# How do you factor 2d^4-32f^4?

Dec 28, 2016

$2 {d}^{4} - 32 {f}^{4} = 2 \left(d - 2 f\right) \left(d + 2 f\right) \left({d}^{2} + 4 {f}^{2}\right)$

#### Explanation:

The difference of squares identity can be written:

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

Hence we find:

$2 {d}^{4} - 32 {f}^{4} = 2 \left({d}^{4} - 16 {f}^{4}\right)$

$\textcolor{w h i t e}{2 {d}^{4} - 32 {f}^{4}} = 2 \left({\left({d}^{2}\right)}^{2} - {\left(4 {f}^{2}\right)}^{2}\right)$

$\textcolor{w h i t e}{2 {d}^{4} - 32 {f}^{4}} = 2 \left({d}^{2} - 4 {f}^{2}\right) \left({d}^{2} + 4 {f}^{2}\right)$

$\textcolor{w h i t e}{2 {d}^{4} - 32 {f}^{4}} = 2 \left({d}^{2} - {\left(2 f\right)}^{2}\right) \left({d}^{2} + 4 {f}^{2}\right)$

$\textcolor{w h i t e}{2 {d}^{4} - 32 {f}^{4}} = 2 \left(d - 2 f\right) \left(d + 2 f\right) \left({d}^{2} + 4 {f}^{2}\right)$

The remaining sum of squares can only be factored further with Complex coefficients.