Question

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

Answer 1

`12x^4-75y^4`

`=3(2x^2-5y^2)(2x^2+5y^2)`

`=3(sqrt(2)x-sqrt(5)y)(sqrt(2)x+sqrt(5)y)(2x^2+5y^2)`

Explanation:

The difference of squares identity is `a^2-b^2 = (a-b)(a+b)`

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

`12x^4-75y^4`

`=3((2x^2)^2-(5y^2)^2)`

`=3(2x^2-5y^2)(2x^2+5y^2)`

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

`=3((sqrt(2)x)^2-(sqrt(5)y)^2)(2x^2+5y^2)`

`=3(sqrt(2)x-sqrt(5)y)(sqrt(2)x+sqrt(5)y)(2x^2+5y^2)`

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