Find the Roots/Zeros Using the Rational Roots Test x^4+2x^3-7x^2-8x+12

$x^{4} + 2 x^{3} - 7 x^{2} - 8 x + 12$

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.

${(1)}^{4} + 2 {(1)}^{3} - 7 {(1)}^{2} - 8 \cdot 1 + 12$

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.

$1 + 2 {(1)}^{3} - 7 {(1)}^{2} - 8 \cdot 1 + 12$

One to any power is one.

$1 + 2 \cdot 1 - 7 {(1)}^{2} - 8 \cdot 1 + 12$

Multiply $2$ by $1$ .

$1 + 2 - 7 {(1)}^{2} - 8 \cdot 1 + 12$

One to any power is one.

$1 + 2 - 7 \cdot 1 - 8 \cdot 1 + 12$

Multiply $- 7$ by $1$ .

$1 + 2 - 7 - 8 \cdot 1 + 12$

Multiply $- 8$ by $1$ .

$1 + 2 - 7 - 8 + 12$

Simplify by adding and subtracting.

Add $1$ and $2$ .

$3 - 7 - 8 + 12$

Subtract $7$ from $3$ .

$- 4 - 8 + 12$

Subtract $8$ from $- 4$ .

$- 12 + 12$

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.

$\frac{x^{4} + 2 x^{3} - 7 x^{2} - 8 x + 12}{x - 1}$

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.

$1$	$1$	$2$	$- 7$	$- 8$	$12$
	\xa0				\xa0

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

$1$	$1$	$2$	$- 7$	$- 8$	$12$
	\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 $(2)$ .

$1$	$1$	$2$	$- 7$	$- 8$	$12$
	\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.

$1$	$1$	$2$	$- 7$	$- 8$	$12$
	\xa0	$1$			\xa0
	$1$	$3$

Multiply the newest entry in the result $(3)$ by the divisor $(1)$ and place the result of $(3)$ under the next term in the dividend $(- 7)$ .

$1$	$1$	$2$	$- 7$	$- 8$	$12$
	\xa0	$1$	$3$		\xa0
	$1$	$3$

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

$1$	$1$	$2$	$- 7$	$- 8$	$12$
	\xa0	$1$	$3$		\xa0
	$1$	$3$	$- 4$

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

$1$	$1$	$2$	$- 7$	$- 8$	$12$
	\xa0	$1$	$3$	$- 4$	\xa0
	$1$	$3$	$- 4$

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

$1$	$1$	$2$	$- 7$	$- 8$	$12$
	\xa0	$1$	$3$	$- 4$	\xa0
	$1$	$3$	$- 4$	$- 12$

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 $(12)$ .

$1$	$1$	$2$	$- 7$	$- 8$	$12$
	\xa0	$1$	$3$	$- 4$	$- 12$
	$1$	$3$	$- 4$	$- 12$

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

$1$	$1$	$2$	$- 7$	$- 8$	$12$
	\xa0	$1$	$3$	$- 4$	$- 12$
	$1$	$3$	$- 4$	$- 12$	$0$

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

$1 x^{3} + 3 x^{2} + (- 4) x - 12$

Simplify the quotient polynomial.

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

Step 7

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x - 1) (x^{3} + 3 x^{2}) - 4 x - 12$

Factor out the greatest common factor (GCF) from each group.

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

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

Step 8

Factor the polynomial by factoring out the greatest common factor, $x + 3$ .

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

Step 9

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

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

Step 10

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 2$ .

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

Remove unnecessary parentheses.

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

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

Step 11

Factor the left side of the equation.

Regroup terms.

$x^{4} + 2 x^{3} - 7 x^{2} - 8 x + 12 = 0$

Factor $x^{3}$ out of $x^{4} + 2 x^{3}$ .

Factor $x^{3}$ out of $x^{4}$ .

$x^{3} x + 2 x^{3} - 7 x^{2} - 8 x + 12 = 0$

Factor $x^{3}$ out of $2 x^{3}$ .

$x^{3} x + x^{3} \cdot 2 - 7 x^{2} - 8 x + 12 = 0$

Factor $x^{3}$ out of $x^{3} x + x^{3} \cdot 2$ .

$x^{3} (x + 2) - 7 x^{2} - 8 x + 12 = 0$

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 7 \cdot 12 = - 84$ and whose sum is $b = - 8$ .

Factor $- 8$ out of $- 8 x$ .

$x^{3} (x + 2) - 7 x^{2} - 8 x + 12 = 0$

Rewrite $- 8$ as $6$ plus $- 14$

$x^{3} (x + 2) - 7 x^{2} + (6 - 14) x + 12 = 0$

Apply the distributive property.

$x^{3} (x + 2) - 7 x^{2} + 6 x - 14 x + 12 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$x^{3} (x + 2) - 7 x^{2} + 6 x - 14 x + 12 = 0$

Factor out the greatest common factor (GCF) from each group.

