Question

factor completely: 16x^4-1

Answer 1

It isn't a perfect binomial so you can first find a binomial that multiplied using FOIL gives you half of the the result so then you find what you need to multiply it by to get 16x^4-1 so I first got (2x-1)(2x+1) then (4x^2+1)
so it's (4x^2+1) (2x-1)(2x+1)

Answer 2

do you need me to explain more?

Answer 3

answers
a) \[(4x ^{2}+1)(4x ^{2}-1)\]
b) \[(2x+1)^{3}(2x-1)\]
c) \[(4x ^{2}+1)(2x+1)(2x-1)\]
d) \[(4x+1)(4x-1)\]

Answer 4

From my work I got C