Find the Roots/Zeros Using the Rational Roots Test 6x^4-23x^3-13x^2+32x+16

$6 x^{4} - 23 x^{3} - 13 x^{2} + 32 x + 16$

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.

$p = \pm 1, \pm 2, \pm 4, \pm 8, \pm 16$

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

Step 2

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

$\pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}, \pm 2, \pm \frac{2}{3}, \pm 4, \pm \frac{4}{3}, \pm 8, \pm \frac{8}{3}, \pm 16, \pm \frac{16}{3}$

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.

$6 {(- 1)}^{4} - 23 {(- 1)}^{3} - 13 {(- 1)}^{2} + 32 (- 1) + 16$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = - 1$ is a root of the polynomial.

Simplify each term.

Raise $- 1$ to the power of $4$ .

$6 \cdot 1 - 23 {(- 1)}^{3} - 13 {(- 1)}^{2} + 32 (- 1) + 16$

Multiply $6$ by $1$ .

$6 - 23 {(- 1)}^{3} - 13 {(- 1)}^{2} + 32 (- 1) + 16$

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

$6 - 23 \cdot - 1 - 13 {(- 1)}^{2} + 32 (- 1) + 16$

Multiply $- 23$ by $- 1$ .

$6 + 23 - 13 {(- 1)}^{2} + 32 (- 1) + 16$

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

$6 + 23 - 13 \cdot 1 + 32 (- 1) + 16$

Multiply $- 13$ by $1$ .

$6 + 23 - 13 + 32 (- 1) + 16$

Multiply $32$ by $- 1$ .

$6 + 23 - 13 - 32 + 16$

Simplify by adding and subtracting.

Add $6$ and $23$ .

$29 - 13 - 32 + 16$

Subtract $13$ from $29$ .

$16 - 32 + 16$

Subtract $32$ from $16$ .

$- 16 + 16$

Add $- 16$ and $16$ .

$0$

Step 5

Since $- 1$ is a known root, divide the polynomial by $x + 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{6 x^{4} - 23 x^{3} - 13 x^{2} + 32 x + 16}{x + 1}$

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.

$- 1$	$6$	$- 23$	$- 13$	$32$	$16$
	\xa0				\xa0

The first number in the dividend $(6)$ is put into the first position of the result area (below the horizontal line).

$- 1$	$6$	$- 23$	$- 13$	$32$	$16$
	\xa0				\xa0
	$6$

Multiply the newest entry in the result $(6)$ by the divisor $(- 1)$ and place the result of $(- 6)$ under the next term in the dividend $(- 23)$ .

$- 1$	$6$	$- 23$	$- 13$	$32$	$16$
	\xa0	$- 6$			\xa0
	$6$

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

$- 1$	$6$	$- 23$	$- 13$	$32$	$16$
	\xa0	$- 6$			\xa0
	$6$	$- 29$

Multiply the newest entry in the result $(- 29)$ by the divisor $(- 1)$ and place the result of $(29)$ under the next term in the dividend $(- 13)$ .

$- 1$	$6$	$- 23$	$- 13$	$32$	$16$
	\xa0	$- 6$	$29$		\xa0
	$6$	$- 29$

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

$- 1$	$6$	$- 23$	$- 13$	$32$	$16$
	\xa0	$- 6$	$29$		\xa0
	$6$	$- 29$	$16$

Multiply the newest entry in the result $(16)$ by the divisor $(- 1)$ and place the result of $(- 16)$ under the next term in the dividend $(32)$ .

$- 1$	$6$	$- 23$	$- 13$	$32$	$16$
	\xa0	$- 6$	$29$	$- 16$	\xa0
	$6$	$- 29$	$16$

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

$- 1$	$6$	$- 23$	$- 13$	$32$	$16$
	\xa0	$- 6$	$29$	$- 16$	\xa0
	$6$	$- 29$	$16$	$16$

Multiply the newest entry in the result $(16)$ by the divisor $(- 1)$ and place the result of $(- 16)$ under the next term in the dividend $(16)$ .

$- 1$	$6$	$- 23$	$- 13$	$32$	$16$
	\xa0	$- 6$	$29$	$- 16$	$- 16$
	$6$	$- 29$	$16$	$16$

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

$- 1$	$6$	$- 23$	$- 13$	$32$	$16$
	\xa0	$- 6$	$29$	$- 16$	$- 16$
	$6$	$- 29$	$16$	$16$	$0$

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

$6 x^{3} + - 29 x^{2} + (16) x + 16$

Simplify the quotient polynomial.

$6 x^{3} - 29 x^{2} + 16 x + 16$

Step 7

Solve the equation to find any remaining roots.

