Factor f(x)=-x^4+x^3+8x^2-12x

$f (x) = - x^{4} + x^{3} + 8 x^{2} - 12 x$

Step 1

Factor $x$ out of $- x^{4} + x^{3} + 8 x^{2} - 12 x$ .

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

$f (x) = x (- x^{3}) + x^{3} + 8 x^{2} - 12 x$

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

$f (x) = x (- x^{3}) + x \cdot x^{2} + 8 x^{2} - 12 x$

Factor $x$ out of $8 x^{2}$ .

$f (x) = x (- x^{3}) + x \cdot x^{2} + x (8 x) - 12 x$

Factor $x$ out of $- 12 x$ .

$f (x) = x (- x^{3}) + x \cdot x^{2} + x (8 x) + x \cdot - 12$

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

$f (x) = x (- x^{3} + x^{2}) + x (8 x) + x \cdot - 12$

Factor $x$ out of $x (- x^{3} + x^{2}) + x (8 x)$ .

$f (x) = x (- x^{3} + x^{2} + 8 x) + x \cdot - 12$

Factor $x$ out of $x (- x^{3} + x^{2} + 8 x) + x \cdot - 12$ .

$f (x) = x (- x^{3} + x^{2} + 8 x - 12)$

Step 2

Factor $- x^{3} + x^{2} + 8 x - 12$ 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.

$p = \pm 1, \pm 12, \pm 2, \pm 6, \pm 3, \pm 4 q = \pm 1$

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

$\pm 1, \pm 12, \pm 2, \pm 6, \pm 3, \pm 4$

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.

$- 2^{3} + 2^{2} + 8 \cdot 2 - 12$

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

$- 1 \cdot 8 + 2^{2} + 8 \cdot 2 - 12$

Multiply $- 1$ by $8$ .

$- 8 + 2^{2} + 8 \cdot 2 - 12$

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

$- 8 + 4 + 8 \cdot 2 - 12$

Add $- 8$ and $4$ .

$- 4 + 8 \cdot 2 - 12$

Multiply $8$ by $2$ .

$- 4 + 16 - 12$

Add $- 4$ and $16$ .

$12 - 12$

Subtract $12$ from $12$ .

$0$

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.

$\frac{- x^{3} + x^{2} + 8 x - 12}{x - 2}$

Divide $- x^{3} + x^{2} + 8 x - 12$ 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 $0$ .

$x$

$2$

$x^{3}$

$x^{2}$

$8 x$

$12$

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

				-	$x^{2}$
	$x$	-	$2$	-	$x^{3}$	+	$x^{2}$	+	$8 x$	-	$12$

Multiply the new quotient term by the divisor.

			-	$x^{2}$
$x$	-	$2$	-	$x^{3}$	+	$x^{2}$	+	$8 x$	-	$12$
			-	$x^{3}$	+	$2 x^{2}$

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

			-	$x^{2}$
$x$	-	$2$	-	$x^{3}$	+	$x^{2}$	+	$8 x$	-	$12$
			+	$x^{3}$	-	$2 x^{2}$

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

			-	$x^{2}$
$x$	-	$2$	-	$x^{3}$	+	$x^{2}$	+	$8 x$	-	$12$
			+	$x^{3}$	-	$2 x^{2}$
					-	$x^{2}$

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

			-	$x^{2}$
$x$	-	$2$	-	$x^{3}$	+	$x^{2}$	+	$8 x$	-	$12$
			+	$x^{3}$	-	$2 x^{2}$
					-	$x^{2}$	+	$8 x$

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

			-	$x^{2}$	-	$x$
$x$	-	$2$	-	$x^{3}$	+	$x^{2}$	+	$8 x$	-	$12$
			+	$x^{3}$	-	$2 x^{2}$
					-	$x^{2}$	+	$8 x$

Multiply the new quotient term by the divisor.

