Find the Roots/Zeros Using the Rational Roots Test f(x)=8x^3-46x^2+31x-5

$f (x) = 8 x^{3} - 46 x^{2} + 31 x - 5$

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.

$p = \pm 1, \pm 5$

$q = \pm 1, \pm 2, \pm 4, \pm 8$

Step 2

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

$\pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm \frac{1}{8}, \pm 5, \pm \frac{5}{2}, \pm \frac{5}{4}, \pm \frac{5}{8}$

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.

$8 {(\frac{1}{2})}^{3} - 46 {(\frac{1}{2})}^{2} + 31 (\frac{1}{2}) - 5$

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 $\frac{1}{2}$ .

$8 \frac{1^{3}}{2^{3}} - 46 {(\frac{1}{2})}^{2} + 31 (\frac{1}{2}) - 5$

One to any power is one.

$8 \frac{1}{2^{3}} - 46 {(\frac{1}{2})}^{2} + 31 (\frac{1}{2}) - 5$

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

$8 (\frac{1}{8}) - 46 {(\frac{1}{2})}^{2} + 31 (\frac{1}{2}) - 5$

Cancel the common factor of $8$ .

Cancel the common factor.

$8 (\frac{1}{8}) - 46 {(\frac{1}{2})}^{2} + 31 (\frac{1}{2}) - 5$

Rewrite the expression.

$1 - 46 {(\frac{1}{2})}^{2} + 31 (\frac{1}{2}) - 5$

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

$1 - 46 \frac{1^{2}}{2^{2}} + 31 (\frac{1}{2}) - 5$

One to any power is one.

$1 - 46 \frac{1}{2^{2}} + 31 (\frac{1}{2}) - 5$

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

$1 - 46 (\frac{1}{4}) + 31 (\frac{1}{2}) - 5$

Cancel the common factor of $2$ .

Factor $2$ out of $- 46$ .

$1 + 2 (- 23) \frac{1}{4} + 31 (\frac{1}{2}) - 5$

Factor $2$ out of $4$ .

$1 + 2 \cdot - 23 \frac{1}{2 \cdot 2} + 31 (\frac{1}{2}) - 5$

Cancel the common factor.

$1 + 2 \cdot - 23 \frac{1}{2 \cdot 2} + 31 (\frac{1}{2}) - 5$

Rewrite the expression.

$1 - 23 (\frac{1}{2}) + 31 (\frac{1}{2}) - 5$

Combine $- 23$ and $\frac{1}{2}$ .

$1 + \frac{- 23}{2} + 31 (\frac{1}{2}) - 5$

Move the negative in front of the fraction.

$1 - \frac{23}{2} + 31 (\frac{1}{2}) - 5$

Combine $31$ and $\frac{1}{2}$ .

$1 - \frac{23}{2} + \frac{31}{2} - 5$

Combine fractions.

Combine the numerators over the common denominator.

$1 - 5 + \frac{- 23 + 31}{2}$

Add $- 23$ and $31$ .

$1 - 5 + \frac{8}{2}$

Find the common denominator.

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

$\frac{1}{1} - 5 + \frac{8}{2}$

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

$\frac{1}{1} \cdot \frac{2}{2} - 5 + \frac{8}{2}$

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

$\frac{2}{2} - 5 + \frac{8}{2}$

Write $- 5$ as a fraction with denominator $1$ .

$\frac{2}{2} + \frac{- 5}{1} + \frac{8}{2}$

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

$\frac{2}{2} + \frac{- 5}{1} \cdot \frac{2}{2} + \frac{8}{2}$

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

$\frac{2}{2} + \frac{- 5 \cdot 2}{2} + \frac{8}{2}$

Combine the numerators over the common denominator.

$\frac{2 - 5 \cdot 2 + 8}{2}$

Simplify the expression.

Multiply $- 5$ by $2$ .

$\frac{2 - 10 + 8}{2}$

Subtract $10$ from $2$ .

$\frac{- 8 + 8}{2}$

Add $- 8$ and $8$ .

$\frac{0}{2}$

Divide $0$ by $2$ .

$0$

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.

$\frac{8 x^{3} - 46 x^{2} + 31 x - 5}{x - \frac{1}{2}}$

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.

