Find the Roots/Zeros Using the Rational Roots Test f(x)=24x^3-8x^2-34x-12

$f (x) = 24 x^{3} - 8 x^{2} - 34 x - 12$

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.

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

$q = \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24$

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

$\pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{4}, \pm \frac{1}{6}, \pm \frac{1}{8}, \pm \frac{1}{12}, \pm \frac{1}{24}, \pm 2, \pm \frac{2}{3}, \pm 3, \pm \frac{3}{2}, \pm \frac{3}{4}, \pm \frac{3}{8}, \pm 4, \pm \frac{4}{3}, \pm 6, \pm 12$

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.

$24 {(- \frac{1}{2})}^{3} - 8 {(- \frac{1}{2})}^{2} - 34 (- \frac{1}{2}) - 12$

Simplify the expression. In this case, the expression is equal to $0$ so $x = - \frac{1}{2}$ is a root of the polynomial.

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Simplify each term.

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

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Apply the product rule to $- \frac{1}{2}$ .

$24 ({(- 1)}^{3} {(\frac{1}{2})}^{3}) - 8 {(- \frac{1}{2})}^{2} - 34 (- \frac{1}{2}) - 12$

Apply the product rule to $\frac{1}{2}$ .

$24 ({(- 1)}^{3} \frac{1^{3}}{2^{3}}) - 8 {(- \frac{1}{2})}^{2} - 34 (- \frac{1}{2}) - 12$

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

$24 (- \frac{1^{3}}{2^{3}}) - 8 {(- \frac{1}{2})}^{2} - 34 (- \frac{1}{2}) - 12$

One to any power is one.

$24 (- \frac{1}{2^{3}}) - 8 {(- \frac{1}{2})}^{2} - 34 (- \frac{1}{2}) - 12$

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

$24 (- \frac{1}{8}) - 8 {(- \frac{1}{2})}^{2} - 34 (- \frac{1}{2}) - 12$

Cancel the common factor of $8$ .

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Move the leading negative in $- \frac{1}{8}$ into the numerator.

$24 (\frac{- 1}{8}) - 8 {(- \frac{1}{2})}^{2} - 34 (- \frac{1}{2}) - 12$

Factor $8$ out of $24$ .

$8 (3) \frac{- 1}{8} - 8 {(- \frac{1}{2})}^{2} - 34 (- \frac{1}{2}) - 12$

Cancel the common factor.

$8 \cdot 3 \frac{- 1}{8} - 8 {(- \frac{1}{2})}^{2} - 34 (- \frac{1}{2}) - 12$

Rewrite the expression.

$3 \cdot - 1 - 8 {(- \frac{1}{2})}^{2} - 34 (- \frac{1}{2}) - 12$

Multiply $3$ by $- 1$ .

$- 3 - 8 {(- \frac{1}{2})}^{2} - 34 (- \frac{1}{2}) - 12$

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

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Apply the product rule to $- \frac{1}{2}$ .

$- 3 - 8 ({(- 1)}^{2} {(\frac{1}{2})}^{2}) - 34 (- \frac{1}{2}) - 12$

Apply the product rule to $\frac{1}{2}$ .

$- 3 - 8 ({(- 1)}^{2} \frac{1^{2}}{2^{2}}) - 34 (- \frac{1}{2}) - 12$

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

$- 3 - 8 (1 \frac{1^{2}}{2^{2}}) - 34 (- \frac{1}{2}) - 12$

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

$- 3 - 8 \frac{1^{2}}{2^{2}} - 34 (- \frac{1}{2}) - 12$

One to any power is one.

$- 3 - 8 \frac{1}{2^{2}} - 34 (- \frac{1}{2}) - 12$

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

$- 3 - 8 (\frac{1}{4}) - 34 (- \frac{1}{2}) - 12$

Cancel the common factor of $4$ .

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Factor $4$ out of $- 8$ .

$- 3 + 4 (- 2) \frac{1}{4} - 34 (- \frac{1}{2}) - 12$

Cancel the common factor.

$- 3 + 4 \cdot - 2 \frac{1}{4} - 34 (- \frac{1}{2}) - 12$

Rewrite the expression.

$- 3 - 2 - 34 (- \frac{1}{2}) - 12$

Cancel the common factor of $2$ .

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Move the leading negative in $- \frac{1}{2}$ into the numerator.

$- 3 - 2 - 34 (\frac{- 1}{2}) - 12$

Factor $2$ out of $- 34$ .

