Find the Roots/Zeros Using the Rational Roots Test f(x)=6x^3-13x^2-41x-12

$f (x) = 6 x^{3} - 13 x^{2} - 41 x - 12$

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.

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

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

Step 2

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

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

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.

$6 {(- \frac{3}{2})}^{3} - 13 {(- \frac{3}{2})}^{2} - 41 (- \frac{3}{2}) - 12$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = - \frac{3}{2}$ is a root of the polynomial.

Simplify each term.

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $- \frac{3}{2}$ .

$6 ({(- 1)}^{3} {(\frac{3}{2})}^{3}) - 13 {(- \frac{3}{2})}^{2} - 41 (- \frac{3}{2}) - 12$

Apply the product rule to $\frac{3}{2}$ .

$6 ({(- 1)}^{3} \frac{3^{3}}{2^{3}}) - 13 {(- \frac{3}{2})}^{2} - 41 (- \frac{3}{2}) - 12$

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

$6 (- \frac{3^{3}}{2^{3}}) - 13 {(- \frac{3}{2})}^{2} - 41 (- \frac{3}{2}) - 12$

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

$6 (- \frac{27}{2^{3}}) - 13 {(- \frac{3}{2})}^{2} - 41 (- \frac{3}{2}) - 12$

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

$6 (- \frac{27}{8}) - 13 {(- \frac{3}{2})}^{2} - 41 (- \frac{3}{2}) - 12$

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{27}{8}$ into the numerator.

$6 (\frac{- 27}{8}) - 13 {(- \frac{3}{2})}^{2} - 41 (- \frac{3}{2}) - 12$

Factor $2$ out of $6$ .

$2 (3) \frac{- 27}{8} - 13 {(- \frac{3}{2})}^{2} - 41 (- \frac{3}{2}) - 12$

Factor $2$ out of $8$ .

$2 \cdot 3 \frac{- 27}{2 \cdot 4} - 13 {(- \frac{3}{2})}^{2} - 41 (- \frac{3}{2}) - 12$

Cancel the common factor.

$2 \cdot 3 \frac{- 27}{2 \cdot 4} - 13 {(- \frac{3}{2})}^{2} - 41 (- \frac{3}{2}) - 12$

Rewrite the expression.

$3 (\frac{- 27}{4}) - 13 {(- \frac{3}{2})}^{2} - 41 (- \frac{3}{2}) - 12$

Combine $3$ and $\frac{- 27}{4}$ .

$\frac{3 \cdot - 27}{4} - 13 {(- \frac{3}{2})}^{2} - 41 (- \frac{3}{2}) - 12$

Multiply $3$ by $- 27$ .

$\frac{- 81}{4} - 13 {(- \frac{3}{2})}^{2} - 41 (- \frac{3}{2}) - 12$

Move the negative in front of the fraction.

$- \frac{81}{4} - 13 {(- \frac{3}{2})}^{2} - 41 (- \frac{3}{2}) - 12$

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $- \frac{3}{2}$ .

$- \frac{81}{4} - 13 ({(- 1)}^{2} {(\frac{3}{2})}^{2}) - 41 (- \frac{3}{2}) - 12$

Apply the product rule to $\frac{3}{2}$ .

$- \frac{81}{4} - 13 ({(- 1)}^{2} \frac{3^{2}}{2^{2}}) - 41 (- \frac{3}{2}) - 12$

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

$- \frac{81}{4} - 13 (1 \frac{3^{2}}{2^{2}}) - 41 (- \frac{3}{2}) - 12$

Multiply $\frac{3^{2}}{2^{2}}$ by $1$ .

$- \frac{81}{4} - 13 \frac{3^{2}}{2^{2}} - 41 (- \frac{3}{2}) - 12$

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

$- \frac{81}{4} - 13 \frac{9}{2^{2}} - 41 (- \frac{3}{2}) - 12$

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

$- \frac{81}{4} - 13 (\frac{9}{4}) - 41 (- \frac{3}{2}) - 12$

Multiply $- 13 (\frac{9}{4})$ .

Combine $- 13$ and $\frac{9}{4}$ .