$\frac{1}{2}$	$8$	$- 46$	$31$	$- 5$
	\xa0			\xa0

The first number in the dividend $(8)$ is put into the first position of the result area (below the horizontal line).

$\frac{1}{2}$	$8$	$- 46$	$31$	$- 5$
	\xa0			\xa0
	$8$

Multiply the newest entry in the result $(8)$ by the divisor $(\frac{1}{2})$ and place the result of $(4)$ under the next term in the dividend $(- 46)$ .

$\frac{1}{2}$	$8$	$- 46$	$31$	$- 5$
	\xa0	$4$		\xa0
	$8$

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

$\frac{1}{2}$	$8$	$- 46$	$31$	$- 5$
	\xa0	$4$		\xa0
	$8$	$- 42$

Multiply the newest entry in the result $(- 42)$ by the divisor $(\frac{1}{2})$ and place the result of $(- 21)$ under the next term in the dividend $(31)$ .

$\frac{1}{2}$	$8$	$- 46$	$31$	$- 5$
	\xa0	$4$	$- 21$	\xa0
	$8$	$- 42$

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

$\frac{1}{2}$	$8$	$- 46$	$31$	$- 5$
	\xa0	$4$	$- 21$	\xa0
	$8$	$- 42$	$10$

Multiply the newest entry in the result $(10)$ by the divisor $(\frac{1}{2})$ and place the result of $(5)$ under the next term in the dividend $(- 5)$ .

$\frac{1}{2}$	$8$	$- 46$	$31$	$- 5$
	\xa0	$4$	$- 21$	$5$
	$8$	$- 42$	$10$

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

$\frac{1}{2}$	$8$	$- 46$	$31$	$- 5$
	\xa0	$4$	$- 21$	$5$
	$8$	$- 42$	$10$	$0$

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

$8 x^{2} + - 42 x + 10$

Simplify the quotient polynomial.

$8 x^{2} - 42 x + 10$

Step 7

Factor $2$ out of $8 x^{2} - 42 x + 10$ .

Factor $2$ out of $8 x^{2}$ .

$(x - \frac{1}{2}) (2 (4 x^{2}) - 42 x + 10)$

Factor $2$ out of $- 42 x$ .

$(x - \frac{1}{2}) (2 (4 x^{2}) + 2 (- 21 x) + 10)$

Factor $2$ out of $10$ .

$(x - \frac{1}{2}) (2 (4 x^{2}) + 2 (- 21 x) + 2 (5))$

Factor $2$ out of $2 (4 x^{2}) + 2 (- 21 x)$ .

$(x - \frac{1}{2}) (2 (4 x^{2} - 21 x) + 2 (5))$

Factor $2$ out of $2 (4 x^{2} - 21 x) + 2 (5)$ .

$(x - \frac{1}{2}) (2 (4 x^{2} - 21 x + 5))$

Step 8

Factor.

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 $a \cdot c = 4 \cdot 5 = 20$ and whose sum is $b = - 21$ .

Factor $- 21$ out of $- 21 x$ .

$(x - \frac{1}{2}) (2 (4 x^{2} - 21 x + 5))$

Rewrite $- 21$ as $- 1$ plus $- 20$

$(x - \frac{1}{2}) (2 (4 x^{2} + (- 1 - 20) x + 5))$

Apply the distributive property.

$(x - \frac{1}{2}) (2 (4 x^{2} - 1 x - 20 x + 5))$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x - \frac{1}{2}) (2 ((4 x^{2} - 1 x) - 20 x + 5))$

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

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

Factor the polynomial by factoring out the greatest common factor, $4 x - 1$ .

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

Remove unnecessary parentheses.

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

Step 9

Factor the left side of the equation.

Factor $8 x^{3} - 46 x^{2} + 31 x - 5$ 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.

$p = \pm 1, \pm 5 q = \pm 1, \pm 8, \pm 2, \pm 4$

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

$\pm 1, \pm 0.125, \pm 0.5, \pm 0.25, \pm 5, \pm 0.625, \pm 2.5, \pm 1.25$

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.

