Find the Roots/Zeros Using the Rational Roots Test x^3-5x^2+12x+18

$x^{3} - 5 x^{2} + 12 x + 18$

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.

$p = \pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18$

$q = \pm 1$

Step 2

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

$\pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18$

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.

${(- 1)}^{3} - 5 {(- 1)}^{2} + 12 (- 1) + 18$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = - 1$ is a root of the polynomial.

Simplify each term.

Raise $- 1$ to the power of $3$ .

$- 1 - 5 {(- 1)}^{2} + 12 (- 1) + 18$

Raise $- 1$ to the power of $2$ .

$- 1 - 5 \cdot 1 + 12 (- 1) + 18$

Multiply $- 5$ by $1$ .

$- 1 - 5 + 12 (- 1) + 18$

Multiply $12$ by $- 1$ .

$- 1 - 5 - 12 + 18$

Simplify by adding and subtracting.

Subtract $5$ from $- 1$ .

$- 6 - 12 + 18$

Subtract $12$ from $- 6$ .

$- 18 + 18$

Add $- 18$ and $18$ .

$0$

Step 5

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

$\frac{x^{3} - 5 x^{2} + 12 x + 18}{x + 1}$

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.

$- 1$	$1$	$- 5$	$12$	$18$
	\xa0			\xa0

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

$- 1$	$1$	$- 5$	$12$	$18$
	\xa0			\xa0
	$1$

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

$- 1$	$1$	$- 5$	$12$	$18$
	\xa0	$- 1$		\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.

$- 1$	$1$	$- 5$	$12$	$18$
	\xa0	$- 1$		\xa0
	$1$	$- 6$

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

$- 1$	$1$	$- 5$	$12$	$18$
	\xa0	$- 1$	$6$	\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.

$- 1$	$1$	$- 5$	$12$	$18$
	\xa0	$- 1$	$6$	\xa0
	$1$	$- 6$	$18$

Multiply the newest entry in the result $(18)$ by the divisor $(- 1)$ and place the result of $(- 18)$ under the next term in the dividend $(18)$ .

$- 1$	$1$	$- 5$	$12$	$18$
	\xa0	$- 1$	$6$	$- 18$
	$1$	$- 6$	$18$

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

$- 1$	$1$	$- 5$	$12$	$18$
	\xa0	$- 1$	$6$	$- 18$
	$1$	$- 6$	$18$	$0$

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

$1 x^{2} + - 6 x + 18$

Simplify the quotient polynomial.

$x^{2} - 6 x + 18$

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 $a = 1$ , $b = - 6$ , and $c = 18$ into the quadratic formula and solve for $x$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (1 \cdot 18)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

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

$x = \frac{6 \pm \sqrt{36 - 4 \cdot 1 \cdot 18}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 18$ .

Multiply $- 4$ by $1$ .

$x = \frac{6 \pm \sqrt{36 - 4 \cdot 18}}{2 \cdot 1}$

Multiply $- 4$ by $18$ .

$x = \frac{6 \pm \sqrt{36 - 72}}{2 \cdot 1}$

Subtract $72$ from $36$ .

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

Rewrite $- 36$ as $- 1 (36)$ .

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

Rewrite $\sqrt{- 1 (36)}$ as $\sqrt{- 1} \cdot \sqrt{36}$ .

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

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

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

Rewrite $36$ as $6^{2}$ .

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

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

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

Move $6$ to the left of $i$ .

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

Multiply $2$ by $1$ .

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

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

$x = 3 \pm 3 i$

The final answer is the combination of both solutions.

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

Step 8

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

$(x + 1) (x - 3 - 3 i) (x - 3 + 3 i)$

Step 9

These are the roots (zeros) of the polynomial $x^{3} - 5 x^{2} + 12 x + 18$ .

$x = - 1, 3 + 3 i, 3 - 3 i$

Step 10

Find the Roots/Zeros Using the Rational Roots Test x^3-5x^2+12x+18

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