Find the Roots/Zeros Using the Rational Roots Test x^3+6x^2+4x+24

$x^{3} + 6 x^{2} + 4 x + 24$

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.

$p = \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24$

$q = \pm 1$

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

$\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24$

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.

${(- 6)}^{3} + 6 {(- 6)}^{2} + 4 (- 6) + 24$

Simplify the expression. In this case, the expression is equal to $0$ so $x = - 6$ is a root of the polynomial.

Simplify each term.

Raise $- 6$ to the power of $3$ .

$- 216 + 6 {(- 6)}^{2} + 4 (- 6) + 24$

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

$- 216 + 6 \cdot 36 + 4 (- 6) + 24$

Multiply $6$ by $36$ .

$- 216 + 216 + 4 (- 6) + 24$

Multiply $4$ by $- 6$ .

$- 216 + 216 - 24 + 24$

Simplify by adding and subtracting.

Add $- 216$ and $216$ .

$0 - 24 + 24$

Subtract $24$ from $0$ .

$- 24 + 24$

Add $- 24$ and $24$ .

$0$

Since $- 6$ is a known root, divide the polynomial by $x + 6$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{3} + 6 x^{2} + 4 x + 24}{x + 6}$

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.

$- 6$	$1$	$6$	$4$	$24$
	\xa0			\xa0

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

$- 6$	$1$	$6$	$4$	$24$
	\xa0			\xa0
	$1$

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

$- 6$	$1$	$6$	$4$	$24$
	\xa0	$- 6$		\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.

$- 6$	$1$	$6$	$4$	$24$
	\xa0	$- 6$		\xa0
	$1$	$0$

Multiply the newest entry in the result $(0)$ by the divisor $(- 6)$ and place the result of $(0)$ under the next term in the dividend $(4)$ .

$- 6$	$1$	$6$	$4$	$24$
	\xa0	$- 6$	$0$	\xa0
	$1$	$0$

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

$- 6$	$1$	$6$	$4$	$24$
	\xa0	$- 6$	$0$	\xa0
	$1$	$0$	$4$

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

$- 6$	$1$	$6$	$4$	$24$
	\xa0	$- 6$	$0$	$- 24$
	$1$	$0$	$4$

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

$- 6$	$1$	$6$	$4$	$24$
	\xa0	$- 6$	$0$	$- 24$
	$1$	$0$	$4$	$0$

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

$1 x^{2} + 0 x + 4$

Simplify the quotient polynomial.

$x^{2} + 4$

Solve the equation to find any remaining roots.

Subtract $4$ from both sides of the equation.

$x^{2} = - 4$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{- 4}$

Simplify $\pm \sqrt{- 4}$ .

Rewrite $- 4$ as $- 1 (4)$ .

$x = \pm \sqrt{- 1 (4)}$

Rewrite $\sqrt{- 1 (4)}$ as $\sqrt{- 1} \cdot \sqrt{4}$ .

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

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

$x = \pm i \cdot \sqrt{4}$

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

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

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

$x = \pm i \cdot 2$

Move $2$ to the left of $i$ .

$x = \pm 2 i$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = 2 i$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 2 i$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 2 i, - 2 i$

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

$(x + 6) (x - 2 i) (x + 2 i)$

These are the roots (zeros) of the polynomial $x^{3} + 6 x^{2} + 4 x + 24$ .

$x = - 6, 2 i, - 2 i$

Find the Roots/Zeros Using the Rational Roots Test x^3+6x^2+4x+24

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