Find the Roots (Zeros) 2x^3-x^2-9x+6

$2 x^{3} - x^{2} - 9 x + 6$

Step 1

Set $2 x^{3} - x^{2} - 9 x + 6$ equal to $0$ .

$2 x^{3} - x^{2} - 9 x + 6 = 0$

Step 2

Solve for $x$ .

Factor $2 x^{3} - x^{2} - 9 x + 6$ 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.

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

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

$\pm 1, \pm 0.5, \pm 6, \pm 3, \pm 2, \pm 1.5$

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.

$2 \cdot 2^{3} - 2^{2} - 9 \cdot 2 + 6$

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

$2 \cdot 8 - 2^{2} - 9 \cdot 2 + 6$

Multiply $2$ by $8$ .

$16 - 2^{2} - 9 \cdot 2 + 6$

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

$16 - 1 \cdot 4 - 9 \cdot 2 + 6$

Multiply $- 1$ by $4$ .

$16 - 4 - 9 \cdot 2 + 6$

Subtract $4$ from $16$ .

$12 - 9 \cdot 2 + 6$

Multiply $- 9$ by $2$ .

$12 - 18 + 6$

Subtract $18$ from $12$ .

$- 6 + 6$

Add $- 6$ and $6$ .

$0$

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.

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

Divide $2 x^{3} - x^{2} - 9 x + 6$ 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 $0$ .

$x$

$2$

$2 x^{3}$

$x^{2}$

$9 x$

$6$

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

					$2 x^{2}$
	$x$	-	$2$		$2 x^{3}$	-	$x^{2}$	-	$9 x$	+	$6$

Multiply the new quotient term by the divisor.

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

The expression needs to be subtracted from the dividend, so change all the signs in $2 x^{3} - 4 x^{2}$

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

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

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

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

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

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

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

Multiply the new quotient term by the divisor.

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

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

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

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

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

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

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

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

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

Multiply the new quotient term by the divisor.

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

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

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

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

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

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

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

Write $2 x^{3} - x^{2} - 9 x + 6$ as a set of factors.

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

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

$x - 2 = 0$

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

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

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

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

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

Use the quadratic formula to find the solutions.

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

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

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

Simplify.

Simplify the numerator.

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

$x = \frac{- 3 \pm \sqrt{9 - 4 \cdot 2 \cdot - 3}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot - 3$ .

Multiply $- 4$ by $2$ .

$x = \frac{- 3 \pm \sqrt{9 - 8 \cdot - 3}}{2 \cdot 2}$

Multiply $- 8$ by $- 3$ .

$x = \frac{- 3 \pm \sqrt{9 + 24}}{2 \cdot 2}$

Add $9$ and $24$ .

$x = \frac{- 3 \pm \sqrt{33}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{- 3 \pm \sqrt{33}}{4}$

The final answer is the combination of both solutions.

$x = - \frac{3 - \sqrt{33}}{4}, - \frac{3 + \sqrt{33}}{4}$

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

$x = 2, - \frac{3 - \sqrt{33}}{4}, - \frac{3 + \sqrt{33}}{4}$

Step 3

The result can be shown in multiple forms.

Exact Form:

$x = 2, - \frac{3 - \sqrt{33}}{4}, - \frac{3 + \sqrt{33}}{4}$

Decimal Form:

$x = 2, 0.68614066 \dots, - 2.18614066 \dots$

Step 4

Find the Roots (Zeros) 2x^3-x^2-9x+6

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