$- 3 - 2 + 2 (- 17) \frac{- 1}{2} - 12$

Cancel the common factor.

$- 3 - 2 + 2 \cdot - 17 \frac{- 1}{2} - 12$

Rewrite the expression.

$- 3 - 2 - 17 \cdot - 1 - 12$

Multiply $- 17$ by $- 1$ .

$- 3 - 2 + 17 - 12$

Simplify by adding and subtracting.

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Subtract $2$ from $- 3$ .

$- 5 + 17 - 12$

Add $- 5$ and $17$ .

$12 - 12$

Subtract $12$ from $12$ .

$0$

Since $- \frac{1}{2}$ is a known root, divide the polynomial by $x + \frac{1}{2}$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{24 x^{3} - 8 x^{2} - 34 x - 12}{x + \frac{1}{2}}$

Next, find the roots of the remaining polynomial. The order of the polynomial has been reduced by $1$ .

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Place the numbers representing the divisor and the dividend into a division-like configuration.

$- \frac{1}{2}$	$24$	$- 8$	$- 34$	$- 12$

The first number in the dividend $(24)$ is put into the first position of the result area (below the horizontal line).

$- \frac{1}{2}$	$24$	$- 8$	$- 34$	$- 12$

	$24$

Multiply the newest entry in the result $(24)$ by the divisor $(- \frac{1}{2})$ and place the result of $(- 12)$ under the next term in the dividend $(- 8)$ .

$- \frac{1}{2}$	$24$	$- 8$	$- 34$	$- 12$
		$- 12$
	$24$

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

$- \frac{1}{2}$	$24$	$- 8$	$- 34$	$- 12$
		$- 12$
	$24$	$- 20$

Multiply the newest entry in the result $(- 20)$ by the divisor $(- \frac{1}{2})$ and place the result of $(10)$ under the next term in the dividend $(- 34)$ .

$- \frac{1}{2}$	$24$	$- 8$	$- 34$	$- 12$
		$- 12$	$10$
	$24$	$- 20$

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

$- \frac{1}{2}$	$24$	$- 8$	$- 34$	$- 12$
		$- 12$	$10$
	$24$	$- 20$	$- 24$

Multiply the newest entry in the result $(- 24)$ by the divisor $(- \frac{1}{2})$ and place the result of $(12)$ under the next term in the dividend $(- 12)$ .

$- \frac{1}{2}$	$24$	$- 8$	$- 34$	$- 12$
		$- 12$	$10$	$12$
	$24$	$- 20$	$- 24$

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

$- \frac{1}{2}$	$24$	$- 8$	$- 34$	$- 12$
		$- 12$	$10$	$12$
	$24$	$- 20$	$- 24$	$0$

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

$24 x^{2} + - 20 x - 24$

Simplify the quotient polynomial.

$24 x^{2} - 20 x - 24$

Factor $4$ out of $24 x^{2} - 20 x - 24$ .

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Factor $4$ out of $24 x^{2}$ .

$(x + \frac{1}{2}) (4 (6 x^{2}) - 20 x - 24)$

Factor $4$ out of $- 20 x$ .

$(x + \frac{1}{2}) (4 (6 x^{2}) + 4 (- 5 x) - 24)$

Factor $4$ out of $- 24$ .

$(x + \frac{1}{2}) (4 (6 x^{2}) + 4 (- 5 x) + 4 (- 6))$

Factor $4$ out of $4 (6 x^{2}) + 4 (- 5 x)$ .

$(x + \frac{1}{2}) (4 (6 x^{2} - 5 x) + 4 (- 6))$

Factor $4$ out of $4 (6 x^{2} - 5 x) + 4 (- 6)$ .

$(x + \frac{1}{2}) (4 (6 x^{2} - 5 x - 6))$

Factor.

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Factor by grouping.

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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 $a \cdot c = 6 \cdot - 6 = - 36$ and whose sum is $b = - 5$ .

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

$(x + \frac{1}{2}) (4 (6 x^{2} - 5 x - 6))$

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

$(x + \frac{1}{2}) (4 (6 x^{2} + (4 - 9) x - 6))$

Apply the distributive property.

$(x + \frac{1}{2}) (4 (6 x^{2} + 4 x - 9 x - 6))$

Factor out the greatest common factor from each group.

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Group the first two terms and the last two terms.

$(x + \frac{1}{2}) (4 ((6 x^{2} + 4 x) - 9 x - 6))$

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

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

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

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

Remove unnecessary parentheses.

