Find the Roots/Zeros Using the Rational Roots Test f(x)=2x^3-3x^2-32x-15

$f (x) = 2 x^{3} - 3 x^{2} - 32 x - 15$

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.

$p = \pm 1, \pm 3, \pm 5, \pm 15$

$q = \pm 1, \pm 2$

Step 2

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

$\pm 1, \pm \frac{1}{2}, \pm 3, \pm \frac{3}{2}, \pm 5, \pm \frac{5}{2}, \pm 15, \pm \frac{15}{2}$

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.

$2 {(- \frac{1}{2})}^{3} - 3 {(- \frac{1}{2})}^{2} - 32 (- \frac{1}{2}) - 15$

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.

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

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

$2 ({(- 1)}^{3} {(\frac{1}{2})}^{3}) - 3 {(- \frac{1}{2})}^{2} - 32 (- \frac{1}{2}) - 15$

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

$2 ({(- 1)}^{3} \frac{1^{3}}{2^{3}}) - 3 {(- \frac{1}{2})}^{2} - 32 (- \frac{1}{2}) - 15$

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

$2 (- \frac{1^{3}}{2^{3}}) - 3 {(- \frac{1}{2})}^{2} - 32 (- \frac{1}{2}) - 15$

One to any power is one.

$2 (- \frac{1}{2^{3}}) - 3 {(- \frac{1}{2})}^{2} - 32 (- \frac{1}{2}) - 15$

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

$2 (- \frac{1}{8}) - 3 {(- \frac{1}{2})}^{2} - 32 (- \frac{1}{2}) - 15$

Cancel the common factor of $2$ .

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

$2 (\frac{- 1}{8}) - 3 {(- \frac{1}{2})}^{2} - 32 (- \frac{1}{2}) - 15$

Factor $2$ out of $8$ .

$2 \frac{- 1}{2 (4)} - 3 {(- \frac{1}{2})}^{2} - 32 (- \frac{1}{2}) - 15$

Cancel the common factor.

$2 \frac{- 1}{2 \cdot 4} - 3 {(- \frac{1}{2})}^{2} - 32 (- \frac{1}{2}) - 15$

Rewrite the expression.

$\frac{- 1}{4} - 3 {(- \frac{1}{2})}^{2} - 32 (- \frac{1}{2}) - 15$

Move the negative in front of the fraction.

$- \frac{1}{4} - 3 {(- \frac{1}{2})}^{2} - 32 (- \frac{1}{2}) - 15$

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

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

$- \frac{1}{4} - 3 ({(- 1)}^{2} {(\frac{1}{2})}^{2}) - 32 (- \frac{1}{2}) - 15$

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

$- \frac{1}{4} - 3 ({(- 1)}^{2} \frac{1^{2}}{2^{2}}) - 32 (- \frac{1}{2}) - 15$

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

$- \frac{1}{4} - 3 (1 \frac{1^{2}}{2^{2}}) - 32 (- \frac{1}{2}) - 15$

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

$- \frac{1}{4} - 3 \frac{1^{2}}{2^{2}} - 32 (- \frac{1}{2}) - 15$

One to any power is one.

$- \frac{1}{4} - 3 \frac{1}{2^{2}} - 32 (- \frac{1}{2}) - 15$

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

$- \frac{1}{4} - 3 (\frac{1}{4}) - 32 (- \frac{1}{2}) - 15$

Combine $- 3$ and $\frac{1}{4}$ .

$- \frac{1}{4} + \frac{- 3}{4} - 32 (- \frac{1}{2}) - 15$

Move the negative in front of the fraction.

$- \frac{1}{4} - \frac{3}{4} - 32 (- \frac{1}{2}) - 15$

Cancel the common factor of $2$ .

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

$- \frac{1}{4} - \frac{3}{4} - 32 (\frac{- 1}{2}) - 15$

Factor $2$ out of $- 32$ .

$- \frac{1}{4} - \frac{3}{4} + 2 (- 16) \frac{- 1}{2} - 15$

Cancel the common factor.

