Find the Roots/Zeros Using the Rational Roots Test 54x^3+63x^2-39x-18

$54 x^{3} + 63 x^{2} - 39 x - 18$

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.

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

$q = \pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18, \pm 27, \pm 54$

Step 2

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

$\pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}, \pm \frac{1}{9}, \pm \frac{1}{18}, \pm \frac{1}{27}, \pm \frac{1}{54}, \pm 2, \pm \frac{2}{3}, \pm \frac{2}{9}, \pm \frac{2}{27}, \pm 3, \pm \frac{3}{2}, \pm 6, \pm 9, \pm \frac{9}{2}, \pm 18$

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.

$54 {(- \frac{3}{2})}^{3} + 63 {(- \frac{3}{2})}^{2} - 39 (- \frac{3}{2}) - 18$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = - \frac{3}{2}$ is a root of the polynomial.

Simplify each term.

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $- \frac{3}{2}$ .

$54 ({(- 1)}^{3} {(\frac{3}{2})}^{3}) + 63 {(- \frac{3}{2})}^{2} - 39 (- \frac{3}{2}) - 18$

Apply the product rule to $\frac{3}{2}$ .

$54 ({(- 1)}^{3} \frac{3^{3}}{2^{3}}) + 63 {(- \frac{3}{2})}^{2} - 39 (- \frac{3}{2}) - 18$

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

$54 (- \frac{3^{3}}{2^{3}}) + 63 {(- \frac{3}{2})}^{2} - 39 (- \frac{3}{2}) - 18$

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

$54 (- \frac{27}{2^{3}}) + 63 {(- \frac{3}{2})}^{2} - 39 (- \frac{3}{2}) - 18$

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

$54 (- \frac{27}{8}) + 63 {(- \frac{3}{2})}^{2} - 39 (- \frac{3}{2}) - 18$

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{27}{8}$ into the numerator.

$54 (\frac{- 27}{8}) + 63 {(- \frac{3}{2})}^{2} - 39 (- \frac{3}{2}) - 18$

Factor $2$ out of $54$ .

$2 (27) \frac{- 27}{8} + 63 {(- \frac{3}{2})}^{2} - 39 (- \frac{3}{2}) - 18$

Factor $2$ out of $8$ .

$2 \cdot 27 \frac{- 27}{2 \cdot 4} + 63 {(- \frac{3}{2})}^{2} - 39 (- \frac{3}{2}) - 18$

Cancel the common factor.

$2 \cdot 27 \frac{- 27}{2 \cdot 4} + 63 {(- \frac{3}{2})}^{2} - 39 (- \frac{3}{2}) - 18$

Rewrite the expression.

$27 (\frac{- 27}{4}) + 63 {(- \frac{3}{2})}^{2} - 39 (- \frac{3}{2}) - 18$

Combine $27$ and $\frac{- 27}{4}$ .

$\frac{27 \cdot - 27}{4} + 63 {(- \frac{3}{2})}^{2} - 39 (- \frac{3}{2}) - 18$

Multiply $27$ by $- 27$ .

$\frac{- 729}{4} + 63 {(- \frac{3}{2})}^{2} - 39 (- \frac{3}{2}) - 18$

Move the negative in front of the fraction.

$- \frac{729}{4} + 63 {(- \frac{3}{2})}^{2} - 39 (- \frac{3}{2}) - 18$

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $- \frac{3}{2}$ .

$- \frac{729}{4} + 63 ({(- 1)}^{2} {(\frac{3}{2})}^{2}) - 39 (- \frac{3}{2}) - 18$

Apply the product rule to $\frac{3}{2}$ .

$- \frac{729}{4} + 63 ({(- 1)}^{2} \frac{3^{2}}{2^{2}}) - 39 (- \frac{3}{2}) - 18$

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

$- \frac{729}{4} + 63 (1 \frac{3^{2}}{2^{2}}) - 39 (- \frac{3}{2}) - 18$

Multiply $\frac{3^{2}}{2^{2}}$ by $1$ .

$- \frac{729}{4} + 63 \frac{3^{2}}{2^{2}} - 39 (- \frac{3}{2}) - 18$

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

$- \frac{729}{4} + 63 \frac{9}{2^{2}} - 39 (- \frac{3}{2}) - 18$

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

$- \frac{729}{4} + 63 (\frac{9}{4}) - 39 (- \frac{3}{2}) - 18$

Multiply $63 (\frac{9}{4})$ .

