# How do you factor f^3+2f^2-64f-128?

Dec 3, 2017

Required factors are : $\left(f + 8\right) \left(f - 8\right) \left(f + 2\right)$

#### Explanation:

The problem is to factor

${f}^{3} + 2 {f}^{2} - 64 f - 128$

Observe the groupings collected for further simplification:

color(green)((f^3 + 2f^2)$\textcolor{b l u e}{- 64 f - 128}$ $\textcolor{red}{E x p r e s s i o n .1}$

The group color(green)((f^3 + 2f^2) can be factored as

$\textcolor{g r e e n}{{f}^{2} \left(f + 2\right)}$ $\text{ } \textcolor{red}{R e s .1}$

The group $\textcolor{b l u e}{- 64 f - 128}$ can be factored as

$\textcolor{b l u e}{- 64 \left(f + 2\right)}$ $\text{ } \textcolor{red}{R e s .2}$

Using our intermediate results $\textcolor{red}{R e s .1}$ and $\textcolor{red}{R e s .2}$ we can write our $\textcolor{red}{E x p r e s s i o n .1}$ as

$\textcolor{g r e e n}{{f}^{2} \left(f + 2\right) - 64 \left(f + 2\right)}$

We can now factor them as

$\left({f}^{2} - 64\right) \left(f + 2\right)$ $\textcolor{red}{E x p r e s s i o n .2}$

Next, we rewrite $\left({f}^{2} - 64\right)$ as

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

Using the factoring rule "Difference of Squares" we get

$\left(f + 8\right) \left(f - 8\right)$

Using the above result and our $\textcolor{red}{E x p r e s s i o n .2}$
we get all of our required factors:

$\textcolor{b l u e}{\left(f + 8\right) \left(f - 8\right) \left(f + 2\right)}$