$- \frac{1}{4} - \frac{3}{4} + 2 \cdot - 16 \frac{- 1}{2} - 15$

Rewrite the expression.

$- \frac{1}{4} - \frac{3}{4} - 16 \cdot - 1 - 15$

Multiply $- 16$ by $- 1$ .

$- \frac{1}{4} - \frac{3}{4} + 16 - 15$

Combine fractions.

Combine the numerators over the common denominator.

$16 - 15 + \frac{- 1 - 3}{4}$

Subtract $3$ from $- 1$ .

$16 - 15 + \frac{- 4}{4}$

Find the common denominator.

Write $16$ as a fraction with denominator $1$ .

$\frac{16}{1} - 15 + \frac{- 4}{4}$

Multiply $\frac{16}{1}$ by $\frac{4}{4}$ .

$\frac{16}{1} \cdot \frac{4}{4} - 15 + \frac{- 4}{4}$

Multiply $\frac{16}{1}$ by $\frac{4}{4}$ .

$\frac{16 \cdot 4}{4} - 15 + \frac{- 4}{4}$

Write $- 15$ as a fraction with denominator $1$ .

$\frac{16 \cdot 4}{4} + \frac{- 15}{1} + \frac{- 4}{4}$

Multiply $\frac{- 15}{1}$ by $\frac{4}{4}$ .

$\frac{16 \cdot 4}{4} + \frac{- 15}{1} \cdot \frac{4}{4} + \frac{- 4}{4}$

Multiply $\frac{- 15}{1}$ by $\frac{4}{4}$ .

$\frac{16 \cdot 4}{4} + \frac{- 15 \cdot 4}{4} + \frac{- 4}{4}$

Combine the numerators over the common denominator.

$\frac{16 \cdot 4 - 15 \cdot 4 - 4}{4}$

Simplify each term.

Multiply $16$ by $4$ .

$\frac{64 - 15 \cdot 4 - 4}{4}$

Multiply $- 15$ by $4$ .

$\frac{64 - 60 - 4}{4}$

Simplify the expression.

Subtract $60$ from $64$ .

$\frac{4 - 4}{4}$

Subtract $4$ from $4$ .

$\frac{0}{4}$

Divide $0$ by $4$ .

$0$

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.

$\frac{2 x^{3} - 3 x^{2} - 32 x - 15}{x + \frac{1}{2}}$

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.

$- \frac{1}{2}$	$2$	$- 3$	$- 32$	$- 15$
	\xa0			\xa0

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

$- \frac{1}{2}$	$2$	$- 3$	$- 32$	$- 15$
	\xa0			\xa0
	$2$

Multiply the newest entry in the result $(2)$ by the divisor $(- \frac{1}{2})$ and place the result of $(- 1)$ under the next term in the dividend $(- 3)$ .

$- \frac{1}{2}$	$2$	$- 3$	$- 32$	$- 15$
	\xa0	$- 1$		\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.

$- \frac{1}{2}$	$2$	$- 3$	$- 32$	$- 15$
	\xa0	$- 1$		\xa0
	$2$	$- 4$

Multiply the newest entry in the result $(- 4)$ by the divisor $(- \frac{1}{2})$ and place the result of $(2)$ under the next term in the dividend $(- 32)$ .

$- \frac{1}{2}$	$2$	$- 3$	$- 32$	$- 15$
	\xa0	$- 1$	$2$	\xa0
	$2$	$- 4$

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

$- \frac{1}{2}$	$2$	$- 3$	$- 32$	$- 15$
	\xa0	$- 1$	$2$	\xa0
	$2$	$- 4$	$- 30$

Multiply the newest entry in the result $(- 30)$ by the divisor $(- \frac{1}{2})$ and place the result of $(15)$ under the next term in the dividend $(- 15)$ .