Combine $63$ and $\frac{9}{4}$ .

$- \frac{729}{4} + \frac{63 \cdot 9}{4} - 39 (- \frac{3}{2}) - 18$

Multiply $63$ by $9$ .

$- \frac{729}{4} + \frac{567}{4} - 39 (- \frac{3}{2}) - 18$

Multiply $- 39 (- \frac{3}{2})$ .

Multiply $- 1$ by $- 39$ .

$- \frac{729}{4} + \frac{567}{4} + 39 (\frac{3}{2}) - 18$

Combine $39$ and $\frac{3}{2}$ .

$- \frac{729}{4} + \frac{567}{4} + \frac{39 \cdot 3}{2} - 18$

Multiply $39$ by $3$ .

$- \frac{729}{4} + \frac{567}{4} + \frac{117}{2} - 18$

Combine fractions.

Combine the numerators over the common denominator.

$- 18 + \frac{- 729 + 567}{4} + \frac{117}{2}$

Add $- 729$ and $567$ .

$- 18 + \frac{- 162}{4} + \frac{117}{2}$

Find the common denominator.

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

$\frac{- 18}{1} + \frac{- 162}{4} + \frac{117}{2}$

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

$\frac{- 18}{1} \cdot \frac{4}{4} + \frac{- 162}{4} + \frac{117}{2}$

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

$\frac{- 18 \cdot 4}{4} + \frac{- 162}{4} + \frac{117}{2}$

Multiply $\frac{117}{2}$ by $\frac{2}{2}$ .

$\frac{- 18 \cdot 4}{4} + \frac{- 162}{4} + \frac{117}{2} \cdot \frac{2}{2}$

Multiply $\frac{117}{2}$ by $\frac{2}{2}$ .

$\frac{- 18 \cdot 4}{4} + \frac{- 162}{4} + \frac{117 \cdot 2}{2 \cdot 2}$

Multiply $2$ by $2$ .

$\frac{- 18 \cdot 4}{4} + \frac{- 162}{4} + \frac{117 \cdot 2}{4}$

Combine the numerators over the common denominator.

$\frac{- 18 \cdot 4 - 162 + 117 \cdot 2}{4}$

Simplify each term.

Multiply $- 18$ by $4$ .

$\frac{- 72 - 162 + 117 \cdot 2}{4}$

Multiply $117$ by $2$ .

$\frac{- 72 - 162 + 234}{4}$

Simplify the expression.

Subtract $162$ from $- 72$ .

$\frac{- 234 + 234}{4}$

Add $- 234$ and $234$ .

$\frac{0}{4}$

Divide $0$ by $4$ .

$0$

Step 5

Since $- \frac{3}{2}$ is a known root, divide the polynomial by $x + \frac{3}{2}$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{54 x^{3} + 63 x^{2} - 39 x - 18}{x + \frac{3}{2}}$

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.

$- \frac{3}{2}$	$54$	$63$	$- 39$	$- 18$
	\xa0			\xa0

The first number in the dividend $(54)$ is put into the first position of the result area (below the horizontal line).

$- \frac{3}{2}$	$54$	$63$	$- 39$	$- 18$
	\xa0			\xa0
	$54$

Multiply the newest entry in the result $(54)$ by the divisor $(- \frac{3}{2})$ and place the result of $(- 81)$ under the next term in the dividend $(63)$ .

$- \frac{3}{2}$	$54$	$63$	$- 39$	$- 18$
	\xa0	$- 81$		\xa0
	$54$

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

$- \frac{3}{2}$	$54$	$63$	$- 39$	$- 18$
	\xa0	$- 81$		\xa0
	$54$	$- 18$

Multiply the newest entry in the result $(- 18)$ by the divisor $(- \frac{3}{2})$ and place the result of $(27)$ under the next term in the dividend $(- 39)$ .

$- \frac{3}{2}$	$54$	$63$	$- 39$	$- 18$
	\xa0	$- 81$	$27$	\xa0
	$54$	$- 18$

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

$- \frac{3}{2}$	$54$	$63$	$- 39$	$- 18$
	\xa0	$- 81$	$27$	\xa0
	$54$	$- 18$	$- 12$

Multiply the newest entry in the result $(- 12)$ by the divisor $(- \frac{3}{2})$ and place the result of $(18)$ under the next term in the dividend $(- 18)$ .

