Find the Roots/Zeros Using the Rational Roots Test f(x)=6x^3-53x^2-11x+18

$f (x) = 6 x^{3} - 53 x^{2} - 11 x + 18$

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.

$p = \pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18$

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

Step 2

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

$\pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}, \pm 2, \pm \frac{2}{3}, \pm 3, \pm \frac{3}{2}, \pm 6, \pm 9, \pm \frac{9}{2}, \pm 18$

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.

$6 {(\frac{1}{2})}^{3} - 53 {(\frac{1}{2})}^{2} - 11 (\frac{1}{2}) + 18$

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 $\frac{1}{2}$ .

$6 \frac{1^{3}}{2^{3}} - 53 {(\frac{1}{2})}^{2} - 11 (\frac{1}{2}) + 18$

One to any power is one.

$6 \frac{1}{2^{3}} - 53 {(\frac{1}{2})}^{2} - 11 (\frac{1}{2}) + 18$

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

$6 (\frac{1}{8}) - 53 {(\frac{1}{2})}^{2} - 11 (\frac{1}{2}) + 18$

Cancel the common factor of $2$ .

Factor $2$ out of $6$ .

$2 (3) \frac{1}{8} - 53 {(\frac{1}{2})}^{2} - 11 (\frac{1}{2}) + 18$

Factor $2$ out of $8$ .

$2 \cdot 3 \frac{1}{2 \cdot 4} - 53 {(\frac{1}{2})}^{2} - 11 (\frac{1}{2}) + 18$

Cancel the common factor.

$2 \cdot 3 \frac{1}{2 \cdot 4} - 53 {(\frac{1}{2})}^{2} - 11 (\frac{1}{2}) + 18$

Rewrite the expression.

$3 (\frac{1}{4}) - 53 {(\frac{1}{2})}^{2} - 11 (\frac{1}{2}) + 18$

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

$\frac{3}{4} - 53 {(\frac{1}{2})}^{2} - 11 (\frac{1}{2}) + 18$

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

$\frac{3}{4} - 53 \frac{1^{2}}{2^{2}} - 11 (\frac{1}{2}) + 18$

One to any power is one.

$\frac{3}{4} - 53 \frac{1}{2^{2}} - 11 (\frac{1}{2}) + 18$

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

$\frac{3}{4} - 53 (\frac{1}{4}) - 11 (\frac{1}{2}) + 18$

Combine $- 53$ and $\frac{1}{4}$ .

$\frac{3}{4} + \frac{- 53}{4} - 11 (\frac{1}{2}) + 18$

Move the negative in front of the fraction.

$\frac{3}{4} - \frac{53}{4} - 11 (\frac{1}{2}) + 18$

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

$\frac{3}{4} - \frac{53}{4} + \frac{- 11}{2} + 18$

Move the negative in front of the fraction.

$\frac{3}{4} - \frac{53}{4} - \frac{11}{2} + 18$

Combine fractions.

Combine the numerators over the common denominator.

$18 + \frac{3 - 53}{4} + \frac{- 11}{2}$

Subtract $53$ from $3$ .

$18 + \frac{- 50}{4} + \frac{- 11}{2}$

Find the common denominator.

Write $18$ as a fraction with denominator $1$ .

$\frac{18}{1} + \frac{- 50}{4} + \frac{- 11}{2}$

Multiply $\frac{18}{1}$ by $\frac{4}{4}$ .

$\frac{18}{1} \cdot \frac{4}{4} + \frac{- 50}{4} + \frac{- 11}{2}$

Multiply $\frac{18}{1}$ by $\frac{4}{4}$ .

$\frac{18 \cdot 4}{4} + \frac{- 50}{4} + \frac{- 11}{2}$

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

$\frac{18 \cdot 4}{4} + \frac{- 50}{4} + \frac{- 11}{2} \cdot \frac{2}{2}$

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

$\frac{18 \cdot 4}{4} + \frac{- 50}{4} + \frac{- 11 \cdot 2}{2 \cdot 2}$

Multiply $2$ by $2$ .

$\frac{18 \cdot 4}{4} + \frac{- 50}{4} + \frac{- 11 \cdot 2}{4}$

Combine the numerators over the common denominator.

$\frac{18 \cdot 4 - 50 - 11 \cdot 2}{4}$

Simplify each term.

