# How do you factor 81x^4-1?

Apr 27, 2018

$81 {x}^{4} - 1 = \left(3 x - 1\right) \left(3 x + 1\right) \left(9 {x}^{2} + 1\right)$

#### Explanation:

The difference of squares identity can be written:

${A}^{2} - {B}^{2} = \left(A - B\right) \left(A + B\right)$

We can use this a couple of times to find:

$81 {x}^{4} - 1 = \left(9 {x}^{2}\right) - {1}^{2}$

$\textcolor{w h i t e}{81 {x}^{4} - 1} = \left(9 {x}^{2} - 1\right) \left(9 {x}^{2} + 1\right)$

$\textcolor{w h i t e}{81 {x}^{4} - 1} = \left({\left(3 x\right)}^{2} - {1}^{2}\right) \left(9 {x}^{2} + 1\right)$

$\textcolor{w h i t e}{81 {x}^{4} - 1} = \left(3 x - 1\right) \left(3 x + 1\right) \left(9 {x}^{2} + 1\right)$

The remaining quadratic factor is always positive for real values of $x$, so has no linear factors with real coefficients.

We can treat it as a difference of squares using the imaginary unit $i$, which satisfies ${i}^{2} = - 1$ as follows:

$9 {x}^{2} + 1 = {\left(3 x\right)}^{2} - {i}^{2} = \left(3 x - i\right) \left(3 x + i\right)$

So if we allow complex coefficients then:

$81 {x}^{4} - 1 = \left(3 x - 1\right) \left(3 x + 1\right) \left(3 x - i\right) \left(3 x + i\right)$