How do you factor #16x^4 - 4y^2#?

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Answer 1 · 2015-06-26T04:33:35

#(16x^4-y^2)=4(2x^2-y)(2x^2+y)#

#16x^4-y^2#

Factor out the GCF #4#.

#4(4x^4-y^2)#

#(4x^4-y^2)# is in the form of the difference of squares:

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

#a=2x^2 and b=y#

#4((2x^2)^2-y)# =

#4(2x^2-y)(2x^2+y)#