$- \frac{1}{2}$	$2$	$- 3$	$- 32$	$- 15$
	\xa0	$- 1$	$2$	$15$
	$2$	$- 4$	$- 30$

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

$- \frac{1}{2}$	$2$	$- 3$	$- 32$	$- 15$
	\xa0	$- 1$	$2$	$15$
	$2$	$- 4$	$- 30$	$0$

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

$2 x^{2} + - 4 x - 30$

Simplify the quotient polynomial.

$2 x^{2} - 4 x - 30$

Step 7

Factor $2$ out of $2 x^{2} - 4 x - 30$ .

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

$(x + \frac{1}{2}) (2 (x^{2}) - 4 x - 30)$

Factor $2$ out of $- 4 x$ .

$(x + \frac{1}{2}) (2 (x^{2}) + 2 (- 2 x) - 30)$

Factor $2$ out of $- 30$ .

$(x + \frac{1}{2}) (2 x^{2} + 2 (- 2 x) + 2 \cdot - 15)$

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

$(x + \frac{1}{2}) (2 (x^{2} - 2 x) + 2 \cdot - 15)$

Factor $2$ out of $2 (x^{2} - 2 x) + 2 \cdot - 15$ .

$(x + \frac{1}{2}) (2 (x^{2} - 2 x - 15))$

Step 8

Factor.

Factor $x^{2} - 2 x - 15$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 15$ and whose sum is $- 2$ .

$- 5, 3$

Write the factored form using these integers.

$(x + \frac{1}{2}) (2 ((x - 5) (x + 3)))$

Remove unnecessary parentheses.

$(x + \frac{1}{2}) (2 (x - 5) (x + 3))$

Step 9

Factor the left side of the equation.

Factor $2 x^{3} - 3 x^{2} - 32 x - 15$ 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.

$p = \pm 1, \pm 15, \pm 3, \pm 5 q = \pm 1, \pm 2$

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

$\pm 1, \pm 0.5, \pm 15, \pm 7.5, \pm 3, \pm 1.5, \pm 5, \pm 2.5$

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.

$2 {(- 0.5)}^{3} - 3 {(- 0.5)}^{2} - 32 \cdot - 0.5 - 15$

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

$2 \cdot - 0.125 - 3 {(- 0.5)}^{2} - 32 \cdot - 0.5 - 15$

Multiply $2$ by $- 0.125$ .

$- 0.25 - 3 {(- 0.5)}^{2} - 32 \cdot - 0.5 - 15$

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

$- 0.25 - 3 \cdot 0.25 - 32 \cdot - 0.5 - 15$

Multiply $- 3$ by $0.25$ .

$- 0.25 - 0.75 - 32 \cdot - 0.5 - 15$

Subtract $0.75$ from $- 0.25$ .

$- 1 - 32 \cdot - 0.5 - 15$

Multiply $- 32$ by $- 0.5$ .

$- 1 + 16 - 15$

Add $- 1$ and $16$ .

$15 - 15$

Subtract $15$ from $15$ .

$0$

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.

$\frac{2 x^{3} - 3 x^{2} - 32 x - 15}{2 x + 1}$

Divide $2 x^{3} - 3 x^{2} - 32 x - 15$ 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 $0$ .

$2 x$

$1$

$2 x^{3}$

$3 x^{2}$

$32 x$

$15$

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

					$x^{2}$
	$2 x$	+	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$32 x$	-	$15$

Multiply the new quotient term by the divisor.

				$x^{2}$
$2 x$	+	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$32 x$	-	$15$
			+	$2 x^{3}$	+	$x^{2}$

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

				$x^{2}$
$2 x$	+	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$32 x$	-	$15$
			-	$2 x^{3}$	-	$x^{2}$

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

				$x^{2}$
$2 x$	+	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$32 x$	-	$15$
			-	$2 x^{3}$	-	$x^{2}$
					-	$4 x^{2}$

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

				$x^{2}$
$2 x$	+	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$32 x$	-	$15$
			-	$2 x^{3}$	-	$x^{2}$
					-	$4 x^{2}$	-	$32 x$

Divide the highest order term in the dividend $- 4 x^{2}$ by the highest order term in divisor $2 x$ .

