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

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

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.

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

$q = \pm 1, \pm 3$

Step 2

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

$\pm 1, \pm \frac{1}{3}, \pm 2, \pm \frac{2}{3}, \pm 3, \pm 4, \pm \frac{4}{3}, \pm 6, \pm 12$

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.

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

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.

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

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

Multiply $3$ by $- 1$ .

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

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

$- 3 + 8 \cdot 1 - 7 \cdot - 1 - 12$

Multiply $8$ by $1$ .

$- 3 + 8 - 7 \cdot - 1 - 12$

Multiply $- 7$ by $- 1$ .

$- 3 + 8 + 7 - 12$

Simplify by adding and subtracting.

Add $- 3$ and $8$ .

$5 + 7 - 12$

Add $5$ and $7$ .

$12 - 12$

Subtract $12$ from $12$ .

$0$

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.

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

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.

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

The first number in the dividend $(3)$ is put into the first position of the result area (below the horizontal line).

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

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 $(8)$ .

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

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

$- 1$	$3$	$8$	$- 7$	$- 12$
	\xa0	$- 3$		\xa0
	$3$	$5$

Multiply the newest entry in the result $(5)$ by the divisor $(- 1)$ and place the result of $(- 5)$ under the next term in the dividend $(- 7)$ .

$- 1$	$3$	$8$	$- 7$	$- 12$
	\xa0	$- 3$	$- 5$	\xa0
	$3$	$5$

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

$- 1$	$3$	$8$	$- 7$	$- 12$
	\xa0	$- 3$	$- 5$	\xa0
	$3$	$5$	$- 12$

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 $(- 12)$ .

$- 1$	$3$	$8$	$- 7$	$- 12$
	\xa0	$- 3$	$- 5$	$12$
	$3$	$5$	$- 12$

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

$- 1$	$3$	$8$	$- 7$	$- 12$
	\xa0	$- 3$	$- 5$	$12$
	$3$	$5$	$- 12$	$0$

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

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

Step 7

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 $a \cdot c = 3 \cdot - 12 = - 36$ and whose sum is $b = 5$ .

Factor $5$ out of $5 x$ .

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

Rewrite $5$ as $- 4$ plus $9$

$(x + 1) + 3 x^{2} + (- 4 + 9) x - 12$

Apply the distributive property.

$(x + 1) + 3 x^{2} - 4 x + 9 x - 12$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x + 1) (3 x^{2} - 4 x) + 9 x - 12$

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

$(x + 1) + x (3 x - 4) + 3 (3 x - 4)$

Factor the polynomial by factoring out the greatest common factor, $3 x - 4$ .

$(x + 1) + (3 x - 4) (x + 3)$

$(x + 1) (3 x - 4) (x + 3)$

Step 8

Factor the left side of the equation.

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

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

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

$\pm 1, \pm 0.\overline{3}, \pm 12, \pm 4, \pm 2, \pm 0.\overline{6}, \pm 6, \pm 3, \pm 1.\overline{3}$

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.

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

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

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

Multiply $3$ by $- 1$ .

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

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

$- 3 + 8 \cdot 1 - 7 \cdot - 1 - 12$

Multiply $8$ by $1$ .

$- 3 + 8 - 7 \cdot - 1 - 12$

Add $- 3$ and $8$ .

$5 - 7 \cdot - 1 - 12$

Multiply $- 7$ by $- 1$ .

$5 + 7 - 12$

Add $5$ and $7$ .

$12 - 12$

Subtract $12$ from $12$ .

$0$

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.

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

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

$x$

$1$

$3 x^{3}$

$8 x^{2}$

$7 x$

$12$

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

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

Multiply the new quotient term by the divisor.

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

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

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

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

				$3 x^{2}$
$x$	+	$1$		$3 x^{3}$	+	$8 x^{2}$	-	$7 x$	-	$12$
			-	$3 x^{3}$	-	$3 x^{2}$
					+	$5 x^{2}$

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

				$3 x^{2}$
$x$	+	$1$		$3 x^{3}$	+	$8 x^{2}$	-	$7 x$	-	$12$
			-	$3 x^{3}$	-	$3 x^{2}$
					+	$5 x^{2}$	-	$7 x$

Divide the highest order term in the dividend $5 x^{2}$ by the highest order term in divisor $x$ .

				$3 x^{2}$	+	$5 x$
$x$	+	$1$		$3 x^{3}$	+	$8 x^{2}$	-	$7 x$	-	$12$
			-	$3 x^{3}$	-	$3 x^{2}$
					+	$5 x^{2}$	-	$7 x$

Multiply the new quotient term by the divisor.

