Find the Roots/Zeros Using the Rational Roots Test 2x^3-5x^2-5x-1

$2 x^{3} - 5 x^{2} - 5 x - 1$

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.

$p = \pm 1$

$q = \pm 1, \pm 2$

Step 2

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

$\pm 1, \pm \frac{1}{2}$

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.

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

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 $- \frac{1}{2}$ .

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

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

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

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

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

One to any power is one.

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

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

$2 (- \frac{1}{8}) - 5 {(- \frac{1}{2})}^{2} - 5 (- \frac{1}{2}) - 1$

Cancel the common factor of $2$ .

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

$2 (\frac{- 1}{8}) - 5 {(- \frac{1}{2})}^{2} - 5 (- \frac{1}{2}) - 1$

Factor $2$ out of $8$ .

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

Cancel the common factor.

$2 \frac{- 1}{2 \cdot 4} - 5 {(- \frac{1}{2})}^{2} - 5 (- \frac{1}{2}) - 1$

Rewrite the expression.

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

Move the negative in front of the fraction.

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

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

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

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

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

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

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

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

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

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

One to any power is one.

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

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

$- \frac{1}{4} - 5 (\frac{1}{4}) - 5 (- \frac{1}{2}) - 1$

Combine $- 5$ and $\frac{1}{4}$ .

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

Move the negative in front of the fraction.

$- \frac{1}{4} - \frac{5}{4} - 5 (- \frac{1}{2}) - 1$

Multiply $- 5 (- \frac{1}{2})$ .

Multiply $- 1$ by $- 5$ .

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

Combine $5$ and $\frac{1}{2}$ .

$- \frac{1}{4} - \frac{5}{4} + \frac{5}{2} - 1$

Combine fractions.

Combine the numerators over the common denominator.

$- 1 + \frac{- 1 - 5}{4} + \frac{5}{2}$

Subtract $5$ from $- 1$ .

$- 1 + \frac{- 6}{4} + \frac{5}{2}$

Find the common denominator.

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

$\frac{- 1}{1} + \frac{- 6}{4} + \frac{5}{2}$

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

$\frac{- 1}{1} \cdot \frac{4}{4} + \frac{- 6}{4} + \frac{5}{2}$

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

$\frac{- 1 \cdot 4}{4} + \frac{- 6}{4} + \frac{5}{2}$

Multiply $\frac{5}{2}$ by $\frac{2}{2}$ .

$\frac{- 1 \cdot 4}{4} + \frac{- 6}{4} + \frac{5}{2} \cdot \frac{2}{2}$

Multiply $\frac{5}{2}$ by $\frac{2}{2}$ .

$\frac{- 1 \cdot 4}{4} + \frac{- 6}{4} + \frac{5 \cdot 2}{2 \cdot 2}$

Multiply $2$ by $2$ .

$\frac{- 1 \cdot 4}{4} + \frac{- 6}{4} + \frac{5 \cdot 2}{4}$

Combine the numerators over the common denominator.

$\frac{- 1 \cdot 4 - 6 + 5 \cdot 2}{4}$

Simplify each term.

Multiply $- 1$ by $4$ .

$\frac{- 4 - 6 + 5 \cdot 2}{4}$

Multiply $5$ by $2$ .

$\frac{- 4 - 6 + 10}{4}$

Simplify the expression.

Subtract $6$ from $- 4$ .

$\frac{- 10 + 10}{4}$

Add $- 10$ and $10$ .

$\frac{0}{4}$

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.

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

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.

$- \frac{1}{2}$	$2$	$- 5$	$- 5$	$- 1$
	\xa0			\xa0

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

$- \frac{1}{2}$	$2$	$- 5$	$- 5$	$- 1$
	\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 $(- 5)$ .

$- \frac{1}{2}$	$2$	$- 5$	$- 5$	$- 1$
	\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.

$- \frac{1}{2}$	$2$	$- 5$	$- 5$	$- 1$
	\xa0	$- 1$		\xa0
	$2$	$- 6$

Multiply the newest entry in the result $(- 6)$ by the divisor $(- \frac{1}{2})$ and place the result of $(3)$ under the next term in the dividend $(- 5)$ .

