How do you multiply #(2-2x)^4#?

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Answer 1 · 2015-10-25T05:26:54

#16x^4 + 48x^2+16x+16#

#(2-2x)^4=(2 - 2x)(2 - 2x)(2 - 2x)(2 - 2x)#
You should separate them into two, you must use FOIL, (first, outside, inside, last)

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

Do it again!

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

#16x^4 - 16x^3 + 16x^2 - 16x^3 + 16x^2 - 16x +16x^2 - 16x + 16#

#16x^4 + 48x^2 + 16x + 16 #

It's a very long and boring problem which can probably be simplified with some rule.