			-	$x^{2}$	-	$x$
$x$	-	$2$	-	$x^{3}$	+	$x^{2}$	+	$8 x$	-	$12$
			+	$x^{3}$	-	$2 x^{2}$
					-	$x^{2}$	+	$8 x$
					-	$x^{2}$	+	$2 x$

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

			-	$x^{2}$	-	$x$
$x$	-	$2$	-	$x^{3}$	+	$x^{2}$	+	$8 x$	-	$12$
			+	$x^{3}$	-	$2 x^{2}$
					-	$x^{2}$	+	$8 x$
					+	$x^{2}$	-	$2 x$

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

			-	$x^{2}$	-	$x$
$x$	-	$2$	-	$x^{3}$	+	$x^{2}$	+	$8 x$	-	$12$
			+	$x^{3}$	-	$2 x^{2}$
					-	$x^{2}$	+	$8 x$
					+	$x^{2}$	-	$2 x$
							+	$6 x$

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

			-	$x^{2}$	-	$x$
$x$	-	$2$	-	$x^{3}$	+	$x^{2}$	+	$8 x$	-	$12$
			+	$x^{3}$	-	$2 x^{2}$
					-	$x^{2}$	+	$8 x$
					+	$x^{2}$	-	$2 x$
							+	$6 x$	-	$12$

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

			-	$x^{2}$	-	$x$	+	$6$
$x$	-	$2$	-	$x^{3}$	+	$x^{2}$	+	$8 x$	-	$12$
			+	$x^{3}$	-	$2 x^{2}$
					-	$x^{2}$	+	$8 x$
					+	$x^{2}$	-	$2 x$
							+	$6 x$	-	$12$

Multiply the new quotient term by the divisor.

			-	$x^{2}$	-	$x$	+	$6$
$x$	-	$2$	-	$x^{3}$	+	$x^{2}$	+	$8 x$	-	$12$
			+	$x^{3}$	-	$2 x^{2}$
					-	$x^{2}$	+	$8 x$
					+	$x^{2}$	-	$2 x$
							+	$6 x$	-	$12$
							+	$6 x$	-	$12$

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

			-	$x^{2}$	-	$x$	+	$6$
$x$	-	$2$	-	$x^{3}$	+	$x^{2}$	+	$8 x$	-	$12$
			+	$x^{3}$	-	$2 x^{2}$
					-	$x^{2}$	+	$8 x$
					+	$x^{2}$	-	$2 x$
							+	$6 x$	-	$12$
							-	$6 x$	+	$12$

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

			-	$x^{2}$	-	$x$	+	$6$
$x$	-	$2$	-	$x^{3}$	+	$x^{2}$	+	$8 x$	-	$12$
			+	$x^{3}$	-	$2 x^{2}$
					-	$x^{2}$	+	$8 x$
					+	$x^{2}$	-	$2 x$
							+	$6 x$	-	$12$
							-	$6 x$	+	$12$
										$0$

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

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

Write $- x^{3} + x^{2} + 8 x - 12$ as a set of factors.

$f (x) = x ((x - 2) (- x^{2} - x + 6))$

Step 3

Factor.

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 1 \cdot 6 = - 6$ and whose sum is $b = - 1$ .

Factor $- 1$ out of $- x$ .

$f (x) = x ((x - 2) (- x^{2} - (x) + 6))$

Rewrite $- 1$ as $2$ plus $- 3$

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

Factor out the greatest common factor (GCF) from each group.

$f (x) = x ((x - 2) (x (- x + 2) + 3 (- x + 2)))$

Factor the polynomial by factoring out the greatest common factor, $- x + 2$ .

$f (x) = x ((x - 2) ((- x + 2) (x + 3)))$

Remove unnecessary parentheses.

$f (x) = x ((x - 2) (- x + 2) (x + 3))$

Remove unnecessary parentheses.

$f (x) = x (x - 2) (- x + 2) (x + 3)$

Step 4

Combine exponents.

Factor $- 1$ out of $x$ .

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

Rewrite $- 2$ as $- 1 (2)$ .

$f (x) = x (- 1 (- x) - 1 \cdot 2) (- x + 2) (x + 3)$

Factor $- 1$ out of $- 1 (- x) - 1 (2)$ .

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

Remove parentheses.

$f (x) = x \cdot (- 1 (- x + 2) (- x + 2) (x + 3))$

Raise $- x + 2$ to the power of $1$ .

$f (x) = x \cdot (- 1 ((- x + 2) (- x + 2)) (x + 3))$

Raise $- x + 2$ to the power of $1$ .

$f (x) = x \cdot (- 1 ((- x + 2) (- x + 2)) (x + 3))$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f (x) = x \cdot (- 1 {(- x + 2)}^{1 + 1} (x + 3))$

Add $1$ and $1$ .

$f (x) = x \cdot (- 1 {(- x + 2)}^{2} (x + 3))$

Factor f(x)=-x^4+x^3+8x^2-12x

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