Find the Roots/Zeros Using the Rational Roots Test f(x)=2x^3-3x^2-11x+6

$f (x) = 2 x^{3} - 3 x^{2} - 11 x + 6$

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.

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

$q = \pm 1, \pm 2$

Step 2

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

$\pm 1, \pm \frac{1}{2}, \pm 2, \pm 3, \pm \frac{3}{2}, \pm 6$

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.

$2 {(\frac{1}{2})}^{3} - 3 {(\frac{1}{2})}^{2} - 11 (\frac{1}{2}) + 6$

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 $\frac{1}{2}$ .

$2 \frac{1^{3}}{2^{3}} - 3 {(\frac{1}{2})}^{2} - 11 (\frac{1}{2}) + 6$

One to any power is one.

$2 \frac{1}{2^{3}} - 3 {(\frac{1}{2})}^{2} - 11 (\frac{1}{2}) + 6$

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

$2 (\frac{1}{8}) - 3 {(\frac{1}{2})}^{2} - 11 (\frac{1}{2}) + 6$

Cancel the common factor of $2$ .

Factor $2$ out of $8$ .

$2 \frac{1}{2 (4)} - 3 {(\frac{1}{2})}^{2} - 11 (\frac{1}{2}) + 6$

Cancel the common factor.

$2 \frac{1}{2 \cdot 4} - 3 {(\frac{1}{2})}^{2} - 11 (\frac{1}{2}) + 6$

Rewrite the expression.

$\frac{1}{4} - 3 {(\frac{1}{2})}^{2} - 11 (\frac{1}{2}) + 6$

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

$\frac{1}{4} - 3 \frac{1^{2}}{2^{2}} - 11 (\frac{1}{2}) + 6$

One to any power is one.

$\frac{1}{4} - 3 \frac{1}{2^{2}} - 11 (\frac{1}{2}) + 6$

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

$\frac{1}{4} - 3 (\frac{1}{4}) - 11 (\frac{1}{2}) + 6$

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

$\frac{1}{4} + \frac{- 3}{4} - 11 (\frac{1}{2}) + 6$

Move the negative in front of the fraction.

$\frac{1}{4} - \frac{3}{4} - 11 (\frac{1}{2}) + 6$

Combine $- 11$ and $\frac{1}{2}$ .

$\frac{1}{4} - \frac{3}{4} + \frac{- 11}{2} + 6$

Move the negative in front of the fraction.

$\frac{1}{4} - \frac{3}{4} - \frac{11}{2} + 6$

Combine fractions.

Combine the numerators over the common denominator.

$6 + \frac{1 - 3}{4} + \frac{- 11}{2}$

Subtract $3$ from $1$ .

$6 + \frac{- 2}{4} + \frac{- 11}{2}$

Find the common denominator.

Write $6$ as a fraction with denominator $1$ .

$\frac{6}{1} + \frac{- 2}{4} + \frac{- 11}{2}$

Multiply $\frac{6}{1}$ by $\frac{4}{4}$ .

$\frac{6}{1} \cdot \frac{4}{4} + \frac{- 2}{4} + \frac{- 11}{2}$

Multiply $\frac{6}{1}$ by $\frac{4}{4}$ .

$\frac{6 \cdot 4}{4} + \frac{- 2}{4} + \frac{- 11}{2}$

Multiply $\frac{- 11}{2}$ by $\frac{2}{2}$ .

$\frac{6 \cdot 4}{4} + \frac{- 2}{4} + \frac{- 11}{2} \cdot \frac{2}{2}$

Multiply $\frac{- 11}{2}$ by $\frac{2}{2}$ .

$\frac{6 \cdot 4}{4} + \frac{- 2}{4} + \frac{- 11 \cdot 2}{2 \cdot 2}$

Multiply $2$ by $2$ .

$\frac{6 \cdot 4}{4} + \frac{- 2}{4} + \frac{- 11 \cdot 2}{4}$

Combine the numerators over the common denominator.

$\frac{6 \cdot 4 - 2 - 11 \cdot 2}{4}$

Simplify each term.

