Find the Roots/Zeros Using the Rational Roots Test f(x)=2x^3+x^2-4x-2

$f (x) = 2 x^{3} + x^{2} - 4 x - 2$

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.

$p = \pm 1, \pm 2$

$q = \pm 1, \pm 2$

Step 2

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

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

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.

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

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

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

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

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

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

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

One to any power is one.

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

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

$2 (- \frac{1}{8}) + {(- \frac{1}{2})}^{2} - 4 (- \frac{1}{2}) - 2$

Cancel the common factor of $2$ .

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

$2 (\frac{- 1}{8}) + {(- \frac{1}{2})}^{2} - 4 (- \frac{1}{2}) - 2$

Factor $2$ out of $8$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Move the negative in front of the fraction.

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

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

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

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

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

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

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

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

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

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

One to any power is one.

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

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

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

Cancel the common factor of $2$ .

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

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

Factor $2$ out of $- 4$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $- 2$ by $- 1$ .

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

Combine fractions.

Combine the numerators over the common denominator.

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

Add $- 1$ and $1$ .

$2 - 2 + \frac{0}{4}$

Find the common denominator.

Write $2$ as a fraction with denominator $1$ .

$\frac{2}{1} - 2 + \frac{0}{4}$

Multiply $\frac{2}{1}$ by $\frac{4}{4}$ .

$\frac{2}{1} \cdot \frac{4}{4} - 2 + \frac{0}{4}$

Multiply $\frac{2}{1}$ by $\frac{4}{4}$ .

$\frac{2 \cdot 4}{4} - 2 + \frac{0}{4}$

Write $- 2$ as a fraction with denominator $1$ .

$\frac{2 \cdot 4}{4} + \frac{- 2}{1} + \frac{0}{4}$

Multiply $\frac{- 2}{1}$ by $\frac{4}{4}$ .

$\frac{2 \cdot 4}{4} + \frac{- 2}{1} \cdot \frac{4}{4} + \frac{0}{4}$

Multiply $\frac{- 2}{1}$ by $\frac{4}{4}$ .

$\frac{2 \cdot 4}{4} + \frac{- 2 \cdot 4}{4} + \frac{0}{4}$

Combine the numerators over the common denominator.

$\frac{2 \cdot 4 - 2 \cdot 4}{4}$

Simplify each term.

Multiply $2$ by $4$ .

$\frac{8 - 2 \cdot 4}{4}$

Multiply $- 2$ by $4$ .

$\frac{8 - 8}{4}$

Simplify the expression.

Subtract $8$ from $8$ .

$\frac{0}{4}$

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.

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

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.

$- \frac{1}{2}$	$2$	$1$	$- 4$	$- 2$
	\xa0			\xa0

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

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

$- \frac{1}{2}$	$2$	$1$	$- 4$	$- 2$
	\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.

$- \frac{1}{2}$	$2$	$1$	$- 4$	$- 2$
	\xa0	$- 1$		\xa0
	$2$	$0$

Multiply the newest entry in the result $(0)$ by the divisor $(- \frac{1}{2})$ and place the result of $(0)$ under the next term in the dividend $(- 4)$ .

$- \frac{1}{2}$	$2$	$1$	$- 4$	$- 2$
	\xa0	$- 1$	$0$	\xa0
	$2$	$0$

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

$- \frac{1}{2}$	$2$	$1$	$- 4$	$- 2$
	\xa0	$- 1$	$0$	\xa0
	$2$	$0$	$- 4$

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 $(- 2)$ .

$- \frac{1}{2}$	$2$	$1$	$- 4$	$- 2$
	\xa0	$- 1$	$0$	$2$
	$2$	$0$	$- 4$

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

$- \frac{1}{2}$	$2$	$1$	$- 4$	$- 2$
	\xa0	$- 1$	$0$	$2$
	$2$	$0$	$- 4$	$0$

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

$2 x^{2} + 0 x - 4$

Simplify the quotient polynomial.

$2 x^{2} - 4$

Step 7

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

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

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

Factor $2$ out of $- 4$ .

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

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

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

Step 8

Factor the left side of the equation.

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

Factor the polynomial by factoring out the greatest common factor, $2 x + 1$ .

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

Step 9

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

$2 x + 1 = 0$

$x^{2} - 2 = 0$

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

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

$x^{2} - 2 = 0$

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

Add $2$ to both sides of the equation.

$x^{2} = 2$

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{2}$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = \sqrt{2}$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - \sqrt{2}$

The complete solution is the result of both the positive and negative portions of the solution.

$x = \sqrt{2}, - \sqrt{2}$

Step 12

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

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

Step 13

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$x = - 0.5, 1.41421356 \dots, - 1.41421356 \dots$

Step 14

Find the Roots/Zeros Using the Rational Roots Test f(x)=2x^3+x^2-4x-2

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