Factor x^4-6x^3+11x^2-6x

$x^{4} - 6 x^{3} + 11 x^{2} - 6 x$

Factor $x$ out of $x^{4} - 6 x^{3} + 11 x^{2} - 6 x$ .

Factor $x$ out of $x^{4}$ .

$x \cdot x^{3} - 6 x^{3} + 11 x^{2} - 6 x$

Factor $x$ out of $- 6 x^{3}$ .

$x \cdot x^{3} + x (- 6 x^{2}) + 11 x^{2} - 6 x$

Factor $x$ out of $11 x^{2}$ .

$x \cdot x^{3} + x (- 6 x^{2}) + x (11 x) - 6 x$

Factor $x$ out of $- 6 x$ .

$x \cdot x^{3} + x (- 6 x^{2}) + x (11 x) + x \cdot - 6$

Factor $x$ out of $x \cdot x^{3} + x (- 6 x^{2})$ .

$x (x^{3} - 6 x^{2}) + x (11 x) + x \cdot - 6$

Factor $x$ out of $x (x^{3} - 6 x^{2}) + x (11 x)$ .

$x (x^{3} - 6 x^{2} + 11 x) + x \cdot - 6$

Factor $x$ out of $x (x^{3} - 6 x^{2} + 11 x) + x \cdot - 6$ .

$x (x^{3} - 6 x^{2} + 11 x - 6)$

Factor $x^{3} - 6 x^{2} + 11 x - 6$ 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.

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

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

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

Substitute $1$ and simplify the expression. In this case, the expression is equal to $0$ so $1$ is a root of the polynomial.

Substitute $1$ into the polynomial.

$1^{3} - 6 \cdot 1^{2} + 11 \cdot 1 - 6$

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

$1 - 6 \cdot 1^{2} + 11 \cdot 1 - 6$

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

$1 - 6 \cdot 1 + 11 \cdot 1 - 6$

Multiply $- 6$ by $1$ .

$1 - 6 + 11 \cdot 1 - 6$

Subtract $6$ from $1$ .

$- 5 + 11 \cdot 1 - 6$

Multiply $11$ by $1$ .

$- 5 + 11 - 6$

Add $- 5$ and $11$ .

$6 - 6$

Subtract $6$ from $6$ .

$0$

Since $1$ is a known root, divide the polynomial by $x - 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{3} - 6 x^{2} + 11 x - 6}{x - 1}$

Divide $x^{3} - 6 x^{2} + 11 x - 6$ by $x - 1$ .

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

$x$

$1$

$x^{3}$

$6 x^{2}$

$11 x$

$6$

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

					$x^{2}$
	$x$	-	$1$		$x^{3}$	-	$6 x^{2}$	+	$11 x$	-	$6$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	-	$1$		$x^{3}$	-	$6 x^{2}$	+	$11 x$	-	$6$
			+	$x^{3}$	-	$x^{2}$

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

				$x^{2}$
$x$	-	$1$		$x^{3}$	-	$6 x^{2}$	+	$11 x$	-	$6$
			-	$x^{3}$	+	$x^{2}$

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

				$x^{2}$
$x$	-	$1$		$x^{3}$	-	$6 x^{2}$	+	$11 x$	-	$6$
			-	$x^{3}$	+	$x^{2}$
					-	$5 x^{2}$

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

				$x^{2}$
$x$	-	$1$		$x^{3}$	-	$6 x^{2}$	+	$11 x$	-	$6$
			-	$x^{3}$	+	$x^{2}$
					-	$5 x^{2}$	+	$11 x$

Divide the highest order term in the dividend $- 5 x^{2}$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$5 x$
$x$	-	$1$		$x^{3}$	-	$6 x^{2}$	+	$11 x$	-	$6$
			-	$x^{3}$	+	$x^{2}$
					-	$5 x^{2}$	+	$11 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$5 x$
$x$	-	$1$		$x^{3}$	-	$6 x^{2}$	+	$11 x$	-	$6$
			-	$x^{3}$	+	$x^{2}$
					-	$5 x^{2}$	+	$11 x$
					-	$5 x^{2}$	+	$5 x$

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

				$x^{2}$	-	$5 x$
$x$	-	$1$		$x^{3}$	-	$6 x^{2}$	+	$11 x$	-	$6$
			-	$x^{3}$	+	$x^{2}$
					-	$5 x^{2}$	+	$11 x$
					+	$5 x^{2}$	-	$5 x$

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

				$x^{2}$	-	$5 x$
$x$	-	$1$		$x^{3}$	-	$6 x^{2}$	+	$11 x$	-	$6$
			-	$x^{3}$	+	$x^{2}$
					-	$5 x^{2}$	+	$11 x$
					+	$5 x^{2}$	-	$5 x$
							+	$6 x$

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

				$x^{2}$	-	$5 x$
$x$	-	$1$		$x^{3}$	-	$6 x^{2}$	+	$11 x$	-	$6$
			-	$x^{3}$	+	$x^{2}$
					-	$5 x^{2}$	+	$11 x$
					+	$5 x^{2}$	-	$5 x$
							+	$6 x$	-	$6$

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

				$x^{2}$	-	$5 x$	+	$6$
$x$	-	$1$		$x^{3}$	-	$6 x^{2}$	+	$11 x$	-	$6$
			-	$x^{3}$	+	$x^{2}$
					-	$5 x^{2}$	+	$11 x$
					+	$5 x^{2}$	-	$5 x$
							+	$6 x$	-	$6$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$5 x$	+	$6$
$x$	-	$1$		$x^{3}$	-	$6 x^{2}$	+	$11 x$	-	$6$
			-	$x^{3}$	+	$x^{2}$
					-	$5 x^{2}$	+	$11 x$
					+	$5 x^{2}$	-	$5 x$
							+	$6 x$	-	$6$
							+	$6 x$	-	$6$

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

				$x^{2}$	-	$5 x$	+	$6$
$x$	-	$1$		$x^{3}$	-	$6 x^{2}$	+	$11 x$	-	$6$
			-	$x^{3}$	+	$x^{2}$
					-	$5 x^{2}$	+	$11 x$
					+	$5 x^{2}$	-	$5 x$
							+	$6 x$	-	$6$
							-	$6 x$	+	$6$

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

				$x^{2}$	-	$5 x$	+	$6$
$x$	-	$1$		$x^{3}$	-	$6 x^{2}$	+	$11 x$	-	$6$
			-	$x^{3}$	+	$x^{2}$
					-	$5 x^{2}$	+	$11 x$
					+	$5 x^{2}$	-	$5 x$
							+	$6 x$	-	$6$
							-	$6 x$	+	$6$
										$0$

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

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

Write $x^{3} - 6 x^{2} + 11 x - 6$ as a set of factors.

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

Factor.

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.

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

Remove unnecessary parentheses.

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

Remove unnecessary parentheses.

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

Factor x^4-6x^3+11x^2-6x

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