2 \pm \sqrt{20}}{2 \cdot 1}$

Rewrite $20$ as $2^{2} \cdot 5$ .

Factor $4$ out of $20$ .

$x = \frac{2 \pm \sqrt{4 (5)}}{2 \cdot 1}$

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

$x = \frac{2 \pm \sqrt{2^{2} \cdot 5}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{2 \pm 2 \sqrt{5}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{2 \pm 2 \sqrt{5}}{2}$

Simplify $\frac{2 \pm 2 \sqrt{5}}{2}$ .

$x = 1 \pm \sqrt{5}$

Change the $\pm$ to $+$ .

$x = 1 + \sqrt{5}$

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

Simplify the numerator.

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

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $- 4$ .

$x = \frac{2 \pm \sqrt{4 + 16}}{2 \cdot 1}$

Add $4$ and $16$ .

$x = \frac{2 \pm \sqrt{20}}{2 \cdot 1}$

Rewrite $20$ as $2^{2} \cdot 5$ .

Factor $4$ out of $20$ .

$x = \frac{2 \pm \sqrt{4 (5)}}{2 \cdot 1}$

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

$x = \frac{2 \pm \sqrt{2^{2} \cdot 5}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{2 \pm 2 \sqrt{5}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{2 \pm 2 \sqrt{5}}{2}$

Simplify $\frac{2 \pm 2 \sqrt{5}}{2}$ .

$x = 1 \pm \sqrt{5}$

Change the $\pm$ to $-$ .

$x = 1 - \sqrt{5}$

The final answer is the combination of both solutions.

$x = 1 + \sqrt{5}, 1 - \sqrt{5}$

Step 8

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

$(x - 3) (x - 1 - \sqrt{5}) (x - 1 + \sqrt{5})$

Step 9

These are the roots (zeros) of the polynomial $x^{3} - 5 x^{2} + 2 x + 12$ .

$x = 3, 1 + \sqrt{5}, 1 - \sqrt{5}$

Step 10

The result can be shown in multiple forms.

Exact Form:

$x = 3, 1 + \sqrt{5}, 1 - \sqrt{5}$

Decimal Form:

$x = 3, 3.23606797 \dots, - 1.23606797 \dots$

Step 11

Find the Roots/Zeros Using the Rational Roots Test h(x)=x^3-5x^2+2x+12

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