$- \frac{81}{4} + \frac{- 13 \cdot 9}{4} - 41 (- \frac{3}{2}) - 12$

Multiply $- 13$ by $9$ .

$- \frac{81}{4} + \frac{- 117}{4} - 41 (- \frac{3}{2}) - 12$

Move the negative in front of the fraction.

$- \frac{81}{4} - \frac{117}{4} - 41 (- \frac{3}{2}) - 12$

Multiply $- 41 (- \frac{3}{2})$ .

Multiply $- 1$ by $- 41$ .

$- \frac{81}{4} - \frac{117}{4} + 41 (\frac{3}{2}) - 12$

Combine $41$ and $\frac{3}{2}$ .

$- \frac{81}{4} - \frac{117}{4} + \frac{41 \cdot 3}{2} - 12$

Multiply $41$ by $3$ .

$- \frac{81}{4} - \frac{117}{4} + \frac{123}{2} - 12$

Combine fractions.

Combine the numerators over the common denominator.

$- 12 + \frac{- 81 - 117}{4} + \frac{123}{2}$

Subtract $117$ from $- 81$ .

$- 12 + \frac{- 198}{4} + \frac{123}{2}$

Find the common denominator.

Write $- 12$ as a fraction with denominator $1$ .

$\frac{- 12}{1} + \frac{- 198}{4} + \frac{123}{2}$

Multiply $\frac{- 12}{1}$ by $\frac{4}{4}$ .

$\frac{- 12}{1} \cdot \frac{4}{4} + \frac{- 198}{4} + \frac{123}{2}$

Multiply $\frac{- 12}{1}$ by $\frac{4}{4}$ .

$\frac{- 12 \cdot 4}{4} + \frac{- 198}{4} + \frac{123}{2}$

Multiply $\frac{123}{2}$ by $\frac{2}{2}$ .

$\frac{- 12 \cdot 4}{4} + \frac{- 198}{4} + \frac{123}{2} \cdot \frac{2}{2}$

Multiply $\frac{123}{2}$ by $\frac{2}{2}$ .

$\frac{- 12 \cdot 4}{4} + \frac{- 198}{4} + \frac{123 \cdot 2}{2 \cdot 2}$

Multiply $2$ by $2$ .

$\frac{- 12 \cdot 4}{4} + \frac{- 198}{4} + \frac{123 \cdot 2}{4}$

Combine the numerators over the common denominator.

$\frac{- 12 \cdot 4 - 198 + 123 \cdot 2}{4}$

Simplify each term.

Multiply $- 12$ by $4$ .

$\frac{- 48 - 198 + 123 \cdot 2}{4}$

Multiply $123$ by $2$ .

$\frac{- 48 - 198 + 246}{4}$

Simplify the expression.

Subtract $198$ from $- 48$ .

$\frac{- 246 + 246}{4}$

Add $- 246$ and $246$ .

$\frac{0}{4}$

Divide $0$ by $4$ .

$0$

Step 5

Since $- \frac{3}{2}$ is a known root, divide the polynomial by $x + \frac{3}{2}$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{6 x^{3} - 13 x^{2} - 41 x - 12}{x + \frac{3}{2}}$

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.

$- \frac{3}{2}$	$6$	$- 13$	$- 41$	$- 12$
	\xa0			\xa0

The first number in the dividend $(6)$ is put into the first position of the result area (below the horizontal line).

$- \frac{3}{2}$	$6$	$- 13$	$- 41$	$- 12$
	\xa0			\xa0
	$6$

Multiply the newest entry in the result $(6)$ by the divisor $(- \frac{3}{2})$ and place the result of $(- 9)$ under the next term in the dividend $(- 13)$ .

$- \frac{3}{2}$	$6$	$- 13$	$- 41$	$- 12$
	\xa0	$- 9$		\xa0
	$6$

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

$- \frac{3}{2}$	$6$	$- 13$	$- 41$	$- 12$
	\xa0	$- 9$		\xa0
	$6$	$- 22$

Multiply the newest entry in the result $(- 22)$ by the divisor $(- \frac{3}{2})$ and place the result of $(33)$ under the next term in the dividend $(- 41)$ .