Multiply $6$ by $4$ .

$\frac{24 - 2 - 11 \cdot 2}{4}$

Multiply $- 11$ by $2$ .

$\frac{24 - 2 - 22}{4}$

Simplify the expression.

Subtract $2$ from $24$ .

$\frac{22 - 22}{4}$

Subtract $22$ from $22$ .

$\frac{0}{4}$

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.

$\frac{2 x^{3} - 3 x^{2} - 11 x + 6}{x - \frac{1}{2}}$

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.

$\frac{1}{2}$	$2$	$- 3$	$- 11$	$6$
	\xa0			\xa0

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

$\frac{1}{2}$	$2$	$- 3$	$- 11$	$6$
	\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 $(- 3)$ .

$\frac{1}{2}$	$2$	$- 3$	$- 11$	$6$
	\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.

$\frac{1}{2}$	$2$	$- 3$	$- 11$	$6$
	\xa0	$1$		\xa0
	$2$	$- 2$

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 $(- 11)$ .

$\frac{1}{2}$	$2$	$- 3$	$- 11$	$6$
	\xa0	$1$	$- 1$	\xa0
	$2$	$- 2$

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

$\frac{1}{2}$	$2$	$- 3$	$- 11$	$6$
	\xa0	$1$	$- 1$	\xa0
	$2$	$- 2$	$- 12$

Multiply the newest entry in the result $(- 12)$ by the divisor $(\frac{1}{2})$ and place the result of $(- 6)$ under the next term in the dividend $(6)$ .

$\frac{1}{2}$	$2$	$- 3$	$- 11$	$6$
	\xa0	$1$	$- 1$	$- 6$
	$2$	$- 2$	$- 12$

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

$\frac{1}{2}$	$2$	$- 3$	$- 11$	$6$
	\xa0	$1$	$- 1$	$- 6$
	$2$	$- 2$	$- 12$	$0$

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

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

Simplify the quotient polynomial.

$2 x^{2} - 2 x - 12$

Step 7

Factor $2$ out of $2 x^{2} - 2 x - 12$ .

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

$(x - \frac{1}{2}) (2 (x^{2}) - 2 x - 12)$

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

$(x - \frac{1}{2}) (2 (x^{2}) + 2 (- x) - 12)$

Factor $2$ out of $- 12$ .

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

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

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

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

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

Step 8

Factor.

Factor $x^{2} - x - 6$ 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 $- 6$ and whose sum is $- 1$ .

$- 3, 2$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

Step 9

Factor the left side of the equation.

Factor $2 x^{3} - 3 x^{2} - 11 x + 6$ 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.

$p = \pm 1, \pm 6, \pm 2, \pm 3 q = \pm 1, \pm 2$

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

$\pm 1, \pm 0.5, \pm 6, \pm 3, \pm 2, \pm 1.5$

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.

$2 \cdot {0.5}^{3} - 3 \cdot {0.5}^{2} - 11 \cdot 0.5 + 6$

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

$2 \cdot 0.125 - 3 \cdot {0.5}^{2} - 11 \cdot 0.5 + 6$

Multiply $2$ by $0.125$ .

$0.25 - 3 \cdot {0.5}^{2} - 11 \cdot 0.5 + 6$

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

$0.25 - 3 \cdot 0.25 - 11 \cdot 0.5 + 6$

Multiply $- 3$ by $0.25$ .

$0.25 - 0.75 - 11 \cdot 0.5 + 6$

Subtract $0.75$ from $0.25$ .

$- 0.5 - 11 \cdot 0.5 + 6$

Multiply $- 11$ by $0.5$ .

$- 0.5 - 5.5 + 6$

Subtract $5.5$ from $- 0.5$ .

$- 6 + 6$

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.

$\frac{2 x^{3} - 3 x^{2} - 11 x + 6}{2 x - 1}$

Divide $2 x^{3} - 3 x^{2} - 11 x + 6$ 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 $0$ .

