Find the Roots/Zeros Using the Rational Roots Test x^4-9x^2-4x+12

$x^{4} - 9 x^{2} - 4 x + 12$

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.

${(1)}^{4} - 9 {(1)}^{2} - 4 \cdot 1 + 12$

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.

One to any power is one.

$1 - 9 {(1)}^{2} - 4 \cdot 1 + 12$

One to any power is one.

$1 - 9 \cdot 1 - 4 \cdot 1 + 12$

Multiply $- 9$ by $1$ .

$1 - 9 - 4 \cdot 1 + 12$

Multiply $- 4$ by $1$ .

$1 - 9 - 4 + 12$

Simplify by adding and subtracting.

Subtract $9$ from $1$ .

$- 8 - 4 + 12$

Subtract $4$ from $- 8$ .

$- 12 + 12$

Add $- 12$ and $12$ .

$0$

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.

$\frac{x^{4} - 9 x^{2} - 4 x + 12}{x - 1}$

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.

$1$	$1$	$0$	$- 9$	$- 4$	$12$
	\xa0				\xa0

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

$1$	$1$	$0$	$- 9$	$- 4$	$12$
	\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 $(0)$ .

$1$	$1$	$0$	$- 9$	$- 4$	$12$
	\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.

$1$	$1$	$0$	$- 9$	$- 4$	$12$
	\xa0	$1$			\xa0
	$1$	$1$

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 $(- 9)$ .

$1$	$1$	$0$	$- 9$	$- 4$	$12$
	\xa0	$1$	$1$		\xa0
	$1$	$1$

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

$1$	$1$	$0$	$- 9$	$- 4$	$12$
	\xa0	$1$	$1$		\xa0
	$1$	$1$	$- 8$

Multiply the newest entry in the result $(- 8)$ by the divisor $(1)$ and place the result of $(- 8)$ under the next term in the dividend $(- 4)$ .

$1$	$1$	$0$	$- 9$	$- 4$	$12$
	\xa0	$1$	$1$	$- 8$	\xa0
	$1$	$1$	$- 8$

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

$1$	$1$	$0$	$- 9$	$- 4$	$12$
	\xa0	$1$	$1$	$- 8$	\xa0
	$1$	$1$	$- 8$	$- 12$

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

$1$	$1$	$0$	$- 9$	$- 4$	$12$
	\xa0	$1$	$1$	$- 8$	$- 12$
	$1$	$1$	$- 8$	$- 12$

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

$1$	$1$	$0$	$- 9$	$- 4$	$12$
	\xa0	$1$	$1$	$- 8$	$- 12$
	$1$	$1$	$- 8$	$- 12$	$0$

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

$1 x^{3} + 1 x^{2} + (- 8) x - 12$

Simplify the quotient polynomial.

$x^{3} + x^{2} - 8 x - 12$

Step 7

Solve the equation to find any remaining roots.

Factor the left side of the equation.

Factor $x^{3} + x^{2} - 8 x - 12$ 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.

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

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

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

Substitute $- 2$ and simplify the expression. In this case, the expression is equal to $0$ so $- 2$ is a root of the polynomial.

Substitute $- 2$ into the polynomial.

${(- 2)}^{3} + {(- 2)}^{2} - 8 \cdot - 2 - 12$

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

$- 8 + {(- 2)}^{2} - 8 \cdot - 2 - 12$

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

$- 8 + 4 - 8 \cdot - 2 - 12$

Add $- 8$ and $4$ .

$- 4 - 8 \cdot - 2 - 12$

Multiply $- 8$ by $- 2$ .

$- 4 + 16 - 12$

Add $- 4$ and $16$ .

$12 - 12$

Subtract $12$ from $12$ .

$0$

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.

$\frac{x^{3} + x^{2} - 8 x - 12}{x + 2}$

Divide $x^{3} + x^{2} - 8 x - 12$ by $x + 2$ .

