Find the Roots/Zeros Using the Rational Roots Test f(x)=6x^4+5x^3-33x^2-12x+20

$f (x) = 6 x^{4} + 5 x^{3} - 33 x^{2} - 12 x + 20$

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.

$p = \pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20$

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

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

$\pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}, \pm 2, \pm \frac{2}{3}, \pm 4, \pm \frac{4}{3}, \pm 5, \pm \frac{5}{2}, \pm \frac{5}{3}, \pm \frac{5}{6}, \pm 10, \pm \frac{10}{3}, \pm 20, \pm \frac{20}{3}$

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.

$6 {(- 1)}^{4} + 5 {(- 1)}^{3} - 33 {(- 1)}^{2} - 12 \cdot - 1 + 20$

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 $4$ .

$6 \cdot 1 + 5 {(- 1)}^{3} - 33 {(- 1)}^{2} - 12 \cdot - 1 + 20$

Multiply $6$ by $1$ .

$6 + 5 {(- 1)}^{3} - 33 {(- 1)}^{2} - 12 \cdot - 1 + 20$

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

$6 + 5 \cdot - 1 - 33 {(- 1)}^{2} - 12 \cdot - 1 + 20$

Multiply $5$ by $- 1$ .

$6 - 5 - 33 {(- 1)}^{2} - 12 \cdot - 1 + 20$

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

$6 - 5 - 33 \cdot 1 - 12 \cdot - 1 + 20$

Multiply $- 33$ by $1$ .

$6 - 5 - 33 - 12 \cdot - 1 + 20$

Multiply $- 12$ by $- 1$ .

$6 - 5 - 33 + 12 + 20$

Simplify by adding and subtracting.

Subtract $5$ from $6$ .

$1 - 33 + 12 + 20$

Subtract $33$ from $1$ .

$- 32 + 12 + 20$

Add $- 32$ and $12$ .

$- 20 + 20$

Add $- 20$ and $20$ .

$0$

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.

$\frac{6 x^{4} + 5 x^{3} - 33 x^{2} - 12 x + 20}{x + 1}$

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.

$- 1$	$6$	$5$	$- 33$	$- 12$	$20$
	\xa0				\xa0

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

$- 1$	$6$	$5$	$- 33$	$- 12$	$20$
	\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 $(5)$ .

$- 1$	$6$	$5$	$- 33$	$- 12$	$20$
	\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.

$- 1$	$6$	$5$	$- 33$	$- 12$	$20$
	\xa0	$- 6$			\xa0
	$6$	$- 1$

Multiply the newest entry in the result $(- 1)$ by the divisor $(- 1)$ and place the result of $(1)$ under the next term in the dividend $(- 33)$ .

$- 1$	$6$	$5$	$- 33$	$- 12$	$20$
	\xa0	$- 6$	$1$		\xa0
	$6$	$- 1$

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

$- 1$	$6$	$5$	$- 33$	$- 12$	$20$
	\xa0	$- 6$	$1$		\xa0
	$6$	$- 1$	$- 32$

Multiply the newest entry in the result $(- 32)$ by the divisor $(- 1)$ and place the result of $(32)$ under the next term in the dividend $(- 12)$ .

$- 1$	$6$	$5$	$- 33$	$- 12$	$20$
	\xa0	$- 6$	$1$	$32$	\xa0
	$6$	$- 1$	$- 32$

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

$- 1$	$6$	$5$	$- 33$	$- 12$	$20$
	\xa0	$- 6$	$1$	$32$	\xa0
	$6$	$- 1$	$- 32$	$20$

Multiply the newest entry in the result $(20)$ by the divisor $(- 1)$ and place the result of $(- 20)$ under the next term in the dividend $(20)$ .

$- 1$	$6$	$5$	$- 33$	$- 12$	$20$
	\xa0	$- 6$	$1$	$32$	$- 20$
	$6$	$- 1$	$- 32$	$20$

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

$- 1$	$6$	$5$	$- 33$	$- 12$	$20$
	\xa0	$- 6$	$1$	$32$	$- 20$
	$6$	$- 1$	$- 32$	$20$	$0$

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

$6 x^{3} + - 1 x^{2} + (- 32) x + 20$

Simplify the quotient polynomial.

$6 x^{3} - x^{2} - 32 x + 20$

Solve the equation to find any remaining roots.

Factor the left side of the equation.

Factor $6 x^{3} - x^{2} - 32 x + 20$ 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.

