# How do you factor 256n^4-c^4?

Feb 5, 2017

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

#### Explanation:

As the two monomials $256 {n}^{4}$ and ${c}^{4}$ are perfect square,

we can use the identity ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$ to factorize it and

$256 {n}^{4} - {c}^{4}$

= ${\left(16 {n}^{2}\right)}^{2} - {\left({c}^{2}\right)}^{2}$

= $\left(16 {n}^{2} + {c}^{2}\right) \left(16 {n}^{2} - {c}^{2}\right)$

= $\left(16 {n}^{2} + {c}^{2}\right) \left({\left(4 n\right)}^{2} - {c}^{2}\right)$

= $\left(16 {n}^{2} + {c}^{2}\right) \left(4 n + c\right) \left(4 n - c\right)$