Find the Roots/Zeros Using the Rational Roots Test f(x)=2x^3+7x^2-10x-24

$f (x) = 2 x^{3} + 7 x^{2} - 10 x - 24$

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.

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

$q = \pm 1, \pm 2$

Step 2

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

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

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.

$2 {(2)}^{3} + 7 {(2)}^{2} - 10 \cdot 2 - 24$

Step 4

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

Simplify each term.

Multiply $2$ by ${(2)}^{3}$ by adding the exponents.

Multiply $2$ by ${(2)}^{3}$ .

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

$2^{1} {(2)}^{3} + 7 {(2)}^{2} - 10 \cdot 2 - 24$

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

$2^{1 + 3} + 7 {(2)}^{2} - 10 \cdot 2 - 24$

Add $1$ and $3$ .

$2^{4} + 7 {(2)}^{2} - 10 \cdot 2 - 24$

Raise $2$ to the power of $4$ .

$16 + 7 {(2)}^{2} - 10 \cdot 2 - 24$

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

$16 + 7 \cdot 4 - 10 \cdot 2 - 24$

Multiply $7$ by $4$ .

$16 + 28 - 10 \cdot 2 - 24$

Multiply $- 10$ by $2$ .

$16 + 28 - 20 - 24$

Simplify by adding and subtracting.

Add $16$ and $28$ .

$44 - 20 - 24$

Subtract $20$ from $44$ .

$24 - 24$

Subtract $24$ from $24$ .

$0$

Step 5

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

$\frac{2 x^{3} + 7 x^{2} - 10 x - 24}{x - 2}$

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.

$2$	$2$	$7$	$- 10$	$- 24$
	\xa0			\xa0

The first number in the dividend $(2)$ is put into the first position of the result area (below the horizontal line).

$2$	$2$	$7$	$- 10$	$- 24$
	\xa0			\xa0
	$2$

Multiply the newest entry in the result $(2)$ by the divisor $(2)$ and place the result of $(4)$ under the next term in the dividend $(7)$ .

$2$	$2$	$7$	$- 10$	$- 24$
	\xa0	$4$		\xa0
	$2$

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

$2$	$2$	$7$	$- 10$	$- 24$
	\xa0	$4$		\xa0
	$2$	$11$

Multiply the newest entry in the result $(11)$ by the divisor $(2)$ and place the result of $(22)$ under the next term in the dividend $(- 10)$ .

$2$	$2$	$7$	$- 10$	$- 24$
	\xa0	$4$	$22$	\xa0
	$2$	$11$

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

$2$	$2$	$7$	$- 10$	$- 24$
	\xa0	$4$	$22$	\xa0
	$2$	$11$	$12$

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

$2$	$2$	$7$	$- 10$	$- 24$
	\xa0	$4$	$22$	$24$
	$2$	$11$	$12$

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

$2$	$2$	$7$	$- 10$	$- 24$
	\xa0	$4$	$22$	$24$
	$2$	$11$	$12$	$0$

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

$2 x^{2} + 11 x + 12$

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 $a \cdot c = 2 \cdot 12 = 24$ and whose sum is $b = 11$ .

Factor $11$ out of $11 x$ .

$(x - 2) + 2 x^{2} + 11 (x) + 12$

Rewrite $11$ as $3$ plus $8$

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

$(x - 2) + x (2 x + 3) + 4 (2 x + 3)$

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

$(x - 2) + (2 x + 3) (x + 4)$

$(x - 2) (2 x + 3) (x + 4)$

Step 8

Factor the left side of the equation.

Factor $2 x^{3} + 7 x^{2} - 10 x - 24$ 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.

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

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

$\pm 1, \pm 0.5, \pm 24, \pm 12, \pm 2, \pm 6, \pm 3, \pm 1.5, \pm 8, \pm 4$

Substitute $2$ and simplify the expression. In this case, the expression is equal to $0$ so $2$ is a root of the polynomial.

Substitute $2$ into the polynomial.

$2 \cdot 2^{3} + 7 \cdot 2^{2} - 10 \cdot 2 - 24$

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

$2 \cdot 8 + 7 \cdot 2^{2} - 10 \cdot 2 - 24$

Multiply $2$ by $8$ .

$16 + 7 \cdot 2^{2} - 10 \cdot 2 - 24$

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

$16 + 7 \cdot 4 - 10 \cdot 2 - 24$

Multiply $7$ by $4$ .

$16 + 28 - 10 \cdot 2 - 24$

Add $16$ and $28$ .

$44 - 10 \cdot 2 - 24$

Multiply $- 10$ by $2$ .

$44 - 20 - 24$

Subtract $20$ from $44$ .

$24 - 24$

Subtract $24$ from $24$ .

$0$

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

$\frac{2 x^{3} + 7 x^{2} - 10 x - 24}{x - 2}$

Divide $2 x^{3} + 7 x^{2} - 10 x - 24$ by $x - 2$ .

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

$x$

$2$

$2 x^{3}$

$7 x^{2}$

$10 x$

$24$

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

					$2 x^{2}$
	$x$	-	$2$		$2 x^{3}$	+	$7 x^{2}$	-	$10 x$	-	$24$

Multiply the new quotient term by the divisor.

