How do you factor #4m^4-64n^4#?
1 Answer
May 12, 2016
Explanation:
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)#