$8 \cdot {0.5}^{3} - 46 \cdot {0.5}^{2} + 31 \cdot 0.5 - 5$

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

$8 \cdot 0.125 - 46 \cdot {0.5}^{2} + 31 \cdot 0.5 - 5$

Multiply $8$ by $0.125$ .

$1 - 46 \cdot {0.5}^{2} + 31 \cdot 0.5 - 5$

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

$1 - 46 \cdot 0.25 + 31 \cdot 0.5 - 5$

Multiply $- 46$ by $0.25$ .

$1 - 11.5 + 31 \cdot 0.5 - 5$

Subtract $11.5$ from $1$ .

$- 10.5 + 31 \cdot 0.5 - 5$

Multiply $31$ by $0.5$ .

$- 10.5 + 15.5 - 5$

Add $- 10.5$ and $15.5$ .

$5 - 5$

Subtract $5$ from $5$ .

$0$

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.

$\frac{8 x^{3} - 46 x^{2} + 31 x - 5}{2 x - 1}$

Divide $8 x^{3} - 46 x^{2} + 31 x - 5$ 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 $0$ .

$2 x$

$1$

$8 x^{3}$

$46 x^{2}$

$31 x$

$5$

Divide the highest order term in the dividend $8 x^{3}$ by the highest order term in divisor $2 x$ .

					$4 x^{2}$
	$2 x$	-	$1$		$8 x^{3}$	-	$46 x^{2}$	+	$31 x$	-	$5$

Multiply the new quotient term by the divisor.

				$4 x^{2}$
$2 x$	-	$1$		$8 x^{3}$	-	$46 x^{2}$	+	$31 x$	-	$5$
			+	$8 x^{3}$	-	$4 x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $8 x^{3} - 4 x^{2}$

				$4 x^{2}$
$2 x$	-	$1$		$8 x^{3}$	-	$46 x^{2}$	+	$31 x$	-	$5$
			-	$8 x^{3}$	+	$4 x^{2}$

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

				$4 x^{2}$
$2 x$	-	$1$		$8 x^{3}$	-	$46 x^{2}$	+	$31 x$	-	$5$
			-	$8 x^{3}$	+	$4 x^{2}$
					-	$42 x^{2}$

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

				$4 x^{2}$
$2 x$	-	$1$		$8 x^{3}$	-	$46 x^{2}$	+	$31 x$	-	$5$
			-	$8 x^{3}$	+	$4 x^{2}$
					-	$42 x^{2}$	+	$31 x$

Divide the highest order term in the dividend $- 42 x^{2}$ by the highest order term in divisor $2 x$ .

				$4 x^{2}$	-	$21 x$
$2 x$	-	$1$		$8 x^{3}$	-	$46 x^{2}$	+	$31 x$	-	$5$
			-	$8 x^{3}$	+	$4 x^{2}$
					-	$42 x^{2}$	+	$31 x$

Multiply the new quotient term by the divisor.

				$4 x^{2}$	-	$21 x$
$2 x$	-	$1$		$8 x^{3}$	-	$46 x^{2}$	+	$31 x$	-	$5$
			-	$8 x^{3}$	+	$4 x^{2}$
					-	$42 x^{2}$	+	$31 x$
					-	$42 x^{2}$	+	$21 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 42 x^{2} + 21 x$

				$4 x^{2}$	-	$21 x$
$2 x$	-	$1$		$8 x^{3}$	-	$46 x^{2}$	+	$31 x$	-	$5$
			-	$8 x^{3}$	+	$4 x^{2}$
					-	$42 x^{2}$	+	$31 x$
					+	$42 x^{2}$	-	$21 x$

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

				$4 x^{2}$	-	$21 x$
$2 x$	-	$1$		$8 x^{3}$	-	$46 x^{2}$	+	$31 x$	-	$5$
			-	$8 x^{3}$	+	$4 x^{2}$
					-	$42 x^{2}$	+	$31 x$
					+	$42 x^{2}$	-	$21 x$
							+	$10 x$

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

				$4 x^{2}$	-	$21 x$
$2 x$	-	$1$		$8 x^{3}$	-	$46 x^{2}$	+	$31 x$	-	$5$
			-	$8 x^{3}$	+	$4 x^{2}$
					-	$42 x^{2}$	+	$31 x$
					+	$42 x^{2}$	-	$21 x$
							+	$10 x$	-	$5$

Divide the highest order term in the dividend $10 x$ by the highest order term in divisor $2 x$ .