$- \frac{3}{2}$	$6$	$- 13$	$- 41$	$- 12$
	\xa0	$- 9$	$33$	\xa0
	$6$	$- 22$

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

$- \frac{3}{2}$	$6$	$- 13$	$- 41$	$- 12$
	\xa0	$- 9$	$33$	\xa0
	$6$	$- 22$	$- 8$

Multiply the newest entry in the result $(- 8)$ by the divisor $(- \frac{3}{2})$ and place the result of $(12)$ under the next term in the dividend $(- 12)$ .

$- \frac{3}{2}$	$6$	$- 13$	$- 41$	$- 12$
	\xa0	$- 9$	$33$	$12$
	$6$	$- 22$	$- 8$

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

$- \frac{3}{2}$	$6$	$- 13$	$- 41$	$- 12$
	\xa0	$- 9$	$33$	$12$
	$6$	$- 22$	$- 8$	$0$

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

$6 x^{2} + - 22 x - 8$

Simplify the quotient polynomial.

$6 x^{2} - 22 x - 8$

Step 7

Factor $2$ out of $6 x^{2} - 22 x - 8$ .

Factor $2$ out of $6 x^{2}$ .

$(x + \frac{3}{2}) (2 (3 x^{2}) - 22 x - 8)$

Factor $2$ out of $- 22 x$ .

$(x + \frac{3}{2}) (2 (3 x^{2}) + 2 (- 11 x) - 8)$

Factor $2$ out of $- 8$ .

$(x + \frac{3}{2}) (2 (3 x^{2}) + 2 (- 11 x) + 2 (- 4))$

Factor $2$ out of $2 (3 x^{2}) + 2 (- 11 x)$ .

$(x + \frac{3}{2}) (2 (3 x^{2} - 11 x) + 2 (- 4))$

Factor $2$ out of $2 (3 x^{2} - 11 x) + 2 (- 4)$ .

$(x + \frac{3}{2}) (2 (3 x^{2} - 11 x - 4))$

Step 8

Factor.

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 $a \cdot c = 3 \cdot - 4 = - 12$ and whose sum is $b = - 11$ .

Factor $- 11$ out of $- 11 x$ .

$(x + \frac{3}{2}) (2 (3 x^{2} - 11 x - 4))$

Rewrite $- 11$ as $1$ plus $- 12$

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

Apply the distributive property.

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

Multiply $x$ by $1$ .

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

Factor the polynomial by factoring out the greatest common factor, $3 x + 1$ .

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

Remove unnecessary parentheses.

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

Step 9

Factor the left side of the equation.

Factor $6 x^{3} - 13 x^{2} - 41 x - 12$ 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.

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

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

$\pm 1, \pm 0.1\overline{6}, \pm 0.5, \pm 0.\overline{3}, \pm 12, \pm 2, \pm 6, \pm 4, \pm 0.\overline{6}, \pm 3, \pm 1.5, \pm 1.\overline{3}$

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

Substitute $- 0.\overline{3}$ into the polynomial.

$6 {(- 0.\overline{3})}^{3} - 13 {(- 0.\overline{3})}^{2} - 41 \cdot - 0.\overline{3} - 12$

Raise $- 0.\overline{3}$ to the power of $3$ .

$6 \cdot - 0.\overline{037} - 13 {(- 0.\overline{3})}^{2} - 41 \cdot - 0.\overline{3} - 12$

Multiply $6$ by $- 0.\overline{037}$ .

$- 0.\overline{2} - 13 {(- 0.\overline{3})}^{2} - 41 \cdot - 0.\overline{3} - 12$

Raise $- 0.\overline{3}$ to the power of $2$ .

$- 0.\overline{2} - 13 \cdot 0.\overline{1} - 41 \cdot - 0.\overline{3} - 12$

Multiply $- 13$ by $0.\overline{1}$ .

$- 0.\overline{2} - 1.\overline{4} - 41 \cdot - 0.\overline{3} - 12$

Subtract $1.\overline{4}$ from $- 0.\overline{2}$ .

