Question

How do you factor `16x^4 - 81y^4`?

Answer 1

`16x^4-81y^4 = (2x-3y)(2x+3y)(4x^2+9y^2)`

Explanation:

The difference of squares identity can be written:

`a^2-b^2 = (a-b)(a+b)`

Hence we find:

`16x^4-81y^4 = (4x^2)^2-(9y^2)^2`

`color(white)(16x^4-81y^4) = (4x^2-9y^2)(4x^2+9y^2)`

`color(white)(16x^4-81y^4) = ((2x)^2-(3y)^2)(4x^2+9y^2)`

`color(white)(16x^4-81y^4) = (2x-3y)(2x+3y)(4x^2+9y^2)`

The remaining quadratic factor has no linear factors with Real coefficients.