				$3 x^{2}$	+	$5 x$
$x$	+	$1$		$3 x^{3}$	+	$8 x^{2}$	-	$7 x$	-	$12$
			-	$3 x^{3}$	-	$3 x^{2}$
					+	$5 x^{2}$	-	$7 x$
					+	$5 x^{2}$	+	$5 x$

The expression needs to be subtracted from the dividend, so change all the signs in $5 x^{2} + 5 x$

				$3 x^{2}$	+	$5 x$
$x$	+	$1$		$3 x^{3}$	+	$8 x^{2}$	-	$7 x$	-	$12$
			-	$3 x^{3}$	-	$3 x^{2}$
					+	$5 x^{2}$	-	$7 x$
					-	$5 x^{2}$	-	$5 x$

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

				$3 x^{2}$	+	$5 x$
$x$	+	$1$		$3 x^{3}$	+	$8 x^{2}$	-	$7 x$	-	$12$
			-	$3 x^{3}$	-	$3 x^{2}$
					+	$5 x^{2}$	-	$7 x$
					-	$5 x^{2}$	-	$5 x$
							-	$12 x$

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

				$3 x^{2}$	+	$5 x$
$x$	+	$1$		$3 x^{3}$	+	$8 x^{2}$	-	$7 x$	-	$12$
			-	$3 x^{3}$	-	$3 x^{2}$
					+	$5 x^{2}$	-	$7 x$
					-	$5 x^{2}$	-	$5 x$
							-	$12 x$	-	$12$

Divide the highest order term in the dividend $- 12 x$ by the highest order term in divisor $x$ .

				$3 x^{2}$	+	$5 x$	-	$12$
$x$	+	$1$		$3 x^{3}$	+	$8 x^{2}$	-	$7 x$	-	$12$
			-	$3 x^{3}$	-	$3 x^{2}$
					+	$5 x^{2}$	-	$7 x$
					-	$5 x^{2}$	-	$5 x$
							-	$12 x$	-	$12$

Multiply the new quotient term by the divisor.

				$3 x^{2}$	+	$5 x$	-	$12$
$x$	+	$1$		$3 x^{3}$	+	$8 x^{2}$	-	$7 x$	-	$12$
			-	$3 x^{3}$	-	$3 x^{2}$
					+	$5 x^{2}$	-	$7 x$
					-	$5 x^{2}$	-	$5 x$
							-	$12 x$	-	$12$
							-	$12 x$	-	$12$

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

				$3 x^{2}$	+	$5 x$	-	$12$
$x$	+	$1$		$3 x^{3}$	+	$8 x^{2}$	-	$7 x$	-	$12$
			-	$3 x^{3}$	-	$3 x^{2}$
					+	$5 x^{2}$	-	$7 x$
					-	$5 x^{2}$	-	$5 x$
							-	$12 x$	-	$12$
							+	$12 x$	+	$12$

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

				$3 x^{2}$	+	$5 x$	-	$12$
$x$	+	$1$		$3 x^{3}$	+	$8 x^{2}$	-	$7 x$	-	$12$
			-	$3 x^{3}$	-	$3 x^{2}$
					+	$5 x^{2}$	-	$7 x$
					-	$5 x^{2}$	-	$5 x$
							-	$12 x$	-	$12$
							+	$12 x$	+	$12$
										$0$

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

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

Write $3 x^{3} + 8 x^{2} - 7 x - 12$ as a set of factors.

$(x + 1) (3 x^{2} + 5 x - 12) = 0$

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 $a \cdot c = 3 \cdot - 12 = - 36$ and whose sum is $b = 5$ .

Factor $5$ out of $5 x$ .

$(x + 1) (3 x^{2} + 5 (x) - 12) = 0$

Rewrite $5$ as $- 4$ plus $9$

$(x + 1) (3 x^{2} + (- 4 + 9) x - 12) = 0$

Apply the distributive property.

$(x + 1) (3 x^{2} - 4 x + 9 x - 12) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x + 1) ((3 x^{2} - 4 x) + 9 x - 12) = 0$

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

$(x + 1) (x (3 x - 4) + 3 (3 x - 4)) = 0$

Factor the polynomial by factoring out the greatest common factor, $3 x - 4$ .

$(x + 1) ((3 x - 4) (x + 3)) = 0$

Remove unnecessary parentheses.

$(x + 1) (3 x - 4) (x + 3) = 0$

Step 9

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

$x + 1 = 0$

$3 x - 4 = 0$

$x + 3 = 0$

Step 10

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

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 11

Set $3 x - 4$ equal to $0$ and solve for $x$ .

Set $3 x - 4$ equal to $0$ .

$3 x - 4 = 0$

Solve $3 x - 4 = 0$ for $x$ .

Add $4$ to both sides of the equation.

$3 x = 4$

Divide each term in $3 x = 4$ by $3$ and simplify.

Divide each term in $3 x = 4$ by $3$ .

$\frac{3 x}{3} = \frac{4}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{4}{3}$

Divide $x$ by $1$ .

$x = \frac{4}{3}$

Step 12

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 13

The final solution is all the values that make $(x + 1) (3 x - 4) (x + 3) = 0$ true.

$x = - 1, \frac{4}{3}, - 3$

Step 14

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

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