Factor 2x^3-9x^2+7x+6

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

Step 1

Factor $2 x^{3} - 9 x^{2} + 7 x + 6$ 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 $- 0.5$ 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.

$2 {(- 0.5)}^{3} - 9 {(- 0.5)}^{2} + 7 \cdot - 0.5 + 6$

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

$2 \cdot - 0.125 - 9 {(- 0.5)}^{2} + 7 \cdot - 0.5 + 6$

Multiply $2$ by $- 0.125$ .

$- 0.25 - 9 {(- 0.5)}^{2} + 7 \cdot - 0.5 + 6$

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

$- 0.25 - 9 \cdot 0.25 + 7 \cdot - 0.5 + 6$

Multiply $- 9$ by $0.25$ .

$- 0.25 - 2.25 + 7 \cdot - 0.5 + 6$

Subtract $2.25$ from $- 0.25$ .

$- 2.5 + 7 \cdot - 0.5 + 6$

Multiply $7$ by $- 0.5$ .

$- 2.5 - 3.5 + 6$

Subtract $3.5$ from $- 2.5$ .

$- 6 + 6$

Add $- 6$ and $6$ .

$0$

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.

$\frac{2 x^{3} - 9 x^{2} + 7 x + 6}{2 x + 1}$

Divide $2 x^{3} - 9 x^{2} + 7 x + 6$ 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 $0$ .

$2 x$

$1$

$2 x^{3}$

$9 x^{2}$

$7 x$

$6$

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

					$x^{2}$
	$2 x$	+	$1$		$2 x^{3}$	-	$9 x^{2}$	+	$7 x$	+	$6$

Multiply the new quotient term by the divisor.

				$x^{2}$
$2 x$	+	$1$		$2 x^{3}$	-	$9 x^{2}$	+	$7 x$	+	$6$
			+	$2 x^{3}$	+	$x^{2}$

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

				$x^{2}$
$2 x$	+	$1$		$2 x^{3}$	-	$9 x^{2}$	+	$7 x$	+	$6$
			-	$2 x^{3}$	-	$x^{2}$

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

				$x^{2}$
$2 x$	+	$1$		$2 x^{3}$	-	$9 x^{2}$	+	$7 x$	+	$6$
			-	$2 x^{3}$	-	$x^{2}$
					-	$10 x^{2}$

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

				$x^{2}$
$2 x$	+	$1$		$2 x^{3}$	-	$9 x^{2}$	+	$7 x$	+	$6$
			-	$2 x^{3}$	-	$x^{2}$
					-	$10 x^{2}$	+	$7 x$

Divide the highest order term in the dividend $- 10 x^{2}$ by the highest order term in divisor $2 x$ .

				$x^{2}$	-	$5 x$
$2 x$	+	$1$		$2 x^{3}$	-	$9 x^{2}$	+	$7 x$	+	$6$
			-	$2 x^{3}$	-	$x^{2}$
					-	$10 x^{2}$	+	$7 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$5 x$
$2 x$	+	$1$		$2 x^{3}$	-	$9 x^{2}$	+	$7 x$	+	$6$
			-	$2 x^{3}$	-	$x^{2}$
					-	$10 x^{2}$	+	$7 x$
					-	$10 x^{2}$	-	$5 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 10 x^{2} - 5 x$

				$x^{2}$	-	$5 x$
$2 x$	+	$1$		$2 x^{3}$	-	$9 x^{2}$	+	$7 x$	+	$6$
			-	$2 x^{3}$	-	$x^{2}$
					-	$10 x^{2}$	+	$7 x$
					+	$10 x^{2}$	+	$5 x$

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

				$x^{2}$	-	$5 x$
$2 x$	+	$1$		$2 x^{3}$	-	$9 x^{2}$	+	$7 x$	+	$6$
			-	$2 x^{3}$	-	$x^{2}$
					-	$10 x^{2}$	+	$7 x$
					+	$10 x^{2}$	+	$5 x$
							+	$12 x$

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

				$x^{2}$	-	$5 x$
$2 x$	+	$1$		$2 x^{3}$	-	$9 x^{2}$	+	$7 x$	+	$6$
			-	$2 x^{3}$	-	$x^{2}$
					-	$10 x^{2}$	+	$7 x$
					+	$10 x^{2}$	+	$5 x$
							+	$12 x$	+	$6$

Divide the highest order term in the dividend $12 x$ by the highest order term in divisor $2 x$ .

				$x^{2}$	-	$5 x$	+	$6$
$2 x$	+	$1$		$2 x^{3}$	-	$9 x^{2}$	+	$7 x$	+	$6$
			-	$2 x^{3}$	-	$x^{2}$
					-	$10 x^{2}$	+	$7 x$
					+	$10 x^{2}$	+	$5 x$
							+	$12 x$	+	$6$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$5 x$	+	$6$
$2 x$	+	$1$		$2 x^{3}$	-	$9 x^{2}$	+	$7 x$	+	$6$
			-	$2 x^{3}$	-	$x^{2}$
					-	$10 x^{2}$	+	$7 x$
					+	$10 x^{2}$	+	$5 x$
							+	$12 x$	+	$6$
							+	$12 x$	+	$6$

The expression needs to be subtracted from the dividend, so change all the signs in $12 x + 6$

				$x^{2}$	-	$5 x$	+	$6$
$2 x$	+	$1$		$2 x^{3}$	-	$9 x^{2}$	+	$7 x$	+	$6$
			-	$2 x^{3}$	-	$x^{2}$
					-	$10 x^{2}$	+	$7 x$
					+	$10 x^{2}$	+	$5 x$
							+	$12 x$	+	$6$
							-	$12 x$	-	$6$

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

				$x^{2}$	-	$5 x$	+	$6$
$2 x$	+	$1$		$2 x^{3}$	-	$9 x^{2}$	+	$7 x$	+	$6$
			-	$2 x^{3}$	-	$x^{2}$
					-	$10 x^{2}$	+	$7 x$
					+	$10 x^{2}$	+	$5 x$
							+	$12 x$	+	$6$
							-	$12 x$	-	$6$
										$0$

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

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

Write $2 x^{3} - 9 x^{2} + 7 x + 6$ as a set of factors.

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

Step 2

Factor $x^{2} - 5 x + 6$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $6$ and whose sum is $- 5$ .

$- 3, - 2$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

Factor 2x^3-9x^2+7x+6

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