Multiply $18$ by $4$ .

$\frac{72 - 50 - 11 \cdot 2}{4}$

Multiply $- 11$ by $2$ .

$\frac{72 - 50 - 22}{4}$

Simplify the expression.

Subtract $50$ from $72$ .

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

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.

$\frac{6 x^{3} - 53 x^{2} - 11 x + 18}{x - \frac{1}{2}}$

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.

$\frac{1}{2}$	$6$	$- 53$	$- 11$	$18$
	\xa0			\xa0

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

$\frac{1}{2}$	$6$	$- 53$	$- 11$	$18$
	\xa0			\xa0
	$6$

Multiply the newest entry in the result $(6)$ by the divisor $(\frac{1}{2})$ and place the result of $(3)$ under the next term in the dividend $(- 53)$ .

$\frac{1}{2}$	$6$	$- 53$	$- 11$	$18$
	\xa0	$3$		\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.

$\frac{1}{2}$	$6$	$- 53$	$- 11$	$18$
	\xa0	$3$		\xa0
	$6$	$- 50$

Multiply the newest entry in the result $(- 50)$ by the divisor $(\frac{1}{2})$ and place the result of $(- 25)$ under the next term in the dividend $(- 11)$ .

$\frac{1}{2}$	$6$	$- 53$	$- 11$	$18$
	\xa0	$3$	$- 25$	\xa0
	$6$	$- 50$

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

$\frac{1}{2}$	$6$	$- 53$	$- 11$	$18$
	\xa0	$3$	$- 25$	\xa0
	$6$	$- 50$	$- 36$

Multiply the newest entry in the result $(- 36)$ by the divisor $(\frac{1}{2})$ and place the result of $(- 18)$ under the next term in the dividend $(18)$ .

$\frac{1}{2}$	$6$	$- 53$	$- 11$	$18$
	\xa0	$3$	$- 25$	$- 18$
	$6$	$- 50$	$- 36$

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

$\frac{1}{2}$	$6$	$- 53$	$- 11$	$18$
	\xa0	$3$	$- 25$	$- 18$
	$6$	$- 50$	$- 36$	$0$

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

$6 x^{2} + - 50 x - 36$

Simplify the quotient polynomial.

$6 x^{2} - 50 x - 36$

Step 7

Factor $2$ out of $6 x^{2} - 50 x - 36$ .

Factor $2$ out of $6 x^{2}$ .

$(x - \frac{1}{2}) (2 (3 x^{2}) - 50 x - 36)$

Factor $2$ out of $- 50 x$ .

$(x - \frac{1}{2}) (2 (3 x^{2}) + 2 (- 25 x) - 36)$

Factor $2$ out of $- 36$ .

$(x - \frac{1}{2}) (2 (3 x^{2}) + 2 (- 25 x) + 2 (- 18))$

Factor $2$ out of $2 (3 x^{2}) + 2 (- 25 x)$ .

$(x - \frac{1}{2}) (2 (3 x^{2} - 25 x) + 2 (- 18))$

Factor $2$ out of $2 (3 x^{2} - 25 x) + 2 (- 18)$ .

$(x - \frac{1}{2}) (2 (3 x^{2} - 25 x - 18))$

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 $a \cdot c = 3 \cdot - 18 = - 54$ and whose sum is $b = - 25$ .

Factor $- 25$ out of $- 25 x$ .

$(x - \frac{1}{2}) (2 (3 x^{2} - 25 x - 18))$

Rewrite $- 25$ as $2$ plus $- 27$

$(x - \frac{1}{2}) (2 (3 x^{2} + (2 - 27) x - 18))$

Apply the distributive property.

$(x - \frac{1}{2}) (2 (3 x^{2} + 2 x - 27 x - 18))$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x - \frac{1}{2}) (2 ((3 x^{2} + 2 x) - 27 x - 18))$

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

$(x - \frac{1}{2}) (2 (x (3 x + 2) - 9 (3 x + 2)))$

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

$(x - \frac{1}{2}) (2 ((3 x + 2) (x - 9)))$

Remove unnecessary parentheses.

$(x - \frac{1}{2}) (2 (3 x + 2) (x - 9))$

Step 9

Factor the left side of the equation.

Factor $6 x^{3} - 53 x^{2} - 11 x + 18$ 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.

