Find the Roots/Zeros Using the Rational Roots Test 24x^3+8x^2-34x+12

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

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.

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

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

Step 2

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

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

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.

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

Step 4

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

Simplify each term.

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

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

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 $8$ .

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.

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

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

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

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 $4$ .

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 $2$ .

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.

$3 + 2 - 17 + 12$

Simplify by adding and subtracting.

Add $3$ and $2$ .

$5 - 17 + 12$

Subtract $17$ from $5$ .

$- 12 + 12$

Add $- 12$ and $12$ .

$0$

Step 5

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.

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

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.

$\frac{1}{2}$	$24$	$8$	$- 34$	$12$
	\xa0			\xa0

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

$\frac{1}{2}$	$24$	$8$	$- 34$	$12$
	\xa0			\xa0
	$24$

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 $(8)$ .

$\frac{1}{2}$	$24$	$8$	$- 34$	$12$
	\xa0	$12$		\xa0
	$24$

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

$\frac{1}{2}$	$24$	$8$	$- 34$	$12$
	\xa0	$12$		\xa0
	$24$	$20$

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 $(- 34)$ .

$\frac{1}{2}$	$24$	$8$	$- 34$	$12$
	\xa0	$12$	$10$	\xa0
	$24$	$20$

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

$\frac{1}{2}$	$24$	$8$	$- 34$	$12$
	\xa0	$12$	$10$	\xa0
	$24$	$20$	$- 24$

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 $(12)$ .

$\frac{1}{2}$	$24$	$8$	$- 34$	$12$
	\xa0	$12$	$10$	$- 12$
	$24$	$20$	$- 24$

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

$\frac{1}{2}$	$24$	$8$	$- 34$	$12$
	\xa0	$12$	$10$	$- 12$
	$24$	$20$	$- 24$	$0$

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

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

Step 7

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

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

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

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

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.

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.

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

Step 9

Factor the left side of the equation.

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

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.

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.

Substitute $0.5$ into the polynomial.

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

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

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

Multiply $12$ by $0.125$ .

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

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

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

Multiply $4$ by $0.25$ .

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

Add $1.5$ and $1$ .

$2.5 - 17 \cdot 0.5 + 6$

Multiply $- 17$ by $0.5$ .

$2.5 - 8.5 + 6$

Subtract $8.5$ from $2.5$ .

$- 6 + 6$

Add $- 6$ and $6$ .

$0$

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 $2 x - 1$ .

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

$2 x$

$1$

$12 x^{3}$

$4 x^{2}$

$17 x$

$6$

Divide the highest order term in the dividend $12 x^{3}$ by the highest order term in divisor $2 x$ .

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

Multiply the new quotient term by the divisor.

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

The expression needs to be subtracted from the dividend, so change all the signs in $12 x^{3} - 6 x^{2}$

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

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

				$6 x^{2}$
$2 x$	-	$1$		$12 x^{3}$	+	$4 x^{2}$	-	$17 x$	+	$6$
			-	$12 x^{3}$	+	$6 x^{2}$
					+	$10 x^{2}$

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

				$6 x^{2}$
$2 x$	-	$1$		$12 x^{3}$	+	$4 x^{2}$	-	$17 x$	+	$6$
			-	$12 x^{3}$	+	$6 x^{2}$
					+	$10 x^{2}$	-	$17 x$

Divide the highest order term in the dividend $10 x^{2}$ by the highest order term in divisor $2 x$ .

				$6 x^{2}$	+	$5 x$
$2 x$	-	$1$		$12 x^{3}$	+	$4 x^{2}$	-	$17 x$	+	$6$
			-	$12 x^{3}$	+	$6 x^{2}$
					+	$10 x^{2}$	-	$17 x$

Multiply the new quotient term by the divisor.

