Find the Roots/Zeros Using the Rational Roots Test x^4+2x^3-3x^2-4x-12

$x^{4} + 2 x^{3} - 3 x^{2} - 4 x - 12$

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.

${(2)}^{4} + 2 {(2)}^{3} - 3 {(2)}^{2} - 4 \cdot 2 - 12$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = 2$ is a root of the polynomial.

Simplify each term.

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

$16 + 2 {(2)}^{3} - 3 {(2)}^{2} - 4 \cdot 2 - 12$

Multiply $2$ by ${(2)}^{3}$ by adding the exponents.

Multiply $2$ by ${(2)}^{3}$ .

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

$16 + 2^{1} {(2)}^{3} - 3 {(2)}^{2} - 4 \cdot 2 - 12$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$16 + 2^{1 + 3} - 3 {(2)}^{2} - 4 \cdot 2 - 12$

Add $1$ and $3$ .

$16 + 2^{4} - 3 {(2)}^{2} - 4 \cdot 2 - 12$

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

$16 + 16 - 3 {(2)}^{2} - 4 \cdot 2 - 12$

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

$16 + 16 - 3 \cdot 4 - 4 \cdot 2 - 12$

Multiply $- 3$ by $4$ .

$16 + 16 - 12 - 4 \cdot 2 - 12$

Multiply $- 4$ by $2$ .

$16 + 16 - 12 - 8 - 12$

Simplify by adding and subtracting.

Add $16$ and $16$ .

$32 - 12 - 8 - 12$

Subtract $12$ from $32$ .

$20 - 8 - 12$

Subtract $8$ from $20$ .

$12 - 12$

Subtract $12$ from $12$ .

$0$

Step 5

Since $2$ is a known root, divide the polynomial by $x - 2$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{4} + 2 x^{3} - 3 x^{2} - 4 x - 12}{x - 2}$

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.

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

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

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

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

$2$	$1$	$2$	$- 3$	$- 4$	$- 12$
	\xa0	$2$			\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.

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

Multiply the newest entry in the result $(4)$ by the divisor $(2)$ and place the result of $(8)$ under the next term in the dividend $(- 3)$ .

$2$	$1$	$2$	$- 3$	$- 4$	$- 12$
	\xa0	$2$	$8$		\xa0
	$1$	$4$

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

$2$	$1$	$2$	$- 3$	$- 4$	$- 12$
	\xa0	$2$	$8$		\xa0
	$1$	$4$	$5$

Multiply the newest entry in the result $(5)$ by the divisor $(2)$ and place the result of $(10)$ under the next term in the dividend $(- 4)$ .

$2$	$1$	$2$	$- 3$	$- 4$	$- 12$
	\xa0	$2$	$8$	$10$	\xa0
	$1$	$4$	$5$

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

$2$	$1$	$2$	$- 3$	$- 4$	$- 12$
	\xa0	$2$	$8$	$10$	\xa0
	$1$	$4$	$5$	$6$

Multiply the newest entry in the result $(6)$ by the divisor $(2)$ and place the result of $(12)$ under the next term in the dividend $(- 12)$ .

$2$	$1$	$2$	$- 3$	$- 4$	$- 12$
	\xa0	$2$	$8$	$10$	$12$
	$1$	$4$	$5$	$6$

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

$2$	$1$	$2$	$- 3$	$- 4$	$- 12$
	\xa0	$2$	$8$	$10$	$12$
	$1$	$4$	$5$	$6$	$0$

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

$1 x^{3} + 4 x^{2} + (5) x + 6$

Simplify the quotient polynomial.

$x^{3} + 4 x^{2} + 5 x + 6$

Step 7

Solve the equation to find any remaining roots.

Factor $x^{3} + 4 x^{2} + 5 x + 6$ 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.

$p = \pm 1, \pm 6, \pm 2, \pm 3 q = \pm 1$

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

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

Substitute $- 3$ and simplify the expression. In this case, the expression is equal to $0$ so $- 3$ is a root of the polynomial.

Substitute $- 3$ into the polynomial.

${(- 3)}^{3} + 4 {(- 3)}^{2} + 5 \cdot - 3 + 6$

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

$- 27 + 4 {(- 3)}^{2} + 5 \cdot - 3 + 6$

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

$- 27 + 4 \cdot 9 + 5 \cdot - 3 + 6$

Multiply $4$ by $9$ .

$- 27 + 36 + 5 \cdot - 3 + 6$

Add $- 27$ and $36$ .

$9 + 5 \cdot - 3 + 6$

Multiply $5$ by $- 3$ .

$9 - 15 + 6$

Subtract $15$ from $9$ .

$- 6 + 6$

Add $- 6$ and $6$ .

$0$

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.

$\frac{x^{3} + 4 x^{2} + 5 x + 6}{x + 3}$

Divide $x^{3} + 4 x^{2} + 5 x + 6$ by $x + 3$ .