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

Factor the left side of the equation.

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Factor $2$ out of $24 x^{3} - 8 x^{2} - 34 x - 12$ .

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Factor $2$ out of $24 x^{3}$ .

$2 (12 x^{3}) - 8 x^{2} - 34 x - 12 = 0$

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

$2 (12 x^{3}) + 2 (- 4 x^{2}) - 34 x - 12 = 0$

Factor $2$ out of $- 34 x$ .

$2 (12 x^{3}) + 2 (- 4 x^{2}) + 2 (- 17 x) - 12 = 0$

Factor $2$ out of $- 12$ .

$2 (12 x^{3}) + 2 (- 4 x^{2}) + 2 (- 17 x) + 2 (- 6) = 0$

Factor $2$ out of $2 (12 x^{3}) + 2 (- 4 x^{2})$ .

$2 (12 x^{3} - 4 x^{2}) + 2 (- 17 x) + 2 (- 6) = 0$

Factor $2$ out of $2 (12 x^{3} - 4 x^{2}) + 2 (- 17 x)$ .

$2 (12 x^{3} - 4 x^{2} - 17 x) + 2 (- 6) = 0$

Factor $2$ out of $2 (12 x^{3} - 4 x^{2} - 17 x) + 2 (- 6)$ .

$2 (12 x^{3} - 4 x^{2} - 17 x - 6) = 0$

Factor $12 x^{3} - 4 x^{2} - 17 x - 6$ using the rational roots test.

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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.

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

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

$\pm 1, \pm 0.08\overline{3}, \pm 0.5, \pm 0.1\overline{6}, \pm 0.\overline{3}, \pm 0.25, \pm 6, \pm 3, \pm 2, \pm 1.5, \pm 0.\overline{6}, \pm 0.75$

Substitute $- 0.5$ and simplify the expression. In this case, the expression is equal to $0$ so $- 0.5$ is a root of the polynomial.

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Substitute $- 0.5$ into the polynomial.

$12 {(- 0.5)}^{3} - 4 {(- 0.5)}^{2} - 17 \cdot - 0.5 - 6$

Raise $- 0.5$ to the power of $3$ .

$12 \cdot - 0.125 - 4 {(- 0.5)}^{2} - 17 \cdot - 0.5 - 6$

Multiply $12$ by $- 0.125$ .

$- 1.5 - 4 {(- 0.5)}^{2} - 17 \cdot - 0.5 - 6$

Raise $- 0.5$ to the power of $2$ .

$- 1.5 - 4 \cdot 0.25 - 17 \cdot - 0.5 - 6$

Multiply $- 4$ by $0.25$ .

$- 1.5 - 1 - 17 \cdot - 0.5 - 6$

Subtract $1$ from $- 1.5$ .

$- 2.5 - 17 \cdot - 0.5 - 6$

Multiply $- 17$ by $- 0.5$ .

$- 2.5 + 8.5 - 6$

Add $- 2.5$ and $8.5$ .

$6 - 6$

Subtract $6$ from $6$ .

$0$

Since $- 0.5$ is a known root, divide the polynomial by $2 x + 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{12 x^{3} - 4 x^{2} - 17 x - 6}{2 x + 1}$

Divide $12 x^{3} - 4 x^{2} - 17 x - 6$ by $2 x + 1$ .

$6 x^{2} - 5 x - 6$

Write $12 x^{3} - 4 x^{2} - 17 x - 6$ as a set of factors.

$2 ((2 x + 1) (6 x^{2} - 5 x - 6)) = 0$

Factor.

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Factor by grouping.

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Factor by grouping.

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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 $a \cdot c = 6 \cdot - 6 = - 36$ and whose sum is $b = - 5$ .

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

$2 ((2 x + 1) (6 x^{2} - 5 x - 6)) = 0$

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

$2 ((2 x + 1) (6 x^{2} + (4 - 9) x - 6)) = 0$

Apply the distributive property.

$2 ((2 x + 1) (6 x^{2} + 4 x - 9 x - 6)) = 0$

Factor out the greatest common factor from each group.

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Group the first two terms and the last two terms.

$2 ((2 x + 1) ((6 x^{2} + 4 x) - 9 x - 6)) = 0$

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

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

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

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

Remove unnecessary parentheses.

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

Remove unnecessary parentheses.

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

Divide each term by $2$ and simplify.

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Divide each term in $2 (2 x + 1) (3 x + 2) (2 x - 3) = 0$ by $2$ .

