# How do you factor 4x^3 + 32?

Aug 25, 2016

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

#### Explanation:

First step is to take out a $\textcolor{b l u e}{\text{common factor}}$ of 4.

$\Rightarrow 4 {x}^{3} + 32 = 4 \left({x}^{3} + 8\right) \ldots \ldots . . \left(A\right)$

Now, ${x}^{3} + 8 \textcolor{b l u e}{\text{ is a sum of cubes}}$ and in general is factorised as follows.

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}} \ldots \ldots . . \left(B\right)$

now, ${\left(x\right)}^{3} = {x}^{3} \text{ and " (2)^3=8rArra=x" and } b = 2$

substitute these values for a and b into (B)

$\Rightarrow {x}^{3} + 8 = \left(x + 2\right) \left({x}^{2} - 2 x + 4\right)$

substitute back into (A)

$\Rightarrow 4 {x}^{3} + 32 = 4 \left(x + 2\right) \left({x}^{2} - 2 x + 4\right)$