Factor 3x^3-4x^2-17x+6

$3 x^{3} - 4 x^{2} - 17 x + 6$

Step 1

Factor $3 x^{3} - 4 x^{2} - 17 x + 6$ 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.

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

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

$\pm 1, \pm 0.\overline{3}, \pm 6, \pm 2, \pm 0.\overline{6}, \pm 3$

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.

$3 \cdot {0.\overline{3}}^{3} - 4 \cdot {0.\overline{3}}^{2} - 17 \cdot 0.\overline{3} + 6$

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

$3 \cdot 0.\overline{037} - 4 \cdot {0.\overline{3}}^{2} - 17 \cdot 0.\overline{3} + 6$

Multiply $3$ by $0.\overline{037}$ .

$0.\overline{1} - 4 \cdot {0.\overline{3}}^{2} - 17 \cdot 0.\overline{3} + 6$

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

$0.\overline{1} - 4 \cdot 0.\overline{1} - 17 \cdot 0.\overline{3} + 6$

Multiply $- 4$ by $0.\overline{1}$ .

$0.\overline{1} - 0.\overline{4} - 17 \cdot 0.\overline{3} + 6$

Subtract $0.\overline{4}$ from $0.\overline{1}$ .

$- 0.\overline{3} - 17 \cdot 0.\overline{3} + 6$

Multiply $- 17$ by $0.\overline{3}$ .

$- 0.\overline{3} - 5.\overline{6} + 6$

Subtract $5.\overline{6}$ from $- 0.\overline{3}$ .

$- 6 + 6$

Add $- 6$ and $6$ .

$0$

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.

$\frac{3 x^{3} - 4 x^{2} - 17 x + 6}{3 x - 1}$

Divide $3 x^{3} - 4 x^{2} - 17 x + 6$ 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 $0$ .

$3 x$

$1$

$3 x^{3}$

$4 x^{2}$

$17 x$

$6$

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

					$x^{2}$
	$3 x$	-	$1$		$3 x^{3}$	-	$4 x^{2}$	-	$17 x$	+	$6$

Multiply the new quotient term by the divisor.

				$x^{2}$
$3 x$	-	$1$		$3 x^{3}$	-	$4 x^{2}$	-	$17 x$	+	$6$
			+	$3 x^{3}$	-	$x^{2}$

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

				$x^{2}$
$3 x$	-	$1$		$3 x^{3}$	-	$4 x^{2}$	-	$17 x$	+	$6$
			-	$3 x^{3}$	+	$x^{2}$

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

				$x^{2}$
$3 x$	-	$1$		$3 x^{3}$	-	$4 x^{2}$	-	$17 x$	+	$6$
			-	$3 x^{3}$	+	$x^{2}$
					-	$3 x^{2}$

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

				$x^{2}$
$3 x$	-	$1$		$3 x^{3}$	-	$4 x^{2}$	-	$17 x$	+	$6$
			-	$3 x^{3}$	+	$x^{2}$
					-	$3 x^{2}$	-	$17 x$

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

				$x^{2}$	-	$x$
$3 x$	-	$1$		$3 x^{3}$	-	$4 x^{2}$	-	$17 x$	+	$6$
			-	$3 x^{3}$	+	$x^{2}$
					-	$3 x^{2}$	-	$17 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$x$
$3 x$	-	$1$		$3 x^{3}$	-	$4 x^{2}$	-	$17 x$	+	$6$
			-	$3 x^{3}$	+	$x^{2}$
					-	$3 x^{2}$	-	$17 x$
					-	$3 x^{2}$	+	$x$

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

				$x^{2}$	-	$x$
$3 x$	-	$1$		$3 x^{3}$	-	$4 x^{2}$	-	$17 x$	+	$6$
			-	$3 x^{3}$	+	$x^{2}$
					-	$3 x^{2}$	-	$17 x$
					+	$3 x^{2}$	-	$x$

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

				$x^{2}$	-	$x$
$3 x$	-	$1$		$3 x^{3}$	-	$4 x^{2}$	-	$17 x$	+	$6$
			-	$3 x^{3}$	+	$x^{2}$
					-	$3 x^{2}$	-	$17 x$
					+	$3 x^{2}$	-	$x$
							-	$18 x$

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

				$x^{2}$	-	$x$
$3 x$	-	$1$		$3 x^{3}$	-	$4 x^{2}$	-	$17 x$	+	$6$
			-	$3 x^{3}$	+	$x^{2}$
					-	$3 x^{2}$	-	$17 x$
					+	$3 x^{2}$	-	$x$
							-	$18 x$	+	$6$

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

				$x^{2}$	-	$x$	-	$6$
$3 x$	-	$1$		$3 x^{3}$	-	$4 x^{2}$	-	$17 x$	+	$6$
			-	$3 x^{3}$	+	$x^{2}$
					-	$3 x^{2}$	-	$17 x$
					+	$3 x^{2}$	-	$x$
							-	$18 x$	+	$6$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$x$	-	$6$
$3 x$	-	$1$		$3 x^{3}$	-	$4 x^{2}$	-	$17 x$	+	$6$
			-	$3 x^{3}$	+	$x^{2}$
					-	$3 x^{2}$	-	$17 x$
					+	$3 x^{2}$	-	$x$
							-	$18 x$	+	$6$
							-	$18 x$	+	$6$

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

				$x^{2}$	-	$x$	-	$6$
$3 x$	-	$1$		$3 x^{3}$	-	$4 x^{2}$	-	$17 x$	+	$6$
			-	$3 x^{3}$	+	$x^{2}$
					-	$3 x^{2}$	-	$17 x$
					+	$3 x^{2}$	-	$x$
							-	$18 x$	+	$6$
							+	$18 x$	-	$6$

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

				$x^{2}$	-	$x$	-	$6$
$3 x$	-	$1$		$3 x^{3}$	-	$4 x^{2}$	-	$17 x$	+	$6$
			-	$3 x^{3}$	+	$x^{2}$
					-	$3 x^{2}$	-	$17 x$
					+	$3 x^{2}$	-	$x$
							-	$18 x$	+	$6$
							+	$18 x$	-	$6$
										$0$

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

$x^{2} - x - 6$

Write $3 x^{3} - 4 x^{2} - 17 x + 6$ as a set of factors.

$(3 x - 1) (x^{2} - x - 6)$

Step 2

Factor $x^{2} - x - 6$ 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 $- 1$ .

$- 3, 2$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

Factor 3x^3-4x^2-17x+6

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