$\frac{2 (2 x + 1) (3 x + 2) (2 x - 3)}{2} = \frac{0}{2}$

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

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Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{2 (2 x + 1) (3 x + 2) (2 x - 3)}{2} = \frac{0}{2}$

Divide $((2 x + 1) (3 x + 2)) (2 x - 3)$ by $1$ .

$((2 x + 1) (3 x + 2)) (2 x - 3) = \frac{0}{2}$

Expand $(2 x + 1) (3 x + 2)$ using the FOIL Method.

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Apply the distributive property.

$(2 x (3 x + 2) + 1 (3 x + 2)) (2 x - 3) = \frac{0}{2}$

Apply the distributive property.

$(2 x (3 x) + 2 x \cdot 2 + 1 (3 x + 2)) (2 x - 3) = \frac{0}{2}$

Apply the distributive property.

$(2 x (3 x) + 2 x \cdot 2 + 1 (3 x) + 1 \cdot 2) (2 x - 3) = \frac{0}{2}$

Simplify and combine like terms.

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Simplify each term.

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Multiply $x$ by $x$ .

$(2 \cdot 3 x^{2} + 2 x \cdot 2 + 1 (3 x) + 1 \cdot 2) (2 x - 3) = \frac{0}{2}$

Multiply $2$ by $3$ .

$(6 x^{2} + 2 x \cdot 2 + 1 (3 x) + 1 \cdot 2) (2 x - 3) = \frac{0}{2}$

Multiply $2$ by $2$ .

$(6 x^{2} + 4 x + 1 (3 x) + 1 \cdot 2) (2 x - 3) = \frac{0}{2}$

Multiply $3 x$ by $1$ .

$(6 x^{2} + 4 x + 3 x + 1 \cdot 2) (2 x - 3) = \frac{0}{2}$

Multiply $2$ by $1$ .

$(6 x^{2} + 4 x + 3 x + 2) (2 x - 3) = \frac{0}{2}$

Add $4 x$ and $3 x$ .

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

Expand $(6 x^{2} + 7 x + 2) (2 x - 3)$ by multiplying each term in the first expression by each term in the second expression.

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

Simplify terms.

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Simplify each term.

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Rewrite using the commutative property of multiplication.

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

Multiply $x^{2}$ by $x$ by adding the exponents.

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Multiply $x^{2}$ by $x$ .

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

$6 \cdot 2 (x^{2} x^{1}) + 6 x^{2} \cdot - 3 + 7 x (2 x) + 7 x \cdot - 3 + 2 (2 x) + 2 \cdot - 3 = \frac{0}{2}$

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

$6 \cdot 2 x^{2 + 1} + 6 x^{2} \cdot - 3 + 7 x (2 x) + 7 x \cdot - 3 + 2 (2 x) + 2 \cdot - 3 = \frac{0}{2}$

Add $2$ and $1$ .

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

Multiply $6$ by $2$ .

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

Multiply $- 3$ by $6$ .

$12 x^{3} - 18 x^{2} + 7 x (2 x) + 7 x \cdot - 3 + 2 (2 x) + 2 \cdot - 3 = \frac{0}{2}$

Multiply $x$ by $x$ .

$12 x^{3} - 18 x^{2} + 7 \cdot 2 x^{2} + 7 x \cdot - 3 + 2 (2 x) + 2 \cdot - 3 = \frac{0}{2}$

Multiply $7$ by $2$ .

$12 x^{3} - 18 x^{2} + 14 x^{2} + 7 x \cdot - 3 + 2 (2 x) + 2 \cdot - 3 = \frac{0}{2}$

Multiply $- 3$ by $7$ .

$12 x^{3} - 18 x^{2} + 14 x^{2} - 21 x + 2 (2 x) + 2 \cdot - 3 = \frac{0}{2}$

Multiply $2$ by $2$ .

$12 x^{3} - 18 x^{2} + 14 x^{2} - 21 x + 4 x + 2 \cdot - 3 = \frac{0}{2}$

Multiply $2$ by $- 3$ .

$12 x^{3} - 18 x^{2} + 14 x^{2} - 21 x + 4 x - 6 = \frac{0}{2}$

Simplify by adding terms.

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Add $- 18 x^{2}$ and $14 x^{2}$ .

$12 x^{3} - 4 x^{2} - 21 x + 4 x - 6 = \frac{0}{2}$

Add $- 21 x$ and $4 x$ .