$- 1.\overline{6} - 41 \cdot - 0.\overline{3} - 12$

Multiply $- 41$ by $- 0.\overline{3}$ .

$- 1.\overline{6} + 13.\overline{6} - 12$

Add $- 1.\overline{6}$ and $13.\overline{6}$ .

$12 - 12$

Subtract $12$ from $12$ .

$0$

Since $- 0.\overline{3}$ is a known root, divide the polynomial by $3 x + 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{6 x^{3} - 13 x^{2} - 41 x - 12}{3 x + 1}$

Divide $6 x^{3} - 13 x^{2} - 41 x - 12$ by $3 x + 1$ .

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

$3 x$

$1$

$6 x^{3}$

$13 x^{2}$

$41 x$

$12$

Divide the highest order term in the dividend $6 x^{3}$ by the highest order term in divisor $3 x$ .

					$2 x^{2}$
	$3 x$	+	$1$		$6 x^{3}$	-	$13 x^{2}$	-	$41 x$	-	$12$

Multiply the new quotient term by the divisor.

				$2 x^{2}$
$3 x$	+	$1$		$6 x^{3}$	-	$13 x^{2}$	-	$41 x$	-	$12$
			+	$6 x^{3}$	+	$2 x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $6 x^{3} + 2 x^{2}$

				$2 x^{2}$
$3 x$	+	$1$		$6 x^{3}$	-	$13 x^{2}$	-	$41 x$	-	$12$
			-	$6 x^{3}$	-	$2 x^{2}$

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

				$2 x^{2}$
$3 x$	+	$1$		$6 x^{3}$	-	$13 x^{2}$	-	$41 x$	-	$12$
			-	$6 x^{3}$	-	$2 x^{2}$
					-	$15 x^{2}$

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

				$2 x^{2}$
$3 x$	+	$1$		$6 x^{3}$	-	$13 x^{2}$	-	$41 x$	-	$12$
			-	$6 x^{3}$	-	$2 x^{2}$
					-	$15 x^{2}$	-	$41 x$

Divide the highest order term in the dividend $- 15 x^{2}$ by the highest order term in divisor $3 x$ .

				$2 x^{2}$	-	$5 x$
$3 x$	+	$1$		$6 x^{3}$	-	$13 x^{2}$	-	$41 x$	-	$12$
			-	$6 x^{3}$	-	$2 x^{2}$
					-	$15 x^{2}$	-	$41 x$

Multiply the new quotient term by the divisor.

				$2 x^{2}$	-	$5 x$
$3 x$	+	$1$		$6 x^{3}$	-	$13 x^{2}$	-	$41 x$	-	$12$
			-	$6 x^{3}$	-	$2 x^{2}$
					-	$15 x^{2}$	-	$41 x$
					-	$15 x^{2}$	-	$5 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 15 x^{2} - 5 x$

				$2 x^{2}$	-	$5 x$
$3 x$	+	$1$		$6 x^{3}$	-	$13 x^{2}$	-	$41 x$	-	$12$
			-	$6 x^{3}$	-	$2 x^{2}$
					-	$15 x^{2}$	-	$41 x$
					+	$15 x^{2}$	+	$5 x$

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

				$2 x^{2}$	-	$5 x$
$3 x$	+	$1$		$6 x^{3}$	-	$13 x^{2}$	-	$41 x$	-	$12$
			-	$6 x^{3}$	-	$2 x^{2}$
					-	$15 x^{2}$	-	$41 x$
					+	$15 x^{2}$	+	$5 x$
							-	$36 x$

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

				$2 x^{2}$	-	$5 x$
$3 x$	+	$1$		$6 x^{3}$	-	$13 x^{2}$	-	$41 x$	-	$12$
			-	$6 x^{3}$	-	$2 x^{2}$
					-	$15 x^{2}$	-	$41 x$
					+	$15 x^{2}$	+	$5 x$
							-	$36 x$	-	$12$

Divide the highest order term in the dividend $- 36 x$ by the highest order term in divisor $3 x$ .

