# How do you factor 12x^4-75y^4?

Jun 26, 2015

$12 {x}^{4} - 75 {y}^{4}$

$= 3 \left(2 {x}^{2} - 5 {y}^{2}\right) \left(2 {x}^{2} + 5 {y}^{2}\right)$

$= 3 \left(\sqrt{2} x - \sqrt{5} y\right) \left(\sqrt{2} x + \sqrt{5} y\right) \left(2 {x}^{2} + 5 {y}^{2}\right)$

#### Explanation:

The difference of squares identity is ${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

First separate out the common factor $3$ then use the difference of squares identity to get:

$12 {x}^{4} - 75 {y}^{4}$

$= 3 \left({\left(2 {x}^{2}\right)}^{2} - {\left(5 {y}^{2}\right)}^{2}\right)$

$= 3 \left(2 {x}^{2} - 5 {y}^{2}\right) \left(2 {x}^{2} + 5 {y}^{2}\right)$

The first quadratic factor can be treated as a difference of squares with irrational coefficients, so...

$= 3 \left({\left(\sqrt{2} x\right)}^{2} - {\left(\sqrt{5} y\right)}^{2}\right) \left(2 {x}^{2} + 5 {y}^{2}\right)$

$= 3 \left(\sqrt{2} x - \sqrt{5} y\right) \left(\sqrt{2} x + \sqrt{5} y\right) \left(2 {x}^{2} + 5 {y}^{2}\right)$

The second quadratic factor is irreducible unless you use complex coefficients.