				$6 x^{2}$	+	$5 x$
$2 x$	-	$1$		$12 x^{3}$	+	$4 x^{2}$	-	$17 x$	+	$6$
			-	$12 x^{3}$	+	$6 x^{2}$
					+	$10 x^{2}$	-	$17 x$
					+	$10 x^{2}$	-	$5 x$

The expression needs to be subtracted from the dividend, so change all the signs in $10 x^{2} - 5 x$

				$6 x^{2}$	+	$5 x$
$2 x$	-	$1$		$12 x^{3}$	+	$4 x^{2}$	-	$17 x$	+	$6$
			-	$12 x^{3}$	+	$6 x^{2}$
					+	$10 x^{2}$	-	$17 x$
					-	$10 x^{2}$	+	$5 x$

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

				$6 x^{2}$	+	$5 x$
$2 x$	-	$1$		$12 x^{3}$	+	$4 x^{2}$	-	$17 x$	+	$6$
			-	$12 x^{3}$	+	$6 x^{2}$
					+	$10 x^{2}$	-	$17 x$
					-	$10 x^{2}$	+	$5 x$
							-	$12 x$

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

				$6 x^{2}$	+	$5 x$
$2 x$	-	$1$		$12 x^{3}$	+	$4 x^{2}$	-	$17 x$	+	$6$
			-	$12 x^{3}$	+	$6 x^{2}$
					+	$10 x^{2}$	-	$17 x$
					-	$10 x^{2}$	+	$5 x$
							-	$12 x$	+	$6$

Divide the highest order term in the dividend $- 12 x$ by the highest order term in divisor $2 x$ .

				$6 x^{2}$	+	$5 x$	-	$6$
$2 x$	-	$1$		$12 x^{3}$	+	$4 x^{2}$	-	$17 x$	+	$6$
			-	$12 x^{3}$	+	$6 x^{2}$
					+	$10 x^{2}$	-	$17 x$
					-	$10 x^{2}$	+	$5 x$
							-	$12 x$	+	$6$

Multiply the new quotient term by the divisor.

				$6 x^{2}$	+	$5 x$	-	$6$
$2 x$	-	$1$		$12 x^{3}$	+	$4 x^{2}$	-	$17 x$	+	$6$
			-	$12 x^{3}$	+	$6 x^{2}$
					+	$10 x^{2}$	-	$17 x$
					-	$10 x^{2}$	+	$5 x$
							-	$12 x$	+	$6$
							-	$12 x$	+	$6$

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

				$6 x^{2}$	+	$5 x$	-	$6$
$2 x$	-	$1$		$12 x^{3}$	+	$4 x^{2}$	-	$17 x$	+	$6$
			-	$12 x^{3}$	+	$6 x^{2}$
					+	$10 x^{2}$	-	$17 x$
					-	$10 x^{2}$	+	$5 x$
							-	$12 x$	+	$6$
							+	$12 x$	-	$6$

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

				$6 x^{2}$	+	$5 x$	-	$6$
$2 x$	-	$1$		$12 x^{3}$	+	$4 x^{2}$	-	$17 x$	+	$6$
			-	$12 x^{3}$	+	$6 x^{2}$
					+	$10 x^{2}$	-	$17 x$
					-	$10 x^{2}$	+	$5 x$
							-	$12 x$	+	$6$
							+	$12 x$	-	$6$
										$0$

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

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

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.

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

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.

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.

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

Step 10

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

$2 x - 1 = 0$

$3 x - 2 = 0$

$2 x + 3 = 0$

Step 11

Set $2 x - 1$ equal to $0$ and solve for $x$ .

Set $2 x - 1$ equal to $0$ .

$2 x - 1 = 0$

Solve $2 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$2 x = 1$

Divide each term in $2 x = 1$ by $2$ and simplify.

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Step 12

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

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

$3 x - 2 = 0$

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

Add $2$ to both sides of the equation.

$3 x = 2$

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

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

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Step 13

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 14

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

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

Step 15

Find the Roots/Zeros Using the Rational Roots Test 24x^3+8x^2-34x+12

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