How do you factor #81-16x^4#?

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Answer 1 · 2015-11-12T07:06:30

Apply the difference of squares formula twice to find
#81-16x^4 = (9 + 4x^2)(3+2x)(3-2x)#

The difference of squares formula states
#x^2 - y^2 = (x+y)(x-y)#

#81 = 9^2# and #16x^4 = (4x^2)^2#

So, applying the formula,
#81 - 16x^4 = 9^2 - (4x^2)^2 = (9 + 4x^2)(9 - 4x^2)#

But we are not quite done.

#9 = 3^2# and #4x^2 = (2x)^2#

So applying the formula to #9-4x^2#, we get the final result of

#81 - 16x^4 = (9 + 4x^2)(9 - 4x^2) = (9 + 4x^2)(3+2x)(3-2x)#