				$2 x^{2}$	-	$5 x$	-	$12$
$3 x$	+	$1$		$6 x^{3}$	-	$13 x^{2}$	-	$41 x$	-	$12$
			-	$6 x^{3}$	-	$2 x^{2}$
					-	$15 x^{2}$	-	$41 x$
					+	$15 x^{2}$	+	$5 x$
							-	$36 x$	-	$12$

Multiply the new quotient term by the divisor.

				$2 x^{2}$	-	$5 x$	-	$12$
$3 x$	+	$1$		$6 x^{3}$	-	$13 x^{2}$	-	$41 x$	-	$12$
			-	$6 x^{3}$	-	$2 x^{2}$
					-	$15 x^{2}$	-	$41 x$
					+	$15 x^{2}$	+	$5 x$
							-	$36 x$	-	$12$
							-	$36 x$	-	$12$

The expression needs to be subtracted from the dividend, so change all the signs in $- 36 x - 12$

				$2 x^{2}$	-	$5 x$	-	$12$
$3 x$	+	$1$		$6 x^{3}$	-	$13 x^{2}$	-	$41 x$	-	$12$
			-	$6 x^{3}$	-	$2 x^{2}$
					-	$15 x^{2}$	-	$41 x$
					+	$15 x^{2}$	+	$5 x$
							-	$36 x$	-	$12$
							+	$36 x$	+	$12$

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

				$2 x^{2}$	-	$5 x$	-	$12$
$3 x$	+	$1$		$6 x^{3}$	-	$13 x^{2}$	-	$41 x$	-	$12$
			-	$6 x^{3}$	-	$2 x^{2}$
					-	$15 x^{2}$	-	$41 x$
					+	$15 x^{2}$	+	$5 x$
							-	$36 x$	-	$12$
							+	$36 x$	+	$12$
										$0$

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

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

Write $6 x^{3} - 13 x^{2} - 41 x - 12$ as a set of factors.

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

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 $a \cdot c = 2 \cdot - 12 = - 24$ and whose sum is $b = - 5$ .

Factor $- 5$ out of $- 5 x$ .

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

Rewrite $- 5$ as $3$ plus $- 8$

$(3 x + 1) (2 x^{2} + (3 - 8) x - 12) = 0$

Apply the distributive property.

$(3 x + 1) (2 x^{2} + 3 x - 8 x - 12) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(3 x + 1) ((2 x^{2} + 3 x) - 8 x - 12) = 0$

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

$(3 x + 1) (x (2 x + 3) - 4 (2 x + 3)) = 0$

Factor the polynomial by factoring out the greatest common factor, $2 x + 3$ .

$(3 x + 1) ((2 x + 3) (x - 4)) = 0$

Remove unnecessary parentheses.

$(3 x + 1) (2 x + 3) (x - 4) = 0$

Step 10

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

$3 x + 1 = 0$

$2 x + 3 = 0$

$x - 4 = 0$

Step 11

Set $3 x + 1$ equal to $0$ and solve for $x$ .

Set $3 x + 1$ equal to $0$ .

$3 x + 1 = 0$

Solve $3 x + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$3 x = - 1$

Divide each term in $3 x = - 1$ by $3$ and simplify.

Divide each term in $3 x = - 1$ by $3$ .

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Step 12

Set $2 x + 3$ equal to $0$ and solve for $x$ .

Set $2 x + 3$ equal to $0$ .

$2 x + 3 = 0$

Solve $2 x + 3 = 0$ for $x$ .

Subtract $3$ from both sides of the equation.

$2 x = - 3$

Divide each term in $2 x = - 3$ by $2$ and simplify.

Divide each term in $2 x = - 3$ by $2$ .

$\frac{2 x}{2} = \frac{- 3}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 3}{2}$

Divide $x$ by $1$ .

$x = \frac{- 3}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{3}{2}$

Step 13

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

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

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Step 14

The final solution is all the values that make $(3 x + 1) (2 x + 3) (x - 4) = 0$ true.

$x = - \frac{1}{3}, - \frac{3}{2}, 4$

Step 15

Find the Roots/Zeros Using the Rational Roots Test f(x)=6x^3-13x^2-41x-12

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