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

Nov 12, 2015

Apply the difference of squares formula twice to find
$81 - 16 {x}^{4} = \left(9 + 4 {x}^{2}\right) \left(3 + 2 x\right) \left(3 - 2 x\right)$

#### Explanation:

The difference of squares formula states
${x}^{2} - {y}^{2} = \left(x + y\right) \left(x - y\right)$

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

So, applying the formula,
$81 - 16 {x}^{4} = {9}^{2} - {\left(4 {x}^{2}\right)}^{2} = \left(9 + 4 {x}^{2}\right) \left(9 - 4 {x}^{2}\right)$

But we are not quite done.

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

So applying the formula to $9 - 4 {x}^{2}$, we get the final result of

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