				$2 x^{2}$
$x$	-	$2$		$2 x^{3}$	+	$7 x^{2}$	-	$10 x$	-	$24$
			+	$2 x^{3}$	-	$4 x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $2 x^{3} - 4 x^{2}$

				$2 x^{2}$
$x$	-	$2$		$2 x^{3}$	+	$7 x^{2}$	-	$10 x$	-	$24$
			-	$2 x^{3}$	+	$4 x^{2}$

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

				$2 x^{2}$
$x$	-	$2$		$2 x^{3}$	+	$7 x^{2}$	-	$10 x$	-	$24$
			-	$2 x^{3}$	+	$4 x^{2}$
					+	$11 x^{2}$

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

				$2 x^{2}$
$x$	-	$2$		$2 x^{3}$	+	$7 x^{2}$	-	$10 x$	-	$24$
			-	$2 x^{3}$	+	$4 x^{2}$
					+	$11 x^{2}$	-	$10 x$

Divide the highest order term in the dividend $11 x^{2}$ by the highest order term in divisor $x$ .

				$2 x^{2}$	+	$11 x$
$x$	-	$2$		$2 x^{3}$	+	$7 x^{2}$	-	$10 x$	-	$24$
			-	$2 x^{3}$	+	$4 x^{2}$
					+	$11 x^{2}$	-	$10 x$

Multiply the new quotient term by the divisor.

				$2 x^{2}$	+	$11 x$
$x$	-	$2$		$2 x^{3}$	+	$7 x^{2}$	-	$10 x$	-	$24$
			-	$2 x^{3}$	+	$4 x^{2}$
					+	$11 x^{2}$	-	$10 x$
					+	$11 x^{2}$	-	$22 x$

The expression needs to be subtracted from the dividend, so change all the signs in $11 x^{2} - 22 x$

				$2 x^{2}$	+	$11 x$
$x$	-	$2$		$2 x^{3}$	+	$7 x^{2}$	-	$10 x$	-	$24$
			-	$2 x^{3}$	+	$4 x^{2}$
					+	$11 x^{2}$	-	$10 x$
					-	$11 x^{2}$	+	$22 x$

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

				$2 x^{2}$	+	$11 x$
$x$	-	$2$		$2 x^{3}$	+	$7 x^{2}$	-	$10 x$	-	$24$
			-	$2 x^{3}$	+	$4 x^{2}$
					+	$11 x^{2}$	-	$10 x$
					-	$11 x^{2}$	+	$22 x$
							+	$12 x$

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

				$2 x^{2}$	+	$11 x$
$x$	-	$2$		$2 x^{3}$	+	$7 x^{2}$	-	$10 x$	-	$24$
			-	$2 x^{3}$	+	$4 x^{2}$
					+	$11 x^{2}$	-	$10 x$
					-	$11 x^{2}$	+	$22 x$
							+	$12 x$	-	$24$

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

				$2 x^{2}$	+	$11 x$	+	$12$
$x$	-	$2$		$2 x^{3}$	+	$7 x^{2}$	-	$10 x$	-	$24$
			-	$2 x^{3}$	+	$4 x^{2}$
					+	$11 x^{2}$	-	$10 x$
					-	$11 x^{2}$	+	$22 x$
							+	$12 x$	-	$24$

Multiply the new quotient term by the divisor.

				$2 x^{2}$	+	$11 x$	+	$12$
$x$	-	$2$		$2 x^{3}$	+	$7 x^{2}$	-	$10 x$	-	$24$
			-	$2 x^{3}$	+	$4 x^{2}$
					+	$11 x^{2}$	-	$10 x$
					-	$11 x^{2}$	+	$22 x$
							+	$12 x$	-	$24$
							+	$12 x$	-	$24$

The expression needs to be subtracted from the dividend, so change all the signs in $12 x - 24$

				$2 x^{2}$	+	$11 x$	+	$12$
$x$	-	$2$		$2 x^{3}$	+	$7 x^{2}$	-	$10 x$	-	$24$
			-	$2 x^{3}$	+	$4 x^{2}$
					+	$11 x^{2}$	-	$10 x$
					-	$11 x^{2}$	+	$22 x$
							+	$12 x$	-	$24$
							-	$12 x$	+	$24$

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

				$2 x^{2}$	+	$11 x$	+	$12$
$x$	-	$2$		$2 x^{3}$	+	$7 x^{2}$	-	$10 x$	-	$24$
			-	$2 x^{3}$	+	$4 x^{2}$
					+	$11 x^{2}$	-	$10 x$
					-	$11 x^{2}$	+	$22 x$
							+	$12 x$	-	$24$
							-	$12 x$	+	$24$
										$0$

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

$2 x^{2} + 11 x + 12$

Write $2 x^{3} + 7 x^{2} - 10 x - 24$ as a set of factors.

$(x - 2) (2 x^{2} + 11 x + 12) = 0$

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 $b = 11$ .

Factor $11$ out of $11 x$ .

$(x - 2) (2 x^{2} + 11 (x) + 12) = 0$

Rewrite $11$ as $3$ plus $8$

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

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

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

Remove unnecessary parentheses.

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

Step 9

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

$x - 2 = 0$

$2 x + 3 = 0$

$x + 4 = 0$

Step 10

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 11

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 12

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

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

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 13

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

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

Step 14

Find the Roots/Zeros Using the Rational Roots Test f(x)=2x^3+7x^2-10x-24

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