# How do you factor 81x^4 -625?

Jan 17, 2017

$81 {x}^{4} - 625 = \left(3 x - 5\right) \left(3 x + 5\right) \left(9 {x}^{2} + 25\right)$

#### Explanation:

The difference of squares identity can be written:

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

So we find:

$81 {x}^{4} - 625 = {\left(9 {x}^{2}\right)}^{2} - {25}^{2}$

$\textcolor{w h i t e}{81 {x}^{4} - 625} = \left(9 {x}^{2} - 25\right) \left(9 {x}^{2} + 25\right)$

$\textcolor{w h i t e}{81 {x}^{4} - 625} = \left({\left(3 x\right)}^{2} - {5}^{2}\right) \left(9 {x}^{2} + 25\right)$

$\textcolor{w h i t e}{81 {x}^{4} - 625} = \left(3 x - 5\right) \left(3 x + 5\right) \left(9 {x}^{2} + 25\right)$

The remaining quadratic factor is irreducible over the reals. That is, it cannot be factorised into linear factors with real coefficients.