$2 x$

$1$

$2 x^{3}$

$3 x^{2}$

$11 x$

$6$

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

					$x^{2}$
	$2 x$	-	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$11 x$	+	$6$

Multiply the new quotient term by the divisor.

				$x^{2}$
$2 x$	-	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$11 x$	+	$6$
			+	$2 x^{3}$	-	$x^{2}$

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

				$x^{2}$
$2 x$	-	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$11 x$	+	$6$
			-	$2 x^{3}$	+	$x^{2}$

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

				$x^{2}$
$2 x$	-	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$11 x$	+	$6$
			-	$2 x^{3}$	+	$x^{2}$
					-	$2 x^{2}$

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

				$x^{2}$
$2 x$	-	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$11 x$	+	$6$
			-	$2 x^{3}$	+	$x^{2}$
					-	$2 x^{2}$	-	$11 x$

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

				$x^{2}$	-	$x$
$2 x$	-	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$11 x$	+	$6$
			-	$2 x^{3}$	+	$x^{2}$
					-	$2 x^{2}$	-	$11 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$x$
$2 x$	-	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$11 x$	+	$6$
			-	$2 x^{3}$	+	$x^{2}$
					-	$2 x^{2}$	-	$11 x$
					-	$2 x^{2}$	+	$x$

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

				$x^{2}$	-	$x$
$2 x$	-	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$11 x$	+	$6$
			-	$2 x^{3}$	+	$x^{2}$
					-	$2 x^{2}$	-	$11 x$
					+	$2 x^{2}$	-	$x$

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

				$x^{2}$	-	$x$
$2 x$	-	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$11 x$	+	$6$
			-	$2 x^{3}$	+	$x^{2}$
					-	$2 x^{2}$	-	$11 x$
					+	$2 x^{2}$	-	$x$
							-	$12 x$

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

				$x^{2}$	-	$x$
$2 x$	-	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$11 x$	+	$6$
			-	$2 x^{3}$	+	$x^{2}$
					-	$2 x^{2}$	-	$11 x$
					+	$2 x^{2}$	-	$x$
							-	$12 x$	+	$6$

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

				$x^{2}$	-	$x$	-	$6$
$2 x$	-	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$11 x$	+	$6$
			-	$2 x^{3}$	+	$x^{2}$
					-	$2 x^{2}$	-	$11 x$
					+	$2 x^{2}$	-	$x$
							-	$12 x$	+	$6$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$x$	-	$6$
$2 x$	-	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$11 x$	+	$6$
			-	$2 x^{3}$	+	$x^{2}$
					-	$2 x^{2}$	-	$11 x$
					+	$2 x^{2}$	-	$x$
							-	$12 x$	+	$6$
							-	$12 x$	+	$6$

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

				$x^{2}$	-	$x$	-	$6$
$2 x$	-	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$11 x$	+	$6$
			-	$2 x^{3}$	+	$x^{2}$
					-	$2 x^{2}$	-	$11 x$
					+	$2 x^{2}$	-	$x$
							-	$12 x$	+	$6$
							+	$12 x$	-	$6$

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

				$x^{2}$	-	$x$	-	$6$
$2 x$	-	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$11 x$	+	$6$
			-	$2 x^{3}$	+	$x^{2}$
					-	$2 x^{2}$	-	$11 x$
					+	$2 x^{2}$	-	$x$
							-	$12 x$	+	$6$
							+	$12 x$	-	$6$
										$0$

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

$x^{2} - x - 6$

Write $2 x^{3} - 3 x^{2} - 11 x + 6$ as a set of factors.

$(2 x - 1) (x^{2} - x - 6) = 0$

Factor $x^{2} - x - 6$ 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 $- 6$ and whose sum is $- 1$ .

$- 3, 2$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

Step 10

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

$2 x - 1 = 0$

$x - 3 = 0$

$x + 2 = 0$

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Step 13

Set $x + 2$ equal to $0$ and solve for $x$ .

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 14

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

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

Step 15

Find the Roots/Zeros Using the Rational Roots Test f(x)=2x^3-3x^2-11x+6

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