$- \frac{1}{2}$	$2$	$- 5$	$- 5$	$- 1$
	\xa0	$- 1$	$3$	\xa0
	$2$	$- 6$

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

$- \frac{1}{2}$	$2$	$- 5$	$- 5$	$- 1$
	\xa0	$- 1$	$3$	\xa0
	$2$	$- 6$	$- 2$

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 $(- 1)$ .

$- \frac{1}{2}$	$2$	$- 5$	$- 5$	$- 1$
	\xa0	$- 1$	$3$	$1$
	$2$	$- 6$	$- 2$

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

$- \frac{1}{2}$	$2$	$- 5$	$- 5$	$- 1$
	\xa0	$- 1$	$3$	$1$
	$2$	$- 6$	$- 2$	$0$

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

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

Simplify the quotient polynomial.

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

Step 7

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

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

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

Factor $2$ out of $- 6 x$ .

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

Factor $2$ out of $- 2$ .

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

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

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

Factor $2$ out of $2 (x^{2} - 3 x) + 2 (- 1)$ .

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

Step 8

Factor $2 x^{3} - 5 x^{2} - 5 x - 1$ 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.

$p = \pm 1 q = \pm 1, \pm 2$

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

$\pm 1, \pm 0.5$

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.

$2 {(- 0.5)}^{3} - 5 {(- 0.5)}^{2} - 5 \cdot - 0.5 - 1$

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

$2 \cdot - 0.125 - 5 {(- 0.5)}^{2} - 5 \cdot - 0.5 - 1$

Multiply $2$ by $- 0.125$ .

$- 0.25 - 5 {(- 0.5)}^{2} - 5 \cdot - 0.5 - 1$

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

$- 0.25 - 5 \cdot 0.25 - 5 \cdot - 0.5 - 1$

Multiply $- 5$ by $0.25$ .

$- 0.25 - 1.25 - 5 \cdot - 0.5 - 1$

Subtract $1.25$ from $- 0.25$ .

$- 1.5 - 5 \cdot - 0.5 - 1$

Multiply $- 5$ by $- 0.5$ .

$- 1.5 + 2.5 - 1$

Add $- 1.5$ and $2.5$ .

$1 - 1$

Subtract $1$ from $1$ .

$0$

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.

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

Divide $2 x^{3} - 5 x^{2} - 5 x - 1$ 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 $0$ .

$2 x$

$1$

$2 x^{3}$

$5 x^{2}$

$5 x$

$1$

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

					$x^{2}$
	$2 x$	+	$1$		$2 x^{3}$	-	$5 x^{2}$	-	$5 x$	-	$1$

Multiply the new quotient term by the divisor.

				$x^{2}$
$2 x$	+	$1$		$2 x^{3}$	-	$5 x^{2}$	-	$5 x$	-	$1$
			+	$2 x^{3}$	+	$x^{2}$

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

				$x^{2}$
$2 x$	+	$1$		$2 x^{3}$	-	$5 x^{2}$	-	$5 x$	-	$1$
			-	$2 x^{3}$	-	$x^{2}$

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

				$x^{2}$
$2 x$	+	$1$		$2 x^{3}$	-	$5 x^{2}$	-	$5 x$	-	$1$
			-	$2 x^{3}$	-	$x^{2}$
					-	$6 x^{2}$

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

				$x^{2}$
$2 x$	+	$1$		$2 x^{3}$	-	$5 x^{2}$	-	$5 x$	-	$1$
			-	$2 x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	-	$5 x$

Divide the highest order term in the dividend $- 6 x^{2}$ by the highest order term in divisor $2 x$ .