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

$x$

$2$

$x^{3}$

$x^{2}$

$8 x$

$12$

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

					$x^{2}$
	$x$	+	$2$		$x^{3}$	+	$x^{2}$	-	$8 x$	-	$12$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	+	$2$		$x^{3}$	+	$x^{2}$	-	$8 x$	-	$12$
			+	$x^{3}$	+	$2 x^{2}$

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

				$x^{2}$
$x$	+	$2$		$x^{3}$	+	$x^{2}$	-	$8 x$	-	$12$
			-	$x^{3}$	-	$2 x^{2}$

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

				$x^{2}$
$x$	+	$2$		$x^{3}$	+	$x^{2}$	-	$8 x$	-	$12$
			-	$x^{3}$	-	$2 x^{2}$
					-	$x^{2}$

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

				$x^{2}$
$x$	+	$2$		$x^{3}$	+	$x^{2}$	-	$8 x$	-	$12$
			-	$x^{3}$	-	$2 x^{2}$
					-	$x^{2}$	-	$8 x$

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

				$x^{2}$	-	$x$
$x$	+	$2$		$x^{3}$	+	$x^{2}$	-	$8 x$	-	$12$
			-	$x^{3}$	-	$2 x^{2}$
					-	$x^{2}$	-	$8 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$x$
$x$	+	$2$		$x^{3}$	+	$x^{2}$	-	$8 x$	-	$12$
			-	$x^{3}$	-	$2 x^{2}$
					-	$x^{2}$	-	$8 x$
					-	$x^{2}$	-	$2 x$

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

				$x^{2}$	-	$x$
$x$	+	$2$		$x^{3}$	+	$x^{2}$	-	$8 x$	-	$12$
			-	$x^{3}$	-	$2 x^{2}$
					-	$x^{2}$	-	$8 x$
					+	$x^{2}$	+	$2 x$

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

				$x^{2}$	-	$x$
$x$	+	$2$		$x^{3}$	+	$x^{2}$	-	$8 x$	-	$12$
			-	$x^{3}$	-	$2 x^{2}$
					-	$x^{2}$	-	$8 x$
					+	$x^{2}$	+	$2 x$
							-	$6 x$

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

				$x^{2}$	-	$x$
$x$	+	$2$		$x^{3}$	+	$x^{2}$	-	$8 x$	-	$12$
			-	$x^{3}$	-	$2 x^{2}$
					-	$x^{2}$	-	$8 x$
					+	$x^{2}$	+	$2 x$
							-	$6 x$	-	$12$

Divide the highest order term in the dividend $- 6 x$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$x$	-	$6$
$x$	+	$2$		$x^{3}$	+	$x^{2}$	-	$8 x$	-	$12$
			-	$x^{3}$	-	$2 x^{2}$
					-	$x^{2}$	-	$8 x$
					+	$x^{2}$	+	$2 x$
							-	$6 x$	-	$12$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$x$	-	$6$
$x$	+	$2$		$x^{3}$	+	$x^{2}$	-	$8 x$	-	$12$
			-	$x^{3}$	-	$2 x^{2}$
					-	$x^{2}$	-	$8 x$
					+	$x^{2}$	+	$2 x$
							-	$6 x$	-	$12$
							-	$6 x$	-	$12$

The expression needs to be subtracted from the dividend, so change all the signs in $- 6 x - 12$

				$x^{2}$	-	$x$	-	$6$
$x$	+	$2$		$x^{3}$	+	$x^{2}$	-	$8 x$	-	$12$
			-	$x^{3}$	-	$2 x^{2}$
					-	$x^{2}$	-	$8 x$
					+	$x^{2}$	+	$2 x$
							-	$6 x$	-	$12$
							+	$6 x$	+	$12$

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

				$x^{2}$	-	$x$	-	$6$
$x$	+	$2$		$x^{3}$	+	$x^{2}$	-	$8 x$	-	$12$
			-	$x^{3}$	-	$2 x^{2}$
					-	$x^{2}$	-	$8 x$
					+	$x^{2}$	+	$2 x$
							-	$6 x$	-	$12$
							+	$6 x$	+	$12$
										$0$

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

$x^{2} - x - 6$

Write $x^{3} + x^{2} - 8 x - 12$ as a set of factors.

$(x + 2) (x^{2} - x - 6) = 0$

Factor $x^{2} - x - 6$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 6$ and whose sum is $- 1$ .

$- 3, 2$

Write the factored form using these integers.

$(x + 2) ((x - 3) (x + 2)) = 0$

Remove unnecessary parentheses.

$(x + 2) (x - 3) (x + 2) = 0$

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

$x + 2 = 0$

$x - 3 = 0$

$x + 2 = 0$

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

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

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

$x = - 2, 3$

Step 8

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

$(x - 1) (x + 2) (x - 3)$

Step 9

These are the roots (zeros) of the polynomial $x^{4} - 9 x^{2} - 4 x + 12$ .

$x = 1, - 2, 3$

Step 10

Find the Roots/Zeros Using the Rational Roots Test x^4-9x^2-4x+12

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