# How do you factor 16x^4 - 81y^4?

Feb 11, 2017

$16 {x}^{4} - 81 {y}^{4} = \left(2 x - 3 y\right) \left(2 x + 3 y\right) \left(4 {x}^{2} + 9 {y}^{2}\right)$

#### Explanation:

The difference of squares identity can be written:

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

Hence we find:

$16 {x}^{4} - 81 {y}^{4} = {\left(4 {x}^{2}\right)}^{2} - {\left(9 {y}^{2}\right)}^{2}$

$\textcolor{w h i t e}{16 {x}^{4} - 81 {y}^{4}} = \left(4 {x}^{2} - 9 {y}^{2}\right) \left(4 {x}^{2} + 9 {y}^{2}\right)$

$\textcolor{w h i t e}{16 {x}^{4} - 81 {y}^{4}} = \left({\left(2 x\right)}^{2} - {\left(3 y\right)}^{2}\right) \left(4 {x}^{2} + 9 {y}^{2}\right)$

$\textcolor{w h i t e}{16 {x}^{4} - 81 {y}^{4}} = \left(2 x - 3 y\right) \left(2 x + 3 y\right) \left(4 {x}^{2} + 9 {y}^{2}\right)$

The remaining quadratic factor has no linear factors with Real coefficients.