Find the Roots/Zeros Using the Rational Roots Test 2x^3+3x^2+18x+27

$2 x^{3} + 3 x^{2} + 18 x + 27$

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.

$p = \pm 1, \pm 3, \pm 9, \pm 27$

$q = \pm 1, \pm 2$

Step 2

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

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

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.

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

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 $- \frac{3}{2}$ .

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

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

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

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

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

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

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

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

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

Cancel the common factor of $2$ .

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

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

Factor $2$ out of $8$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Move the negative in front of the fraction.

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

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

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

$- \frac{27}{4} + 3 ({(- 1)}^{2} {(\frac{3}{2})}^{2}) + 18 (- \frac{3}{2}) + 27$

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

$- \frac{27}{4} + 3 ({(- 1)}^{2} \frac{3^{2}}{2^{2}}) + 18 (- \frac{3}{2}) + 27$

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

$- \frac{27}{4} + 3 (1 \frac{3^{2}}{2^{2}}) + 18 (- \frac{3}{2}) + 27$

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

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

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

$- \frac{27}{4} + 3 \frac{9}{2^{2}} + 18 (- \frac{3}{2}) + 27$

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

$- \frac{27}{4} + 3 (\frac{9}{4}) + 18 (- \frac{3}{2}) + 27$

Multiply $3 (\frac{9}{4})$ .

Combine $3$ and $\frac{9}{4}$ .

$- \frac{27}{4} + \frac{3 \cdot 9}{4} + 18 (- \frac{3}{2}) + 27$

Multiply $3$ by $9$ .

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

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{3}{2}$ into the numerator.

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

Factor $2$ out of $18$ .

$- \frac{27}{4} + \frac{27}{4} + 2 (9) \frac{- 3}{2} + 27$

Cancel the common factor.

$- \frac{27}{4} + \frac{27}{4} + 2 \cdot 9 \frac{- 3}{2} + 27$

Rewrite the expression.

$- \frac{27}{4} + \frac{27}{4} + 9 \cdot - 3 + 27$

Multiply $9$ by $- 3$ .

$- \frac{27}{4} + \frac{27}{4} - 27 + 27$

Combine fractions.

Combine the numerators over the common denominator.

$- 27 + 27 + \frac{- 27 + 27}{4}$

Add $- 27$ and $27$ .

$- 27 + 27 + \frac{0}{4}$

Find the common denominator.

Write $- 27$ as a fraction with denominator $1$ .

$\frac{- 27}{1} + 27 + \frac{0}{4}$

Multiply $\frac{- 27}{1}$ by $\frac{4}{4}$ .

$\frac{- 27}{1} \cdot \frac{4}{4} + 27 + \frac{0}{4}$

Multiply $\frac{- 27}{1}$ by $\frac{4}{4}$ .

$\frac{- 27 \cdot 4}{4} + 27 + \frac{0}{4}$

Write $27$ as a fraction with denominator $1$ .

$\frac{- 27 \cdot 4}{4} + \frac{27}{1} + \frac{0}{4}$

Multiply $\frac{27}{1}$ by $\frac{4}{4}$ .

$\frac{- 27 \cdot 4}{4} + \frac{27}{1} \cdot \frac{4}{4} + \frac{0}{4}$

Multiply $\frac{27}{1}$ by $\frac{4}{4}$ .

$\frac{- 27 \cdot 4}{4} + \frac{27 \cdot 4}{4} + \frac{0}{4}$

Combine the numerators over the common denominator.

$\frac{- 27 \cdot 4 + 27 \cdot 4}{4}$

Simplify each term.

Multiply $- 27$ by $4$ .

$\frac{- 108 + 27 \cdot 4}{4}$

Multiply $27$ by $4$ .

$\frac{- 108 + 108}{4}$

Simplify the expression.

Add $- 108$ and $108$ .

$\frac{0}{4}$

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.

$\frac{2 x^{3} + 3 x^{2} + 18 x + 27}{x + \frac{3}{2}}$

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.

$- \frac{3}{2}$	$2$	$3$	$18$	$27$
	\xa0			\xa0

The first number in the dividend $(2)$ is put into the first position of the result area (below the horizontal line).

$- \frac{3}{2}$	$2$	$3$	$18$	$27$
	\xa0			\xa0
	$2$

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

$- \frac{3}{2}$	$2$	$3$	$18$	$27$
	\xa0	$- 3$		\xa0
	$2$

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

$- \frac{3}{2}$	$2$	$3$	$18$	$27$
	\xa0	$- 3$		\xa0
	$2$	$0$

Multiply the newest entry in the result $(0)$ by the divisor $(- \frac{3}{2})$ and place the result of $(0)$ under the next term in the dividend $(18)$ .

$- \frac{3}{2}$	$2$	$3$	$18$	$27$
	\xa0	$- 3$	$0$	\xa0
	$2$	$0$

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

$- \frac{3}{2}$	$2$	$3$	$18$	$27$
	\xa0	$- 3$	$0$	\xa0
	$2$	$0$	$18$

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 $(27)$ .

$- \frac{3}{2}$	$2$	$3$	$18$	$27$
	\xa0	$- 3$	$0$	$- 27$
	$2$	$0$	$18$

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

$- \frac{3}{2}$	$2$	$3$	$18$	$27$
	\xa0	$- 3$	$0$	$- 27$
	$2$	$0$	$18$	$0$

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

$2 x^{2} + 0 x + 18$

Simplify the quotient polynomial.

$2 x^{2} + 18$

Step 7

Factor $2$ out of $2 x^{2} + 18$ .

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

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

Factor $2$ out of $18$ .

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

Factor $2$ out of $2 x^{2} + 2 \cdot 9$ .

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

Step 8

Factor the left side of the equation.

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

$x^{2} (2 x + 3) + 9 (2 x + 3) = 0$

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

$(2 x + 3) (x^{2} + 9) = 0$

Step 9

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

$2 x + 3 = 0$

$x^{2} + 9 = 0$

Step 10

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 11

Set $x^{2} + 9$ equal to $0$ and solve for $x$ .

Set $x^{2} + 9$ equal to $0$ .

$x^{2} + 9 = 0$

Solve $x^{2} + 9 = 0$ for $x$ .

Subtract $9$ from both sides of the equation.

$x^{2} = - 9$

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{- 9}$

Simplify $\pm \sqrt{- 9}$ .

Rewrite $- 9$ as $- 1 (9)$ .

$x = \pm \sqrt{- 1 (9)}$

Rewrite $\sqrt{- 1 (9)}$ as $\sqrt{- 1} \cdot \sqrt{9}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{9}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i \cdot \sqrt{9}$

Rewrite $9$ as $3^{2}$ .

$x = \pm i \cdot \sqrt{3^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm i \cdot 3$

Move $3$ to the left of $i$ .

$x = \pm 3 i$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = 3 i$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 3 i$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 3 i, - 3 i$

Step 12

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

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

Step 13

Find the Roots/Zeros Using the Rational Roots Test 2x^3+3x^2+18x+27

Related Questions

Support LCMGCF.com by disabling your adblocker.