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

May 21, 2018

See a solution process below:

#### Explanation:

There common factor for each term is: $\textcolor{red}{8 c}$:

$8 {c}^{3} = \textcolor{red}{8 c} \cdot {c}^{2}$

$8 c = \textcolor{red}{8 c} \cdot 1$

We can then factor as:

$8 {c}^{3} - 8 c \implies \left(\textcolor{red}{8 c} \cdot {c}^{2}\right) - \left(\textcolor{red}{8 c} \cdot 1\right) \implies \textcolor{red}{8 c} \left({c}^{2} - 1\right)$

${c}^{2} - 1$ is a special form of the quadratic where:

${c}^{2} - 1 = \left(c + 1\right) \left(c - 1\right)$

$\textcolor{red}{8 c} \left({c}^{2} - 1\right) \implies \textcolor{red}{8 c} \left(c + 1\right) \left(c - 1\right)$