Factor the left side of the equation.

Factor $6 x^{3} - 29 x^{2} + 16 x + 16$ 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.

$p = \pm 1, \pm 16, \pm 2, \pm 8, \pm 4 q = \pm 1, \pm 6, \pm 2, \pm 3$

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

$\pm 1, \pm 0.1\overline{6}, \pm 0.5, \pm 0.\overline{3}, \pm 16, \pm 2.\overline{6}, \pm 8, \pm 5.\overline{3}, \pm 2, \pm 0.\overline{6}, \pm 1.\overline{3}, \pm 4$

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.

$6 {(- 0.5)}^{3} - 29 {(- 0.5)}^{2} + 16 \cdot - 0.5 + 16$

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

$6 \cdot - 0.125 - 29 {(- 0.5)}^{2} + 16 \cdot - 0.5 + 16$

Multiply $6$ by $- 0.125$ .

$- 0.75 - 29 {(- 0.5)}^{2} + 16 \cdot - 0.5 + 16$

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

$- 0.75 - 29 \cdot 0.25 + 16 \cdot - 0.5 + 16$

Multiply $- 29$ by $0.25$ .

$- 0.75 - 7.25 + 16 \cdot - 0.5 + 16$

Subtract $7.25$ from $- 0.75$ .

$- 8 + 16 \cdot - 0.5 + 16$

Multiply $16$ by $- 0.5$ .

$- 8 - 8 + 16$

Subtract $8$ from $- 8$ .

$- 16 + 16$

Add $- 16$ and $16$ .

$0$

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.

$\frac{6 x^{3} - 29 x^{2} + 16 x + 16}{2 x + 1}$

Divide $6 x^{3} - 29 x^{2} + 16 x + 16$ 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 $0$ .

$2 x$

$1$

$6 x^{3}$

$29 x^{2}$

$16 x$

$16$

Divide the highest order term in the dividend $6 x^{3}$ by the highest order term in divisor $2 x$ .

					$3 x^{2}$
	$2 x$	+	$1$		$6 x^{3}$	-	$29 x^{2}$	+	$16 x$	+	$16$

Multiply the new quotient term by the divisor.

				$3 x^{2}$
$2 x$	+	$1$		$6 x^{3}$	-	$29 x^{2}$	+	$16 x$	+	$16$
			+	$6 x^{3}$	+	$3 x^{2}$

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

				$3 x^{2}$
$2 x$	+	$1$		$6 x^{3}$	-	$29 x^{2}$	+	$16 x$	+	$16$
			-	$6 x^{3}$	-	$3 x^{2}$

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

				$3 x^{2}$
$2 x$	+	$1$		$6 x^{3}$	-	$29 x^{2}$	+	$16 x$	+	$16$
			-	$6 x^{3}$	-	$3 x^{2}$
					-	$32 x^{2}$

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

				$3 x^{2}$
$2 x$	+	$1$		$6 x^{3}$	-	$29 x^{2}$	+	$16 x$	+	$16$
			-	$6 x^{3}$	-	$3 x^{2}$
					-	$32 x^{2}$	+	$16 x$

Divide the highest order term in the dividend $- 32 x^{2}$ by the highest order term in divisor $2 x$ .

				$3 x^{2}$	-	$16 x$
$2 x$	+	$1$		$6 x^{3}$	-	$29 x^{2}$	+	$16 x$	+	$16$
			-	$6 x^{3}$	-	$3 x^{2}$
					-	$32 x^{2}$	+	$16 x$

Multiply the new quotient term by the divisor.

				$3 x^{2}$	-	$16 x$
$2 x$	+	$1$		$6 x^{3}$	-	$29 x^{2}$	+	$16 x$	+	$16$
			-	$6 x^{3}$	-	$3 x^{2}$
					-	$32 x^{2}$	+	$16 x$
					-	$32 x^{2}$	-	$16 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 32 x^{2} - 16 x$

				$3 x^{2}$	-	$16 x$
$2 x$	+	$1$		$6 x^{3}$	-	$29 x^{2}$	+	$16 x$	+	$16$
			-	$6 x^{3}$	-	$3 x^{2}$
					-	$32 x^{2}$	+	$16 x$
					+	$32 x^{2}$	+	$16 x$

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

				$3 x^{2}$	-	$16 x$
$2 x$	+	$1$		$6 x^{3}$	-	$29 x^{2}$	+	$16 x$	+	$16$
			-	$6 x^{3}$	-	$3 x^{2}$
					-	$32 x^{2}$	+	$16 x$
					+	$32 x^{2}$	+	$16 x$
							+	$32 x$

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

				$3 x^{2}$	-	$16 x$
$2 x$	+	$1$		$6 x^{3}$	-	$29 x^{2}$	+	$16 x$	+	$16$
			-	$6 x^{3}$	-	$3 x^{2}$
					-	$32 x^{2}$	+	$16 x$
					+	$32 x^{2}$	+	$16 x$
							+	$32 x$	+	$16$

Divide the highest order term in the dividend $32 x$ by the highest order term in divisor $2 x$ .