$- \frac{3}{2}$	$54$	$63$	$- 39$	$- 18$
	\xa0	$- 81$	$27$	$18$
	$54$	$- 18$	$- 12$

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

$- \frac{3}{2}$	$54$	$63$	$- 39$	$- 18$
	\xa0	$- 81$	$27$	$18$
	$54$	$- 18$	$- 12$	$0$

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

$54 x^{2} + - 18 x - 12$

Simplify the quotient polynomial.

$54 x^{2} - 18 x - 12$

Step 7

Factor $6$ out of $54 x^{2} - 18 x - 12$ .

Factor $6$ out of $54 x^{2}$ .

$(x + \frac{3}{2}) (6 (9 x^{2}) - 18 x - 12)$

Factor $6$ out of $- 18 x$ .

$(x + \frac{3}{2}) (6 (9 x^{2}) + 6 (- 3 x) - 12)$

Factor $6$ out of $- 12$ .

$(x + \frac{3}{2}) (6 (9 x^{2}) + 6 (- 3 x) + 6 (- 2))$

Factor $6$ out of $6 (9 x^{2}) + 6 (- 3 x)$ .

$(x + \frac{3}{2}) (6 (9 x^{2} - 3 x) + 6 (- 2))$

Factor $6$ out of $6 (9 x^{2} - 3 x) + 6 (- 2)$ .

$(x + \frac{3}{2}) (6 (9 x^{2} - 3 x - 2))$

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 $a \cdot c = 9 \cdot - 2 = - 18$ and whose sum is $b = - 3$ .

Factor $- 3$ out of $- 3 x$ .

$(x + \frac{3}{2}) (6 (9 x^{2} - 3 x - 2))$

Rewrite $- 3$ as $3$ plus $- 6$

$(x + \frac{3}{2}) (6 (9 x^{2} + (3 - 6) x - 2))$

Apply the distributive property.

$(x + \frac{3}{2}) (6 (9 x^{2} + 3 x - 6 x - 2))$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x + \frac{3}{2}) (6 ((9 x^{2} + 3 x) - 6 x - 2))$

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

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

Factor the polynomial by factoring out the greatest common factor, $3 x + 1$ .

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

Remove unnecessary parentheses.

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

Step 9

Factor the left side of the equation.

Factor $3$ out of $54 x^{3} + 63 x^{2} - 39 x - 18$ .

Factor $3$ out of $54 x^{3}$ .

$3 (18 x^{3}) + 63 x^{2} - 39 x - 18 = 0$

Factor $3$ out of $63 x^{2}$ .

$3 (18 x^{3}) + 3 (21 x^{2}) - 39 x - 18 = 0$

Factor $3$ out of $- 39 x$ .

$3 (18 x^{3}) + 3 (21 x^{2}) + 3 (- 13 x) - 18 = 0$

Factor $3$ out of $- 18$ .

$3 (18 x^{3}) + 3 (21 x^{2}) + 3 (- 13 x) + 3 (- 6) = 0$

Factor $3$ out of $3 (18 x^{3}) + 3 (21 x^{2})$ .

$3 (18 x^{3} + 21 x^{2}) + 3 (- 13 x) + 3 (- 6) = 0$

Factor $3$ out of $3 (18 x^{3} + 21 x^{2}) + 3 (- 13 x)$ .

$3 (18 x^{3} + 21 x^{2} - 13 x) + 3 (- 6) = 0$

Factor $3$ out of $3 (18 x^{3} + 21 x^{2} - 13 x) + 3 (- 6)$ .

$3 (18 x^{3} + 21 x^{2} - 13 x - 6) = 0$

Factor $18 x^{3} + 21 x^{2} - 13 x - 6$ 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.

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

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

$\pm 1, \pm 0.0\overline{5}, \pm 0.5, \pm 0.\overline{1}, \pm 0.\overline{3}, \pm 0.1\overline{6}, \pm 6, \pm 3, \pm 0.\overline{6}, \pm 2, \pm 0.\overline{2}, \pm 1.5$

Substitute $- 0.\overline{3}$ and simplify the expression. In this case, the expression is equal to $0$ so $- 0.\overline{3}$ is a root of the polynomial.