				$4 x^{2}$	-	$21 x$	+	$5$
$2 x$	-	$1$		$8 x^{3}$	-	$46 x^{2}$	+	$31 x$	-	$5$
			-	$8 x^{3}$	+	$4 x^{2}$
					-	$42 x^{2}$	+	$31 x$
					+	$42 x^{2}$	-	$21 x$
							+	$10 x$	-	$5$

Multiply the new quotient term by the divisor.

				$4 x^{2}$	-	$21 x$	+	$5$
$2 x$	-	$1$		$8 x^{3}$	-	$46 x^{2}$	+	$31 x$	-	$5$
			-	$8 x^{3}$	+	$4 x^{2}$
					-	$42 x^{2}$	+	$31 x$
					+	$42 x^{2}$	-	$21 x$
							+	$10 x$	-	$5$
							+	$10 x$	-	$5$

The expression needs to be subtracted from the dividend, so change all the signs in $10 x - 5$

				$4 x^{2}$	-	$21 x$	+	$5$
$2 x$	-	$1$		$8 x^{3}$	-	$46 x^{2}$	+	$31 x$	-	$5$
			-	$8 x^{3}$	+	$4 x^{2}$
					-	$42 x^{2}$	+	$31 x$
					+	$42 x^{2}$	-	$21 x$
							+	$10 x$	-	$5$
							-	$10 x$	+	$5$

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

				$4 x^{2}$	-	$21 x$	+	$5$
$2 x$	-	$1$		$8 x^{3}$	-	$46 x^{2}$	+	$31 x$	-	$5$
			-	$8 x^{3}$	+	$4 x^{2}$
					-	$42 x^{2}$	+	$31 x$
					+	$42 x^{2}$	-	$21 x$
							+	$10 x$	-	$5$
							-	$10 x$	+	$5$
										$0$

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

$4 x^{2} - 21 x + 5$

Write $8 x^{3} - 46 x^{2} + 31 x - 5$ as a set of factors.

$(2 x - 1) (4 x^{2} - 21 x + 5) = 0$

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 $a \cdot c = 4 \cdot 5 = 20$ and whose sum is $b = - 21$ .

Factor $- 21$ out of $- 21 x$ .

$(2 x - 1) (4 x^{2} - 21 x + 5) = 0$

Rewrite $- 21$ as $- 1$ plus $- 20$

$(2 x - 1) (4 x^{2} + (- 1 - 20) x + 5) = 0$

Apply the distributive property.

$(2 x - 1) (4 x^{2} - 1 x - 20 x + 5) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(2 x - 1) ((4 x^{2} - 1 x) - 20 x + 5) = 0$

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

$(2 x - 1) (x (4 x - 1) - 5 (4 x - 1)) = 0$

Factor the polynomial by factoring out the greatest common factor, $4 x - 1$ .

$(2 x - 1) ((4 x - 1) (x - 5)) = 0$

Remove unnecessary parentheses.

$(2 x - 1) (4 x - 1) (x - 5) = 0$

Step 10

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

$2 x - 1 = 0$

$4 x - 1 = 0$

$x - 5 = 0$

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 $4 x - 1$ equal to $0$ and solve for $x$ .

Set $4 x - 1$ equal to $0$ .

$4 x - 1 = 0$

Solve $4 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$4 x = 1$

Divide each term in $4 x = 1$ by $4$ and simplify.

Divide each term in $4 x = 1$ by $4$ .

$\frac{4 x}{4} = \frac{1}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{1}{4}$

Divide $x$ by $1$ .

$x = \frac{1}{4}$

Step 13

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 14

The final solution is all the values that make $(2 x - 1) (4 x - 1) (x - 5) = 0$ true.

$x = \frac{1}{2}, \frac{1}{4}, 5$

Step 15

Find the Roots/Zeros Using the Rational Roots Test f(x)=8x^3-46x^2+31x-5

Related Questions

Support LCMGCF.com by disabling your adblocker.