				$3 x^{2}$	-	$16 x$	+	$16$
$2 x$	+	$1$		$6 x^{3}$	-	$29 x^{2}$	+	$16 x$	+	$16$
			-	$6 x^{3}$	-	$3 x^{2}$
					-	$32 x^{2}$	+	$16 x$
					+	$32 x^{2}$	+	$16 x$
							+	$32 x$	+	$16$

Multiply the new quotient term by the divisor.

				$3 x^{2}$	-	$16 x$	+	$16$
$2 x$	+	$1$		$6 x^{3}$	-	$29 x^{2}$	+	$16 x$	+	$16$
			-	$6 x^{3}$	-	$3 x^{2}$
					-	$32 x^{2}$	+	$16 x$
					+	$32 x^{2}$	+	$16 x$
							+	$32 x$	+	$16$
							+	$32 x$	+	$16$

The expression needs to be subtracted from the dividend, so change all the signs in $32 x + 16$

				$3 x^{2}$	-	$16 x$	+	$16$
$2 x$	+	$1$		$6 x^{3}$	-	$29 x^{2}$	+	$16 x$	+	$16$
			-	$6 x^{3}$	-	$3 x^{2}$
					-	$32 x^{2}$	+	$16 x$
					+	$32 x^{2}$	+	$16 x$
							+	$32 x$	+	$16$
							-	$32 x$	-	$16$

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

				$3 x^{2}$	-	$16 x$	+	$16$
$2 x$	+	$1$		$6 x^{3}$	-	$29 x^{2}$	+	$16 x$	+	$16$
			-	$6 x^{3}$	-	$3 x^{2}$
					-	$32 x^{2}$	+	$16 x$
					+	$32 x^{2}$	+	$16 x$
							+	$32 x$	+	$16$
							-	$32 x$	-	$16$
										$0$

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

$3 x^{2} - 16 x + 16$

Write $6 x^{3} - 29 x^{2} + 16 x + 16$ as a set of factors.

$(2 x + 1) (3 x^{2} - 16 x + 16) = 0$

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 $a \cdot c = 3 \cdot 16 = 48$ and whose sum is $b = - 16$ .

Factor $- 16$ out of $- 16 x$ .

$(2 x + 1) (3 x^{2} - 16 x + 16) = 0$

Rewrite $- 16$ as $- 4$ plus $- 12$

$(2 x + 1) (3 x^{2} + (- 4 - 12) x + 16) = 0$

Apply the distributive property.

$(2 x + 1) (3 x^{2} - 4 x - 12 x + 16) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(2 x + 1) ((3 x^{2} - 4 x) - 12 x + 16) = 0$

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

$(2 x + 1) (x (3 x - 4) - 4 (3 x - 4)) = 0$

Factor the polynomial by factoring out the greatest common factor, $3 x - 4$ .

$(2 x + 1) ((3 x - 4) (x - 4)) = 0$

Remove unnecessary parentheses.

$(2 x + 1) (3 x - 4) (x - 4) = 0$

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

$2 x + 1 = 0$

$3 x - 4 = 0$

$x - 4 = 0$

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.

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

Set $3 x - 4$ equal to $0$ and solve for $x$ .

Set $3 x - 4$ equal to $0$ .

$3 x - 4 = 0$

Solve $3 x - 4 = 0$ for $x$ .

Add $4$ to both sides of the equation.

$3 x = 4$

Divide each term in $3 x = 4$ by $3$ and simplify.

Divide each term in $3 x = 4$ by $3$ .

$\frac{3 x}{3} = \frac{4}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{4}{3}$

Divide $x$ by $1$ .

$x = \frac{4}{3}$

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

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

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

The final solution is all the values that make $(2 x + 1) (3 x - 4) (x - 4) = 0$ true.

$x = - \frac{1}{2}, \frac{4}{3}, 4$

Step 8

The polynomial can be written as a set of linear factors.

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

Step 9

These are the roots (zeros) of the polynomial $6 x^{4} - 23 x^{3} - 13 x^{2} + 32 x + 16$ .

$x = - 1, - \frac{1}{2}, \frac{4}{3}, 4$

Step 10

Find the Roots/Zeros Using the Rational Roots Test 6x^4-23x^3-13x^2+32x+16

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