Substitute $- 0.\overline{3}$ into the polynomial.

$18 {(- 0.\overline{3})}^{3} + 21 {(- 0.\overline{3})}^{2} - 13 \cdot - 0.\overline{3} - 6$

Raise $- 0.\overline{3}$ to the power of $3$ .

$18 \cdot - 0.\overline{037} + 21 {(- 0.\overline{3})}^{2} - 13 \cdot - 0.\overline{3} - 6$

Multiply $18$ by $- 0.\overline{037}$ .

$- 0.\overline{6} + 21 {(- 0.\overline{3})}^{2} - 13 \cdot - 0.\overline{3} - 6$

Raise $- 0.\overline{3}$ to the power of $2$ .

$- 0.\overline{6} + 21 \cdot 0.\overline{1} - 13 \cdot - 0.\overline{3} - 6$

Multiply $21$ by $0.\overline{1}$ .

$- 0.\overline{6} + 2.\overline{3} - 13 \cdot - 0.\overline{3} - 6$

Add $- 0.\overline{6}$ and $2.\overline{3}$ .

$1.\overline{6} - 13 \cdot - 0.\overline{3} - 6$

Multiply $- 13$ by $- 0.\overline{3}$ .

$1.\overline{6} + 4.33333333 - 6$

Add $1.\overline{6}$ and $4.33333333$ .

$6 - 6$

Subtract $6$ from $6$ .

$0$

Since $- 0.\overline{3}$ is a known root, divide the polynomial by $3 x + 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{18 x^{3} + 21 x^{2} - 13 x - 6}{3 x + 1}$

Divide $18 x^{3} + 21 x^{2} - 13 x - 6$ by $3 x + 1$ .

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

$3 x$

$1$

$18 x^{3}$

$21 x^{2}$

$13 x$

$6$

Divide the highest order term in the dividend $18 x^{3}$ by the highest order term in divisor $3 x$ .

					$6 x^{2}$
	$3 x$	+	$1$		$18 x^{3}$	+	$21 x^{2}$	-	$13 x$	-	$6$

Multiply the new quotient term by the divisor.

				$6 x^{2}$
$3 x$	+	$1$		$18 x^{3}$	+	$21 x^{2}$	-	$13 x$	-	$6$
			+	$18 x^{3}$	+	$6 x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $18 x^{3} + 6 x^{2}$

				$6 x^{2}$
$3 x$	+	$1$		$18 x^{3}$	+	$21 x^{2}$	-	$13 x$	-	$6$
			-	$18 x^{3}$	-	$6 x^{2}$

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

				$6 x^{2}$
$3 x$	+	$1$		$18 x^{3}$	+	$21 x^{2}$	-	$13 x$	-	$6$
			-	$18 x^{3}$	-	$6 x^{2}$
					+	$15 x^{2}$

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

				$6 x^{2}$
$3 x$	+	$1$		$18 x^{3}$	+	$21 x^{2}$	-	$13 x$	-	$6$
			-	$18 x^{3}$	-	$6 x^{2}$
					+	$15 x^{2}$	-	$13 x$

Divide the highest order term in the dividend $15 x^{2}$ by the highest order term in divisor $3 x$ .

				$6 x^{2}$	+	$5 x$
$3 x$	+	$1$		$18 x^{3}$	+	$21 x^{2}$	-	$13 x$	-	$6$
			-	$18 x^{3}$	-	$6 x^{2}$
					+	$15 x^{2}$	-	$13 x$

Multiply the new quotient term by the divisor.

				$6 x^{2}$	+	$5 x$
$3 x$	+	$1$		$18 x^{3}$	+	$21 x^{2}$	-	$13 x$	-	$6$
			-	$18 x^{3}$	-	$6 x^{2}$
					+	$15 x^{2}$	-	$13 x$
					+	$15 x^{2}$	+	$5 x$

The expression needs to be subtracted from the dividend, so change all the signs in $15 x^{2} + 5 x$

				$6 x^{2}$	+	$5 x$
$3 x$	+	$1$		$18 x^{3}$	+	$21 x^{2}$	-	$13 x$	-	$6$
			-	$18 x^{3}$	-	$6 x^{2}$
					+	$15 x^{2}$	-	$13 x$
					-	$15 x^{2}$	-	$5 x$