$p = \pm 1, \pm 20, \pm 2, \pm 10, \pm 4, \pm 5 q = \pm 1, \pm 6, \pm 2, \pm 3$

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

$\pm 1, \pm 0.1\overline{6}, \pm 0.5, \pm 0.\overline{3}, \pm 20, \pm 3.\overline{3}, \pm 10, \pm 6.\overline{6}, \pm 2, \pm 0.\overline{6}, \pm 1.\overline{6}, \pm 5, \pm 4, \pm 1.\overline{3}, \pm 0.8\overline{3}, \pm 2.5$

Substitute $2$ and simplify the expression. In this case, the expression is equal to $0$ so $2$ is a root of the polynomial.

Substitute $2$ into the polynomial.

$6 \cdot 2^{3} - 2^{2} - 32 \cdot 2 + 20$

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

$6 \cdot 8 - 2^{2} - 32 \cdot 2 + 20$

Multiply $6$ by $8$ .

$48 - 2^{2} - 32 \cdot 2 + 20$

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

$48 - 1 \cdot 4 - 32 \cdot 2 + 20$

Multiply $- 1$ by $4$ .

$48 - 4 - 32 \cdot 2 + 20$

Subtract $4$ from $48$ .

$44 - 32 \cdot 2 + 20$

Multiply $- 32$ by $2$ .

$44 - 64 + 20$

Subtract $64$ from $44$ .

$- 20 + 20$

Add $- 20$ and $20$ .

$0$

Since $2$ is a known root, divide the polynomial by $x - 2$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{6 x^{3} - x^{2} - 32 x + 20}{x - 2}$

Divide $6 x^{3} - x^{2} - 32 x + 20$ by $x - 2$ .

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$x$

$2$

$6 x^{3}$

$x^{2}$

$32 x$

$20$

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

					$6 x^{2}$
	$x$	-	$2$		$6 x^{3}$	-	$x^{2}$	-	$32 x$	+	$20$

Multiply the new quotient term by the divisor.

				$6 x^{2}$
$x$	-	$2$		$6 x^{3}$	-	$x^{2}$	-	$32 x$	+	$20$
			+	$6 x^{3}$	-	$12 x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $6 x^{3} - 12 x^{2}$

				$6 x^{2}$
$x$	-	$2$		$6 x^{3}$	-	$x^{2}$	-	$32 x$	+	$20$
			-	$6 x^{3}$	+	$12 x^{2}$

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

				$6 x^{2}$
$x$	-	$2$		$6 x^{3}$	-	$x^{2}$	-	$32 x$	+	$20$
			-	$6 x^{3}$	+	$12 x^{2}$
					+	$11 x^{2}$

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

				$6 x^{2}$
$x$	-	$2$		$6 x^{3}$	-	$x^{2}$	-	$32 x$	+	$20$
			-	$6 x^{3}$	+	$12 x^{2}$
					+	$11 x^{2}$	-	$32 x$

Divide the highest order term in the dividend $11 x^{2}$ by the highest order term in divisor $x$ .

				$6 x^{2}$	+	$11 x$
$x$	-	$2$		$6 x^{3}$	-	$x^{2}$	-	$32 x$	+	$20$
			-	$6 x^{3}$	+	$12 x^{2}$
					+	$11 x^{2}$	-	$32 x$

Multiply the new quotient term by the divisor.

				$6 x^{2}$	+	$11 x$
$x$	-	$2$		$6 x^{3}$	-	$x^{2}$	-	$32 x$	+	$20$
			-	$6 x^{3}$	+	$12 x^{2}$
					+	$11 x^{2}$	-	$32 x$
					+	$11 x^{2}$	-	$22 x$

The expression needs to be subtracted from the dividend, so change all the signs in $11 x^{2} - 22 x$

				$6 x^{2}$	+	$11 x$
$x$	-	$2$		$6 x^{3}$	-	$x^{2}$	-	$32 x$	+	$20$
			-	$6 x^{3}$	+	$12 x^{2}$
					+	$11 x^{2}$	-	$32 x$
					-	$11 x^{2}$	+	$22 x$

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

				$6 x^{2}$	+	$11 x$
$x$	-	$2$		$6 x^{3}$	-	$x^{2}$	-	$32 x$	+	$20$
			-	$6 x^{3}$	+	$12 x^{2}$
					+	$11 x^{2}$	-	$32 x$
					-	$11 x^{2}$	+	$22 x$
							-	$10 x$

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

				$6 x^{2}$	+	$11 x$
$x$	-	$2$		$6 x^{3}$	-	$x^{2}$	-	$32 x$	+	$20$
			-	$6 x^{3}$	+	$12 x^{2}$
					+	$11 x^{2}$	-	$32 x$
					-	$11 x^{2}$	+	$22 x$
							-	$10 x$	+	$20$

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

				$6 x^{2}$	+	$11 x$	-	$10$
$x$	-	$2$		$6 x^{3}$	-	$x^{2}$	-	$32 x$	+	$20$
			-	$6 x^{3}$	+	$12 x^{2}$
					+	$11 x^{2}$	-	$32 x$
					-	$11 x^{2}$	+	$22 x$
							-	$10 x$	+	$20$

Multiply the new quotient term by the divisor.

