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

Mar 18, 2016

Use the difference of squares identity to find:

$y = 16 {x}^{4} - 81$

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

$= \left(2 x - 3\right) \left(2 x + 3\right) \left(2 x - 3 i\right) \left(2 x + 3 i\right)$

#### Explanation:

The difference of squares identity can be written:

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

Use this twice as follows:

$y = 16 {x}^{4} - 81$

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

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

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

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

The remaining quadratic factor has no linear factors with Real coefficients, but if you allow Complex coefficients you can find:

$= \left(2 x - 3\right) \left(2 x + 3\right) \left({\left(2 x\right)}^{2} - {\left(3 i\right)}^{2}\right)$

$= \left(2 x - 3\right) \left(2 x + 3\right) \left(2 x - 3 i\right) \left(2 x + 3 i\right)$