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

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Answer 1 · 2018-06-27T17:02:28

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

We can write #(4x+2y+2)(4x+2y-2)# as #(z+2)(z-2)#,

where #z=4x+2y# and

#(4x+2y+2)(4x+2y-2)=(z+2)(z-2)#

= #z^2-4#

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

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

= #16x^2+16xy+4y^2-4#