				$6 x^{2}$	+	$11 x$	-	$10$
$x$	-	$2$		$6 x^{3}$	-	$x^{2}$	-	$32 x$	+	$20$
			-	$6 x^{3}$	+	$12 x^{2}$
					+	$11 x^{2}$	-	$32 x$
					-	$11 x^{2}$	+	$22 x$
							-	$10 x$	+	$20$
							-	$10 x$	+	$20$

The expression needs to be subtracted from the dividend, so change all the signs in $- 10 x + 20$

				$6 x^{2}$	+	$11 x$	-	$10$
$x$	-	$2$		$6 x^{3}$	-	$x^{2}$	-	$32 x$	+	$20$
			-	$6 x^{3}$	+	$12 x^{2}$
					+	$11 x^{2}$	-	$32 x$
					-	$11 x^{2}$	+	$22 x$
							-	$10 x$	+	$20$
							+	$10 x$	-	$20$

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

				$6 x^{2}$	+	$11 x$	-	$10$
$x$	-	$2$		$6 x^{3}$	-	$x^{2}$	-	$32 x$	+	$20$
			-	$6 x^{3}$	+	$12 x^{2}$
					+	$11 x^{2}$	-	$32 x$
					-	$11 x^{2}$	+	$22 x$
							-	$10 x$	+	$20$
							+	$10 x$	-	$20$
										$0$

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

$6 x^{2} + 11 x - 10$

Write $6 x^{3} - x^{2} - 32 x + 20$ as a set of factors.

$(x - 2) (6 x^{2} + 11 x - 10) = 0$

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 $a \cdot c = 6 \cdot - 10 = - 60$ and whose sum is $b = 11$ .

Factor $11$ out of $11 x$ .

$(x - 2) (6 x^{2} + 11 (x) - 10) = 0$

Rewrite $11$ as $- 4$ plus $15$

$(x - 2) (6 x^{2} + (- 4 + 15) x - 10) = 0$

Apply the distributive property.

$(x - 2) (6 x^{2} - 4 x + 15 x - 10) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x - 2) ((6 x^{2} - 4 x) + 15 x - 10) = 0$

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

$(x - 2) (2 x (3 x - 2) + 5 (3 x - 2)) = 0$

Factor the polynomial by factoring out the greatest common factor, $3 x - 2$ .

$(x - 2) ((3 x - 2) (2 x + 5)) = 0$

Remove unnecessary parentheses.

$(x - 2) (3 x - 2) (2 x + 5) = 0$

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

$x - 2 = 0$

$3 x - 2 = 0$

$2 x + 5 = 0$

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

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Set $3 x - 2$ equal to $0$ and solve for $x$ .

Set $3 x - 2$ equal to $0$ .

$3 x - 2 = 0$

Solve $3 x - 2 = 0$ for $x$ .

Add $2$ to both sides of the equation.

$3 x = 2$

Divide each term in $3 x = 2$ by $3$ and simplify.

Divide each term in $3 x = 2$ by $3$ .

$\frac{3 x}{3} = \frac{2}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{2}{3}$

Divide $x$ by $1$ .

$x = \frac{2}{3}$

Set $2 x + 5$ equal to $0$ and solve for $x$ .

Set $2 x + 5$ equal to $0$ .

$2 x + 5 = 0$

Solve $2 x + 5 = 0$ for $x$ .

Subtract $5$ from both sides of the equation.

$2 x = - 5$

Divide each term in $2 x = - 5$ by $2$ and simplify.

Divide each term in $2 x = - 5$ by $2$ .

$\frac{2 x}{2} = \frac{- 5}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 5}{2}$

Divide $x$ by $1$ .

$x = \frac{- 5}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{5}{2}$

The final solution is all the values that make $(x - 2) (3 x - 2) (2 x + 5) = 0$ true.

$x = 2, \frac{2}{3}, - \frac{5}{2}$

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

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

These are the roots (zeros) of the polynomial $6 x^{4} + 5 x^{3} - 33 x^{2} - 12 x + 20$ .

$x = - 1, 2, \frac{2}{3}, - \frac{5}{2}$

Find the Roots/Zeros Using the Rational Roots Test f(x)=6x^4+5x^3-33x^2-12x+20

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