Find the Roots (Zeros) 2x^3+7x^2+12x+9

$2 x^{3} + 7 x^{2} + 12 x + 9$

Step 1

Set $2 x^{3} + 7 x^{2} + 12 x + 9$ equal to $0$ .

$2 x^{3} + 7 x^{2} + 12 x + 9 = 0$

Step 2

Solve for $x$ .

Factor $2 x^{3} + 7 x^{2} + 12 x + 9$ 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.

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

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

$\pm 1, \pm 0.5, \pm 9, \pm 4.5, \pm 3, \pm 1.5$

Substitute $- 1.5$ and simplify the expression. In this case, the expression is equal to $0$ so $- 1.5$ is a root of the polynomial.

Substitute $- 1.5$ into the polynomial.

$2 {(- 1.5)}^{3} + 7 {(- 1.5)}^{2} + 12 \cdot - 1.5 + 9$

Raise $- 1.5$ to the power of $3$ .

$2 \cdot - 3.375 + 7 {(- 1.5)}^{2} + 12 \cdot - 1.5 + 9$

Multiply $2$ by $- 3.375$ .

$- 6.75 + 7 {(- 1.5)}^{2} + 12 \cdot - 1.5 + 9$

Raise $- 1.5$ to the power of $2$ .

$- 6.75 + 7 \cdot 2.25 + 12 \cdot - 1.5 + 9$

Multiply $7$ by $2.25$ .

$- 6.75 + 15.75 + 12 \cdot - 1.5 + 9$

Add $- 6.75$ and $15.75$ .

$9 + 12 \cdot - 1.5 + 9$

Multiply $12$ by $- 1.5$ .

$9 - 18 + 9$

Subtract $18$ from $9$ .

$- 9 + 9$

Add $- 9$ and $9$ .

$0$

Since $- 1.5$ is a known root, divide the polynomial by $2 x + 3$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{2 x^{3} + 7 x^{2} + 12 x + 9}{2 x + 3}$

Divide $2 x^{3} + 7 x^{2} + 12 x + 9$ by $2 x + 3$ .

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

$2 x$

$3$

$2 x^{3}$

$7 x^{2}$

$12 x$

$9$

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

					$x^{2}$
	$2 x$	+	$3$		$2 x^{3}$	+	$7 x^{2}$	+	$12 x$	+	$9$

Multiply the new quotient term by the divisor.

				$x^{2}$
$2 x$	+	$3$		$2 x^{3}$	+	$7 x^{2}$	+	$12 x$	+	$9$
			+	$2 x^{3}$	+	$3 x^{2}$

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

				$x^{2}$
$2 x$	+	$3$		$2 x^{3}$	+	$7 x^{2}$	+	$12 x$	+	$9$
			-	$2 x^{3}$	-	$3 x^{2}$

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

				$x^{2}$
$2 x$	+	$3$		$2 x^{3}$	+	$7 x^{2}$	+	$12 x$	+	$9$
			-	$2 x^{3}$	-	$3 x^{2}$
					+	$4 x^{2}$

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

				$x^{2}$
$2 x$	+	$3$		$2 x^{3}$	+	$7 x^{2}$	+	$12 x$	+	$9$
			-	$2 x^{3}$	-	$3 x^{2}$
					+	$4 x^{2}$	+	$12 x$

Divide the highest order term in the dividend $4 x^{2}$ by the highest order term in divisor $2 x$ .

				$x^{2}$	+	$2 x$
$2 x$	+	$3$		$2 x^{3}$	+	$7 x^{2}$	+	$12 x$	+	$9$
			-	$2 x^{3}$	-	$3 x^{2}$
					+	$4 x^{2}$	+	$12 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$2 x$
$2 x$	+	$3$		$2 x^{3}$	+	$7 x^{2}$	+	$12 x$	+	$9$
			-	$2 x^{3}$	-	$3 x^{2}$
					+	$4 x^{2}$	+	$12 x$
					+	$4 x^{2}$	+	$6 x$

The expression needs to be subtracted from the dividend, so change all the signs in $4 x^{2} + 6 x$

				$x^{2}$	+	$2 x$
$2 x$	+	$3$		$2 x^{3}$	+	$7 x^{2}$	+	$12 x$	+	$9$
			-	$2 x^{3}$	-	$3 x^{2}$
					+	$4 x^{2}$	+	$12 x$
					-	$4 x^{2}$	-	$6 x$

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

				$x^{2}$	+	$2 x$
$2 x$	+	$3$		$2 x^{3}$	+	$7 x^{2}$	+	$12 x$	+	$9$
			-	$2 x^{3}$	-	$3 x^{2}$
					+	$4 x^{2}$	+	$12 x$
					-	$4 x^{2}$	-	$6 x$
							+	$6 x$

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

				$x^{2}$	+	$2 x$
$2 x$	+	$3$		$2 x^{3}$	+	$7 x^{2}$	+	$12 x$	+	$9$
			-	$2 x^{3}$	-	$3 x^{2}$
					+	$4 x^{2}$	+	$12 x$
					-	$4 x^{2}$	-	$6 x$
							+	$6 x$	+	$9$

Divide the highest order term in the dividend $6 x$ by the highest order term in divisor $2 x$ .