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

$x$

$3$

$x^{3}$

$4 x^{2}$

$5 x$

$6$

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

					$x^{2}$
	$x$	+	$3$		$x^{3}$	+	$4 x^{2}$	+	$5 x$	+	$6$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	+	$3$		$x^{3}$	+	$4 x^{2}$	+	$5 x$	+	$6$
			+	$x^{3}$	+	$3 x^{2}$

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

				$x^{2}$
$x$	+	$3$		$x^{3}$	+	$4 x^{2}$	+	$5 x$	+	$6$
			-	$x^{3}$	-	$3 x^{2}$

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

				$x^{2}$
$x$	+	$3$		$x^{3}$	+	$4 x^{2}$	+	$5 x$	+	$6$
			-	$x^{3}$	-	$3 x^{2}$
					+	$x^{2}$

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

				$x^{2}$
$x$	+	$3$		$x^{3}$	+	$4 x^{2}$	+	$5 x$	+	$6$
			-	$x^{3}$	-	$3 x^{2}$
					+	$x^{2}$	+	$5 x$

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

				$x^{2}$	+	$x$
$x$	+	$3$		$x^{3}$	+	$4 x^{2}$	+	$5 x$	+	$6$
			-	$x^{3}$	-	$3 x^{2}$
					+	$x^{2}$	+	$5 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$x$
$x$	+	$3$		$x^{3}$	+	$4 x^{2}$	+	$5 x$	+	$6$
			-	$x^{3}$	-	$3 x^{2}$
					+	$x^{2}$	+	$5 x$
					+	$x^{2}$	+	$3 x$

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

				$x^{2}$	+	$x$
$x$	+	$3$		$x^{3}$	+	$4 x^{2}$	+	$5 x$	+	$6$
			-	$x^{3}$	-	$3 x^{2}$
					+	$x^{2}$	+	$5 x$
					-	$x^{2}$	-	$3 x$

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

				$x^{2}$	+	$x$
$x$	+	$3$		$x^{3}$	+	$4 x^{2}$	+	$5 x$	+	$6$
			-	$x^{3}$	-	$3 x^{2}$
					+	$x^{2}$	+	$5 x$
					-	$x^{2}$	-	$3 x$
							+	$2 x$

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

				$x^{2}$	+	$x$
$x$	+	$3$		$x^{3}$	+	$4 x^{2}$	+	$5 x$	+	$6$
			-	$x^{3}$	-	$3 x^{2}$
					+	$x^{2}$	+	$5 x$
					-	$x^{2}$	-	$3 x$
							+	$2 x$	+	$6$

Divide the highest order term in the dividend $2 x$ by the highest order term in divisor $x$ .

				$x^{2}$	+	$x$	+	$2$
$x$	+	$3$		$x^{3}$	+	$4 x^{2}$	+	$5 x$	+	$6$
			-	$x^{3}$	-	$3 x^{2}$
					+	$x^{2}$	+	$5 x$
					-	$x^{2}$	-	$3 x$
							+	$2 x$	+	$6$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$x$	+	$2$
$x$	+	$3$		$x^{3}$	+	$4 x^{2}$	+	$5 x$	+	$6$
			-	$x^{3}$	-	$3 x^{2}$
					+	$x^{2}$	+	$5 x$
					-	$x^{2}$	-	$3 x$
							+	$2 x$	+	$6$
							+	$2 x$	+	$6$

The expression needs to be subtracted from the dividend, so change all the signs in $2 x + 6$

				$x^{2}$	+	$x$	+	$2$
$x$	+	$3$		$x^{3}$	+	$4 x^{2}$	+	$5 x$	+	$6$
			-	$x^{3}$	-	$3 x^{2}$
					+	$x^{2}$	+	$5 x$
					-	$x^{2}$	-	$3 x$
							+	$2 x$	+	$6$
							-	$2 x$	-	$6$

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

				$x^{2}$	+	$x$	+	$2$
$x$	+	$3$		$x^{3}$	+	$4 x^{2}$	+	$5 x$	+	$6$
			-	$x^{3}$	-	$3 x^{2}$
					+	$x^{2}$	+	$5 x$
					-	$x^{2}$	-	$3 x$
							+	$2 x$	+	$6$
							-	$2 x$	-	$6$
										$0$

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

$x^{2} + x + 2$

Write $x^{3} + 4 x^{2} + 5 x + 6$ as a set of factors.

$(x + 3) (x^{2} + x + 2) = 0$

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

$x + 3 = 0$

$x^{2} + x + 2 = 0$

Set $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

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

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

$x^{2} + x + 2 = 0$

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

Use the quadratic formula to find the solutions.

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

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

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

Simplify.

Simplify the numerator.

One to any power is one.

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

Multiply $- 4 \cdot 1 \cdot 2$ .

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $2$ .

$x = \frac{- 1 \pm \sqrt{1 - 8}}{2 \cdot 1}$

Subtract $8$ from $1$ .

$x = \frac{- 1 \pm \sqrt{- 7}}{2 \cdot 1}$

Rewrite $- 7$ as $- 1 (7)$ .

$x = \frac{- 1 \pm \sqrt{- 1 \cdot 7}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (7)}$ as $\sqrt{- 1} \cdot \sqrt{7}$ .

$x = \frac{- 1 \pm \sqrt{- 1} \cdot \sqrt{7}}{2 \cdot 1}$

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

$x = \frac{- 1 \pm i \sqrt{7}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 1 \pm i \sqrt{7}}{2}$

The final answer is the combination of both solutions.

$x = - \frac{1 - i \sqrt{7}}{2}, - \frac{1 + i \sqrt{7}}{2}$

The final solution is all the values that make $(x + 3) (x^{2} + x + 2) = 0$ true.

$x = - 3, - \frac{1 - i \sqrt{7}}{2}, - \frac{1 + i \sqrt{7}}{2}$

Step 8

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

$(x - 2) (x + 3) (x + \frac{1 - i \sqrt{7}}{2}) (x + \frac{1 + i \sqrt{7}}{2})$

Step 9

These are the roots (zeros) of the polynomial $x^{4} + 2 x^{3} - 3 x^{2} - 4 x - 12$ .

$x = 2, - 3, - \frac{1 - i \sqrt{7}}{2}, - \frac{1 + i \sqrt{7}}{2}$

Step 10

Find the Roots/Zeros Using the Rational Roots Test x^4+2x^3-3x^2-4x-12

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