# How do you factor 4m^4-64n^4?

May 12, 2016

$4 {m}^{4} - 64 {n}^{4} = 4 \left(m - 2 n\right) \left(m + 2 n\right) \left({m}^{2} + 4 {n}^{2}\right)$

#### Explanation:

We can use the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

twice as follows:

$4 {m}^{4} - 64 {n}^{4}$

$= 4 \left({\left({m}^{2}\right)}^{2} - {\left(4 {n}^{2}\right)}^{2}\right)$

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

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

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