4m^12-81n^8?

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factor

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Answer 1 · 2017-11-01T21:18:06

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

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 :)