				$x^{2}$	-	$2 x$
$2 x$	+	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$32 x$	-	$15$
			-	$2 x^{3}$	-	$x^{2}$
					-	$4 x^{2}$	-	$32 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$2 x$
$2 x$	+	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$32 x$	-	$15$
			-	$2 x^{3}$	-	$x^{2}$
					-	$4 x^{2}$	-	$32 x$
					-	$4 x^{2}$	-	$2 x$

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

				$x^{2}$	-	$2 x$
$2 x$	+	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$32 x$	-	$15$
			-	$2 x^{3}$	-	$x^{2}$
					-	$4 x^{2}$	-	$32 x$
					+	$4 x^{2}$	+	$2 x$

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

				$x^{2}$	-	$2 x$
$2 x$	+	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$32 x$	-	$15$
			-	$2 x^{3}$	-	$x^{2}$
					-	$4 x^{2}$	-	$32 x$
					+	$4 x^{2}$	+	$2 x$
							-	$30 x$

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

				$x^{2}$	-	$2 x$
$2 x$	+	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$32 x$	-	$15$
			-	$2 x^{3}$	-	$x^{2}$
					-	$4 x^{2}$	-	$32 x$
					+	$4 x^{2}$	+	$2 x$
							-	$30 x$	-	$15$

Divide the highest order term in the dividend $- 30 x$ by the highest order term in divisor $2 x$ .

				$x^{2}$	-	$2 x$	-	$15$
$2 x$	+	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$32 x$	-	$15$
			-	$2 x^{3}$	-	$x^{2}$
					-	$4 x^{2}$	-	$32 x$
					+	$4 x^{2}$	+	$2 x$
							-	$30 x$	-	$15$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$2 x$	-	$15$
$2 x$	+	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$32 x$	-	$15$
			-	$2 x^{3}$	-	$x^{2}$
					-	$4 x^{2}$	-	$32 x$
					+	$4 x^{2}$	+	$2 x$
							-	$30 x$	-	$15$
							-	$30 x$	-	$15$

The expression needs to be subtracted from the dividend, so change all the signs in $- 30 x - 15$

				$x^{2}$	-	$2 x$	-	$15$
$2 x$	+	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$32 x$	-	$15$
			-	$2 x^{3}$	-	$x^{2}$
					-	$4 x^{2}$	-	$32 x$
					+	$4 x^{2}$	+	$2 x$
							-	$30 x$	-	$15$
							+	$30 x$	+	$15$

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

				$x^{2}$	-	$2 x$	-	$15$
$2 x$	+	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$32 x$	-	$15$
			-	$2 x^{3}$	-	$x^{2}$
					-	$4 x^{2}$	-	$32 x$
					+	$4 x^{2}$	+	$2 x$
							-	$30 x$	-	$15$
							+	$30 x$	+	$15$
										$0$

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

$x^{2} - 2 x - 15$

Write $2 x^{3} - 3 x^{2} - 32 x - 15$ as a set of factors.

$(2 x + 1) (x^{2} - 2 x - 15) = 0$

Factor $x^{2} - 2 x - 15$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 15$ and whose sum is $- 2$ .

$- 5, 3$

Write the factored form using these integers.

$(2 x + 1) ((x - 5) (x + 3)) = 0$

Remove unnecessary parentheses.

$(2 x + 1) (x - 5) (x + 3) = 0$

Step 10

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

$2 x + 1 = 0$

$x - 5 = 0$

$x + 3 = 0$

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 $x$ .

Subtract $1$ from 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 $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Step 12

Set $x - 5$ equal to $0$ and solve for $x$ .

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

Step 13

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 14

The final solution is all the values that make $(2 x + 1) (x - 5) (x + 3) = 0$ true.

$x = - \frac{1}{2}, 5, - 3$

Step 15

Find the Roots/Zeros Using the Rational Roots Test f(x)=2x^3-3x^2-32x-15

Related Questions

Support LCMGCF.com by disabling your adblocker.