How do you factor #(4y-2)^2-(3y)^2#?

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Answer 1 · 2015-06-27T07:37:06

This is a difference of squares:

#(4y-2)^2-(3y)^2#

#= ((4y-2)-3y)((4y-2)+3y)#

#=(y-2)(7y-2)#

For any #a# and #b#, we have #a^2-b^2 = (a-b)(a+b)#

Let #a=4y-2# and #b=3y#

Then

#(4y-2)^2-(3y)^2#

#=a^2 - b^2#

#=(a-b)(a+b)#

#= ((4y-2)-3y)((4y-2)+3y)#

#=(y-2)(7y-2)#