$12 x^{3} - 4 x^{2} - 17 x - 6 = \frac{0}{2}$

Divide $0$ by $2$ .

$12 x^{3} - 4 x^{2} - 17 x - 6 = 0$

Factor the left side of the equation.

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Factor $12 x^{3} - 4 x^{2} - 17 x - 6$ using the rational roots test.

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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.

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

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

$\pm 1, \pm 0.08\overline{3}, \pm 0.5, \pm 0.1\overline{6}, \pm 0.\overline{3}, \pm 0.25, \pm 6, \pm 3, \pm 2, \pm 1.5, \pm 0.\overline{6}, \pm 0.75$

Substitute $- 0.5$ and simplify the expression. In this case, the expression is equal to $0$ so $- 0.5$ is a root of the polynomial.

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Substitute $- 0.5$ into the polynomial.

$12 {(- 0.5)}^{3} - 4 {(- 0.5)}^{2} - 17 \cdot - 0.5 - 6$

Raise $- 0.5$ to the power of $3$ .

$12 \cdot - 0.125 - 4 {(- 0.5)}^{2} - 17 \cdot - 0.5 - 6$

Multiply $12$ by $- 0.125$ .

$- 1.5 - 4 {(- 0.5)}^{2} - 17 \cdot - 0.5 - 6$

Raise $- 0.5$ to the power of $2$ .

$- 1.5 - 4 \cdot 0.25 - 17 \cdot - 0.5 - 6$

Multiply $- 4$ by $0.25$ .

$- 1.5 - 1 - 17 \cdot - 0.5 - 6$

Subtract $1$ from $- 1.5$ .

$- 2.5 - 17 \cdot - 0.5 - 6$

Multiply $- 17$ by $- 0.5$ .

$- 2.5 + 8.5 - 6$

Add $- 2.5$ and $8.5$ .

$6 - 6$

Subtract $6$ from $6$ .

$0$

Since $- 0.5$ is a known root, divide the polynomial by $2 x + 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{12 x^{3} - 4 x^{2} - 17 x - 6}{2 x + 1}$

Divide $12 x^{3} - 4 x^{2} - 17 x - 6$ by $2 x + 1$ .

$6 x^{2} - 5 x - 6$

Write $12 x^{3} - 4 x^{2} - 17 x - 6$ as a set of factors.

$(2 x + 1) (6 x^{2} - 5 x - 6) = 0$

Factor by grouping.

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Factor by grouping.

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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 $a \cdot c = 6 \cdot - 6 = - 36$ and whose sum is $b = - 5$ .

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

$(2 x + 1) (6 x^{2} - 5 x - 6) = 0$

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

$(2 x + 1) (6 x^{2} + (4 - 9) x - 6) = 0$

Apply the distributive property.

$(2 x + 1) (6 x^{2} + 4 x - 9 x - 6) = 0$

Factor out the greatest common factor from each group.

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Group the first two terms and the last two terms.

$(2 x + 1) ((6 x^{2} + 4 x) - 9 x - 6) = 0$

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

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

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

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

Remove unnecessary parentheses.

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

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

$2 x + 1 = 0$

$3 x + 2 = 0$

$2 x - 3 = 0$

Set the first factor equal to $0$ and solve.

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Set the first factor equal to $0$ .

$2 x + 1 = 0$

Subtract $1$ from both sides of the equation.

$2 x = - 1$

Divide each term by $2$ and simplify.

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Divide each term in $2 x = - 1$ by $2$ .

$\frac{2 x}{2} = \frac{- 1}{2}$

Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{2 x}{2} = \frac{- 1}{2}$

Divide $x$ by $1$ .

$x = \frac{- 1}{2}$

Move the negative in front of the fraction.

$x = - \frac{1}{2}$

Set the next factor equal to $0$ and solve.

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Set the next factor equal to $0$ .

$3 x + 2 = 0$

Subtract $2$ from both sides of the equation.

$3 x = - 2$

Divide each term by $3$ and simplify.

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

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

Cancel the common factor of $3$ .

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Cancel the common factor.

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

Divide $x$ by $1$ .

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

Move the negative in front of the fraction.

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

Set the next factor equal to $0$ and solve.

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Set the next factor equal to $0$ .

$2 x - 3 = 0$

Add $3$ to both sides of the equation.

$2 x = 3$

Divide each term by $2$ and simplify.

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

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

Cancel the common factor of $2$ .

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Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

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

Find the Roots/Zeros Using the Rational Roots Test f(x)=24x^3-8x^2-34x-12