# How do you factor c^3 - 2c^2 - 8c?

Oct 19, 2015

The factorization of ${c}^{3} - 2 {c}^{2} - 8 c$ is

$c \left(c + 2\right) \left(c - 4\right)$

#### Explanation:

To factor ${c}^{3} - 2 {c}^{2} - 8 c$
begin by factoring out the variable $c$ as it is a factor of each unit of the tri-nomial.

$c \left({c}^{2} - 2 c - 8\right)$

Next, factor the tri-nomial $\left({c}^{2} - 2 c - 8\right)$
by finding the factor of 8 that will subtract to get 2.

1 and 8 2 and 4

Since the second sign of the tri-nomial is subtraction the factors must have different signs.

$- 4 \left(2\right) = - 8$
$- 4 + 2 = - 2$

Now factor the tri-nomial into two binomials.

${c}^{2} - 2 c - 8$
$\left(c + 2\right) \left(c - 4\right)$

Now complete the factors.

$c \left(c + 2\right) \left(c - 4\right)$

Here is a video on factoring.

Oct 19, 2015

$c \cdot \left(c + 2\right) \cdot \left(c - 4\right)$

#### Explanation:

Your starting expression looks like this

${c}^{3} - 2 {c}^{2} - 8 c$

Notice that you can use $c$ as a common factor for all three terms. This will get you

$c \cdot \left({c}^{2} - 2 c - 8\right)$

Now focus on the paranthesis. Notice that you can rewrite that expression as

${c}^{2} - 2 c - 8 = {c}^{2} + 2 c - 4 c - 8$

$= c \cdot \left(c + 2\right) \cdot \left(- 4\right) \cdot \left(c + 2\right)$

$= \left(c + 2\right) \cdot \left(c - 4\right)$

The expression can thus be factored as

${c}^{3} - 2 {c}^{2} - 8 c = \textcolor{g r e e n}{c \cdot \left(c + 2\right) \cdot \left(c - 4\right)}$