$p = \pm 1, \pm 18, \pm 2, \pm 9, \pm 3, \pm 6 q = \pm 1, \pm 6, \pm 2, \pm 3$

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

$\pm 1, \pm 0.1\overline{6}, \pm 0.5, \pm 0.\overline{3}, \pm 18, \pm 3, \pm 9, \pm 6, \pm 2, \pm 0.\overline{6}, \pm 1.5, \pm 4.5$

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.

$6 \cdot {0.5}^{3} - 53 \cdot {0.5}^{2} - 11 \cdot 0.5 + 18$

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

$6 \cdot 0.125 - 53 \cdot {0.5}^{2} - 11 \cdot 0.5 + 18$

Multiply $6$ by $0.125$ .

$0.75 - 53 \cdot {0.5}^{2} - 11 \cdot 0.5 + 18$

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

$0.75 - 53 \cdot 0.25 - 11 \cdot 0.5 + 18$

Multiply $- 53$ by $0.25$ .

$0.75 - 13.25 - 11 \cdot 0.5 + 18$

Subtract $13.25$ from $0.75$ .

$- 12.5 - 11 \cdot 0.5 + 18$

Multiply $- 11$ by $0.5$ .

$- 12.5 - 5.5 + 18$

Subtract $5.5$ from $- 12.5$ .

$- 18 + 18$

Add $- 18$ and $18$ .

$0$

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.

$\frac{6 x^{3} - 53 x^{2} - 11 x + 18}{2 x - 1}$

Divide $6 x^{3} - 53 x^{2} - 11 x + 18$ 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 $0$ .

$2 x$

$1$

$6 x^{3}$

$53 x^{2}$

$11 x$

$18$

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

					$3 x^{2}$
	$2 x$	-	$1$		$6 x^{3}$	-	$53 x^{2}$	-	$11 x$	+	$18$

Multiply the new quotient term by the divisor.

				$3 x^{2}$
$2 x$	-	$1$		$6 x^{3}$	-	$53 x^{2}$	-	$11 x$	+	$18$
			+	$6 x^{3}$	-	$3 x^{2}$

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

				$3 x^{2}$
$2 x$	-	$1$		$6 x^{3}$	-	$53 x^{2}$	-	$11 x$	+	$18$
			-	$6 x^{3}$	+	$3 x^{2}$

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

				$3 x^{2}$
$2 x$	-	$1$		$6 x^{3}$	-	$53 x^{2}$	-	$11 x$	+	$18$
			-	$6 x^{3}$	+	$3 x^{2}$
					-	$50 x^{2}$

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

				$3 x^{2}$
$2 x$	-	$1$		$6 x^{3}$	-	$53 x^{2}$	-	$11 x$	+	$18$
			-	$6 x^{3}$	+	$3 x^{2}$
					-	$50 x^{2}$	-	$11 x$

Divide the highest order term in the dividend $- 50 x^{2}$ by the highest order term in divisor $2 x$ .

				$3 x^{2}$	-	$25 x$
$2 x$	-	$1$		$6 x^{3}$	-	$53 x^{2}$	-	$11 x$	+	$18$
			-	$6 x^{3}$	+	$3 x^{2}$
					-	$50 x^{2}$	-	$11 x$

Multiply the new quotient term by the divisor.

				$3 x^{2}$	-	$25 x$
$2 x$	-	$1$		$6 x^{3}$	-	$53 x^{2}$	-	$11 x$	+	$18$
			-	$6 x^{3}$	+	$3 x^{2}$
					-	$50 x^{2}$	-	$11 x$
					-	$50 x^{2}$	+	$25 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 50 x^{2} + 25 x$

				$3 x^{2}$	-	$25 x$
$2 x$	-	$1$		$6 x^{3}$	-	$53 x^{2}$	-	$11 x$	+	$18$
			-	$6 x^{3}$	+	$3 x^{2}$
					-	$50 x^{2}$	-	$11 x$
					+	$50 x^{2}$	-	$25 x$

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

				$3 x^{2}$	-	$25 x$
$2 x$	-	$1$		$6 x^{3}$	-	$53 x^{2}$	-	$11 x$	+	$18$
			-	$6 x^{3}$	+	$3 x^{2}$
					-	$50 x^{2}$	-	$11 x$
					+	$50 x^{2}$	-	$25 x$
							-	$36 x$

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

				$3 x^{2}$	-	$25 x$
$2 x$	-	$1$		$6 x^{3}$	-	$53 x^{2}$	-	$11 x$	+	$18$
			-	$6 x^{3}$	+	$3 x^{2}$
					-	$50 x^{2}$	-	$11 x$
					+	$50 x^{2}$	-	$25 x$
							-	$36 x$	+	$18$

Divide the highest order term in the dividend $- 36 x$ by the highest order term in divisor $2 x$ .