				$x^{2}$	+	$2 x$	+	$3$
$2 x$	+	$3$		$2 x^{3}$	+	$7 x^{2}$	+	$12 x$	+	$9$
			-	$2 x^{3}$	-	$3 x^{2}$
					+	$4 x^{2}$	+	$12 x$
					-	$4 x^{2}$	-	$6 x$
							+	$6 x$	+	$9$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$2 x$	+	$3$
$2 x$	+	$3$		$2 x^{3}$	+	$7 x^{2}$	+	$12 x$	+	$9$
			-	$2 x^{3}$	-	$3 x^{2}$
					+	$4 x^{2}$	+	$12 x$
					-	$4 x^{2}$	-	$6 x$
							+	$6 x$	+	$9$
							+	$6 x$	+	$9$

The expression needs to be subtracted from the dividend, so change all the signs in $6 x + 9$

				$x^{2}$	+	$2 x$	+	$3$
$2 x$	+	$3$		$2 x^{3}$	+	$7 x^{2}$	+	$12 x$	+	$9$
			-	$2 x^{3}$	-	$3 x^{2}$
					+	$4 x^{2}$	+	$12 x$
					-	$4 x^{2}$	-	$6 x$
							+	$6 x$	+	$9$
							-	$6 x$	-	$9$

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

				$x^{2}$	+	$2 x$	+	$3$
$2 x$	+	$3$		$2 x^{3}$	+	$7 x^{2}$	+	$12 x$	+	$9$
			-	$2 x^{3}$	-	$3 x^{2}$
					+	$4 x^{2}$	+	$12 x$
					-	$4 x^{2}$	-	$6 x$
							+	$6 x$	+	$9$
							-	$6 x$	-	$9$
										$0$

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

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

Write $2 x^{3} + 7 x^{2} + 12 x + 9$ as a set of factors.

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

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

$2 x + 3 = 0$

$x^{2} + 2 x + 3 = 0$

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.

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

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

Set $x^{2} + 2 x + 3$ equal to $0$ .

$x^{2} + 2 x + 3 = 0$

Solve $x^{2} + 2 x + 3 = 0$ for $x$ .

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = 2$ , and $c = 3$ into the quadratic formula and solve for $x$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (1 \cdot 3)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

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

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 1 \cdot 3}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 3$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 3}}{2 \cdot 1}$

Multiply $- 4$ by $3$ .

$x = \frac{- 2 \pm \sqrt{4 - 12}}{2 \cdot 1}$

Subtract $12$ from $4$ .

$x = \frac{- 2 \pm \sqrt{- 8}}{2 \cdot 1}$

Rewrite $- 8$ as $- 1 (8)$ .

$x = \frac{- 2 \pm \sqrt{- 1 \cdot 8}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (8)}$ as $\sqrt{- 1} \cdot \sqrt{8}$ .

$x = \frac{- 2 \pm \sqrt{- 1} \cdot \sqrt{8}}{2 \cdot 1}$

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

$x = \frac{- 2 \pm i \cdot \sqrt{8}}{2 \cdot 1}$

Rewrite $8$ as $2^{2} \cdot 2$ .

Factor $4$ out of $8$ .

$x = \frac{- 2 \pm i \cdot \sqrt{4 (2)}}{2 \cdot 1}$

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 2 \pm i \cdot \sqrt{2^{2} \cdot 2}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{- 2 \pm i \cdot (2 \sqrt{2})}{2 \cdot 1}$

Move $2$ to the left of $i$ .

$x = \frac{- 2 \pm 2 i \sqrt{2}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 2 \pm 2 i \sqrt{2}}{2}$

Simplify $\frac{- 2 \pm 2 i \sqrt{2}}{2}$ .

$x = - 1 \pm i \sqrt{2}$

The final answer is the combination of both solutions.

$x = - 1 + i \sqrt{2}, - 1 - i \sqrt{2}$

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

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

Step 3

Find the Roots (Zeros) 2x^3+7x^2+12x+9

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