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

				$6 x^{2}$	+	$5 x$
$3 x$	+	$1$		$18 x^{3}$	+	$21 x^{2}$	-	$13 x$	-	$6$
			-	$18 x^{3}$	-	$6 x^{2}$
					+	$15 x^{2}$	-	$13 x$
					-	$15 x^{2}$	-	$5 x$
							-	$18 x$

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

				$6 x^{2}$	+	$5 x$
$3 x$	+	$1$		$18 x^{3}$	+	$21 x^{2}$	-	$13 x$	-	$6$
			-	$18 x^{3}$	-	$6 x^{2}$
					+	$15 x^{2}$	-	$13 x$
					-	$15 x^{2}$	-	$5 x$
							-	$18 x$	-	$6$

Divide the highest order term in the dividend $- 18 x$ by the highest order term in divisor $3 x$ .

				$6 x^{2}$	+	$5 x$	-	$6$
$3 x$	+	$1$		$18 x^{3}$	+	$21 x^{2}$	-	$13 x$	-	$6$
			-	$18 x^{3}$	-	$6 x^{2}$
					+	$15 x^{2}$	-	$13 x$
					-	$15 x^{2}$	-	$5 x$
							-	$18 x$	-	$6$

Multiply the new quotient term by the divisor.

				$6 x^{2}$	+	$5 x$	-	$6$
$3 x$	+	$1$		$18 x^{3}$	+	$21 x^{2}$	-	$13 x$	-	$6$
			-	$18 x^{3}$	-	$6 x^{2}$
					+	$15 x^{2}$	-	$13 x$
					-	$15 x^{2}$	-	$5 x$
							-	$18 x$	-	$6$
							-	$18 x$	-	$6$

The expression needs to be subtracted from the dividend, so change all the signs in $- 18 x - 6$

				$6 x^{2}$	+	$5 x$	-	$6$
$3 x$	+	$1$		$18 x^{3}$	+	$21 x^{2}$	-	$13 x$	-	$6$
			-	$18 x^{3}$	-	$6 x^{2}$
					+	$15 x^{2}$	-	$13 x$
					-	$15 x^{2}$	-	$5 x$
							-	$18 x$	-	$6$
							+	$18 x$	+	$6$

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

				$6 x^{2}$	+	$5 x$	-	$6$
$3 x$	+	$1$		$18 x^{3}$	+	$21 x^{2}$	-	$13 x$	-	$6$
			-	$18 x^{3}$	-	$6 x^{2}$
					+	$15 x^{2}$	-	$13 x$
					-	$15 x^{2}$	-	$5 x$
							-	$18 x$	-	$6$
							+	$18 x$	+	$6$
										$0$

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

$6 x^{2} + 5 x - 6$

Write $18 x^{3} + 21 x^{2} - 13 x - 6$ as a set of factors.

$3 ((3 x + 1) (6 x^{2} + 5 x - 6)) = 0$

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 $a \cdot c = 6 \cdot - 6 = - 36$ and whose sum is $b = 5$ .

Factor $5$ out of $5 x$ .

$3 ((3 x + 1) (6 x^{2} + 5 (x) - 6)) = 0$

Rewrite $5$ as $- 4$ plus $9$

$3 ((3 x + 1) (6 x^{2} + (- 4 + 9) x - 6)) = 0$

Apply the distributive property.

$3 ((3 x + 1) (6 x^{2} - 4 x + 9 x - 6)) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$3 ((3 x + 1) ((6 x^{2} - 4 x) + 9 x - 6)) = 0$

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

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

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

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

Remove unnecessary parentheses.

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

Remove unnecessary parentheses.

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

Step 10

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

$3 x + 1 = 0$

$3 x - 2 = 0$

$2 x + 3 = 0$

Step 11

Set $3 x + 1$ equal to $0$ and solve for $x$ .

Set $3 x + 1$ equal to $0$ .

$3 x + 1 = 0$

Solve $3 x + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$3 x = - 1$

Divide each term in $3 x = - 1$ by $3$ and simplify.

Divide each term in $3 x = - 1$ by $3$ .

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

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 $x$ .

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 $1$ .

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

Step 13

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

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

$2 x + 3 = 0$

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

Subtract $3$ from both sides of the equation.

$2 x = - 3$

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

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Step 14

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

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

Step 15

Find the Roots/Zeros Using the Rational Roots Test 54x^3+63x^2-39x-18

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