				$3 x^{2}$	-	$25 x$	-	$18$
$2 x$	-	$1$		$6 x^{3}$	-	$53 x^{2}$	-	$11 x$	+	$18$
			-	$6 x^{3}$	+	$3 x^{2}$
					-	$50 x^{2}$	-	$11 x$
					+	$50 x^{2}$	-	$25 x$
							-	$36 x$	+	$18$

Multiply the new quotient term by the divisor.

				$3 x^{2}$	-	$25 x$	-	$18$
$2 x$	-	$1$		$6 x^{3}$	-	$53 x^{2}$	-	$11 x$	+	$18$
			-	$6 x^{3}$	+	$3 x^{2}$
					-	$50 x^{2}$	-	$11 x$
					+	$50 x^{2}$	-	$25 x$
							-	$36 x$	+	$18$
							-	$36 x$	+	$18$

The expression needs to be subtracted from the dividend, so change all the signs in $- 36 x + 18$

				$3 x^{2}$	-	$25 x$	-	$18$
$2 x$	-	$1$		$6 x^{3}$	-	$53 x^{2}$	-	$11 x$	+	$18$
			-	$6 x^{3}$	+	$3 x^{2}$
					-	$50 x^{2}$	-	$11 x$
					+	$50 x^{2}$	-	$25 x$
							-	$36 x$	+	$18$
							+	$36 x$	-	$18$

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

				$3 x^{2}$	-	$25 x$	-	$18$
$2 x$	-	$1$		$6 x^{3}$	-	$53 x^{2}$	-	$11 x$	+	$18$
			-	$6 x^{3}$	+	$3 x^{2}$
					-	$50 x^{2}$	-	$11 x$
					+	$50 x^{2}$	-	$25 x$
							-	$36 x$	+	$18$
							+	$36 x$	-	$18$
										$0$

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

$3 x^{2} - 25 x - 18$

Write $6 x^{3} - 53 x^{2} - 11 x + 18$ as a set of factors.

$(2 x - 1) (3 x^{2} - 25 x - 18) = 0$

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 $a \cdot c = 3 \cdot - 18 = - 54$ and whose sum is $b = - 25$ .

Factor $- 25$ out of $- 25 x$ .

$(2 x - 1) (3 x^{2} - 25 x - 18) = 0$

Rewrite $- 25$ as $2$ plus $- 27$

$(2 x - 1) (3 x^{2} + (2 - 27) x - 18) = 0$

Apply the distributive property.

$(2 x - 1) (3 x^{2} + 2 x - 27 x - 18) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(2 x - 1) ((3 x^{2} + 2 x) - 27 x - 18) = 0$

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

$(2 x - 1) (x (3 x + 2) - 9 (3 x + 2)) = 0$

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

$(2 x - 1) ((3 x + 2) (x - 9)) = 0$

Remove unnecessary parentheses.

$(2 x - 1) (3 x + 2) (x - 9) = 0$

Step 10

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

$2 x - 1 = 0$

$3 x + 2 = 0$

$x - 9 = 0$

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 $3 x + 2$ equal to $0$ and solve for $x$ .

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

$3 x + 2 = 0$

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

Subtract $2$ from 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 $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Step 13

Set $x - 9$ equal to $0$ and solve for $x$ .

Set $x - 9$ equal to $0$ .

$x - 9 = 0$

Add $9$ to both sides of the equation.

$x = 9$

Step 14

The final solution is all the values that make $(2 x - 1) (3 x + 2) (x - 9) = 0$ true.

$x = \frac{1}{2}, - \frac{2}{3}, 9$

Step 15

Find the Roots/Zeros Using the Rational Roots Test f(x)=6x^3-53x^2-11x+18

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