4m^12-81n^8?

factor

1 Answer
Nov 1, 2017

(2m^6-9n^4)(2m^6+9n^4)

Explanation:

This is a difference of squares. They factor like this:

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

4m^12-81n^8
=(2m^6-9n^4)(2m^6+9n^4)

There are no common factors/not a difference of squares anymore,
so, if you only want rational coefficients, you are finished :)