How do you factor $4m^4-64n^4$ ?

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Answer 1 · 2016-05-12T22:34:46

$4m^4-64n^4=4(m-2n)(m+2n)(m^2+4n^2)$

We can use the difference of squares identity:

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

twice as follows:

$4m^4-64n^4$

$=4((m^2)^2-(4n^2)^2)$

$=4(m^2-4n^2)(m^2+4n^2)$

$=4(m^2-(2n)^2)(m^2+4n^2)$

$=4(m-2n)(m+2n)(m^2+4n^2)$

How do you factor 4m^4-64n^44m^4-64n^4?