				$x^{2}$	-	$3 x$
$2 x$	+	$1$		$2 x^{3}$	-	$5 x^{2}$	-	$5 x$	-	$1$
			-	$2 x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	-	$5 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$3 x$
$2 x$	+	$1$		$2 x^{3}$	-	$5 x^{2}$	-	$5 x$	-	$1$
			-	$2 x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	-	$5 x$
					-	$6 x^{2}$	-	$3 x$

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

				$x^{2}$	-	$3 x$
$2 x$	+	$1$		$2 x^{3}$	-	$5 x^{2}$	-	$5 x$	-	$1$
			-	$2 x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	-	$5 x$
					+	$6 x^{2}$	+	$3 x$

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

				$x^{2}$	-	$3 x$
$2 x$	+	$1$		$2 x^{3}$	-	$5 x^{2}$	-	$5 x$	-	$1$
			-	$2 x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	-	$5 x$
					+	$6 x^{2}$	+	$3 x$
							-	$2 x$

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

				$x^{2}$	-	$3 x$
$2 x$	+	$1$		$2 x^{3}$	-	$5 x^{2}$	-	$5 x$	-	$1$
			-	$2 x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	-	$5 x$
					+	$6 x^{2}$	+	$3 x$
							-	$2 x$	-	$1$

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

				$x^{2}$	-	$3 x$	-	$1$
$2 x$	+	$1$		$2 x^{3}$	-	$5 x^{2}$	-	$5 x$	-	$1$
			-	$2 x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	-	$5 x$
					+	$6 x^{2}$	+	$3 x$
							-	$2 x$	-	$1$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$3 x$	-	$1$
$2 x$	+	$1$		$2 x^{3}$	-	$5 x^{2}$	-	$5 x$	-	$1$
			-	$2 x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	-	$5 x$
					+	$6 x^{2}$	+	$3 x$
							-	$2 x$	-	$1$
							-	$2 x$	-	$1$

The expression needs to be subtracted from the dividend, so change all the signs in $- 2 x - 1$

				$x^{2}$	-	$3 x$	-	$1$
$2 x$	+	$1$		$2 x^{3}$	-	$5 x^{2}$	-	$5 x$	-	$1$
			-	$2 x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	-	$5 x$
					+	$6 x^{2}$	+	$3 x$
							-	$2 x$	-	$1$
							+	$2 x$	+	$1$

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

				$x^{2}$	-	$3 x$	-	$1$
$2 x$	+	$1$		$2 x^{3}$	-	$5 x^{2}$	-	$5 x$	-	$1$
			-	$2 x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	-	$5 x$
					+	$6 x^{2}$	+	$3 x$
							-	$2 x$	-	$1$
							+	$2 x$	+	$1$
										$0$

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

$x^{2} - 3 x - 1$

Write $2 x^{3} - 5 x^{2} - 5 x - 1$ as a set of factors.

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

Step 9

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

$2 x + 1 = 0$

$x^{2} - 3 x - 1 = 0$

Step 10

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 11

Set $x^{2} - 3 x - 1$ equal to $0$ and solve for $x$ .

Set $x^{2} - 3 x - 1$ equal to $0$ .

$x^{2} - 3 x - 1 = 0$

Solve $x^{2} - 3 x - 1 = 0$ for $x$ .

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = - 3$ , and $c = - 1$ into the quadratic formula and solve for $x$ .

$\frac{3 \pm \sqrt{{(- 3)}^{2} - 4 \cdot (1 \cdot - 1)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

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

$x = \frac{3 \pm \sqrt{9 - 4 \cdot 1 \cdot - 1}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 1$ .

Multiply $- 4$ by $1$ .

$x = \frac{3 \pm \sqrt{9 - 4 \cdot - 1}}{2 \cdot 1}$

Multiply $- 4$ by $- 1$ .

$x = \frac{3 \pm \sqrt{9 + 4}}{2 \cdot 1}$

Add $9$ and $4$ .

$x = \frac{3 \pm \sqrt{13}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{3 \pm \sqrt{13}}{2}$

The final answer is the combination of both solutions.

$x = \frac{3 + \sqrt{13}}{2}, \frac{3 - \sqrt{13}}{2}$

Step 12

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

$x = - \frac{1}{2}, \frac{3 + \sqrt{13}}{2}, \frac{3 - \sqrt{13}}{2}$

Step 13

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{1}{2}, \frac{3 + \sqrt{13}}{2}, \frac{3 - \sqrt{13}}{2}$

Decimal Form:

$x = - 0.5, 3.30277563 \dots, - 0.30277563 \dots$

Step 14

Find the Roots/Zeros Using the Rational Roots Test 2x^3-5x^2-5x-1

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