$x^{3} (x + 2) + x (- 7 x + 6) + 2 (- 7 x + 6) = 0$

Factor the polynomial by factoring out the greatest common factor, $- 7 x + 6$ .

$x^{3} (x + 2) + (- 7 x + 6) (x + 2) = 0$

Factor $x + 2$ out of $x^{3} (x + 2) + (- 7 x + 6) (x + 2)$ .

Factor $x + 2$ out of $x^{3} (x + 2)$ .

$(x + 2) x^{3} + (- 7 x + 6) (x + 2) = 0$

Factor $x + 2$ out of $(- 7 x + 6) (x + 2)$ .

$(x + 2) x^{3} + (x + 2) (- 7 x + 6) = 0$

Factor $x + 2$ out of $(x + 2) x^{3} + (x + 2) (- 7 x + 6)$ .

$(x + 2) (x^{3} - 7 x + 6) = 0$

Factor.

Rewrite $x^{3} - 7 x + 6$ in a factored form.

Factor $x^{3} - 7 x + 6$ 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 $1$ and simplify the expression. In this case, the expression is equal to $0$ so $1$ is a root of the polynomial.

Substitute $1$ into the polynomial.

$1^{3} - 7 \cdot 1 + 6$

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

$1 - 7 \cdot 1 + 6$

Multiply $- 7$ by $1$ .

$1 - 7 + 6$

Subtract $7$ from $1$ .

$- 6 + 6$

Add $- 6$ and $6$ .

$0$

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.

$\frac{x^{3} - 7 x + 6}{x - 1}$

Divide $x^{3} - 7 x + 6$ by $x - 1$ .

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

$x$

$1$

$x^{3}$

$0 x^{2}$

$7 x$

$6$

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

					$x^{2}$
	$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$7 x$	+	$6$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$7 x$	+	$6$
			+	$x^{3}$	-	$x^{2}$

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

				$x^{2}$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$7 x$	+	$6$
			-	$x^{3}$	+	$x^{2}$

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

				$x^{2}$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$7 x$	+	$6$
			-	$x^{3}$	+	$x^{2}$
					+	$x^{2}$

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

				$x^{2}$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$7 x$	+	$6$
			-	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$7 x$

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

				$x^{2}$	+	$x$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$7 x$	+	$6$
			-	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$7 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$x$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$7 x$	+	$6$
			-	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$7 x$
					+	$x^{2}$	-	$x$

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

				$x^{2}$	+	$x$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$7 x$	+	$6$
			-	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$7 x$
					-	$x^{2}$	+	$x$

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

				$x^{2}$	+	$x$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$7 x$	+	$6$
			-	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$7 x$
					-	$x^{2}$	+	$x$
							-	$6 x$

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

				$x^{2}$	+	$x$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$7 x$	+	$6$
			-	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$7 x$
					-	$x^{2}$	+	$x$
							-	$6 x$	+	$6$

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

				$x^{2}$	+	$x$	-	$6$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$7 x$	+	$6$
			-	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$7 x$
					-	$x^{2}$	+	$x$
							-	$6 x$	+	$6$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$x$	-	$6$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$7 x$	+	$6$
			-	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$7 x$
					-	$x^{2}$	+	$x$
							-	$6 x$	+	$6$
							-	$6 x$	+	$6$

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

				$x^{2}$	+	$x$	-	$6$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$7 x$	+	$6$
			-	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$7 x$
					-	$x^{2}$	+	$x$
							-	$6 x$	+	$6$
							+	$6 x$	-	$6$

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

				$x^{2}$	+	$x$	-	$6$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$7 x$	+	$6$
			-	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$7 x$
					-	$x^{2}$	+	$x$
							-	$6 x$	+	$6$
							+	$6 x$	-	$6$
										$0$

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

$x^{2} + x - 6$

Write $x^{3} - 7 x + 6$ as a set of factors.

$(x + 2) ((x - 1) (x^{2} + x - 6)) = 0$

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 $1$ .

$- 2, 3$

Write the factored form using these integers.

$(x + 2) ((x - 1) ((x - 2) (x + 3))) = 0$

Remove unnecessary parentheses.

$(x + 2) ((x - 1) (x - 2) (x + 3)) = 0$

Remove unnecessary parentheses.

$(x + 2) (x - 1) (x - 2) (x + 3) = 0$

Step 12

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

$x + 2 = 0$

$x - 1 = 0$

$x - 2 = 0$

$x + 3 = 0$

Step 13

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.

$x = - 2$

Step 14

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Step 15

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

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Step 16

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.

$x = - 3$

Step 17

The final solution is all the values that make $(x + 2) (x - 1) (x - 2) (x + 3) = 0$ true.

$x = - 2, 1, 2, - 3$

Step 18

Find the Roots/Zeros Using the Rational Roots Test x^4+2x^3-7x^2-8x+12

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