# How do you factor completely 4b^4-1024?

Apr 18, 2016

$4 \left({b}^{2} + 16\right) \left(b - 4\right) \left(b + 4\right)$

#### Explanation:

The first step is to take out the common factor of 4.

$\Rightarrow 4 \left({b}^{4} - 256\right)$

now $\left({b}^{4} - 256\right) \text{ is a difference of squares }$

since ${b}^{4} = {b}^{2} \times {b}^{2} \text{ and } 256 = 16 \times 16$

a difference of squares factors as follows

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

here a = b^2" and b = 16

$\Rightarrow {b}^{4} - 256 = \left({b}^{2} - 16\right) \left({b}^{2} + 16\right)$

now ${b}^{2} - 16 \text{ is also a difference of squares }$

and factors as $\left(b - 4\right) \left(b + 4\right)$

Putting this together gives

$4 {b}^{4} - 1024 = 4 \left({b}^{2} + 16\right) \left(b - 4\right) \left(b + 4\right)$