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

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Answer 1 · 2018-06-27T17:02:28

$(4 x + 2 y + 2) (4 x + 2 y - 2) = 16 x^{2} + 16 x y + 4 y^{2} - 4$ $(4x+2y+2)(4x+2y-2)=16x^2+16xy+4y^2-4$

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

where $z = 4 x + 2 y$ $z=4x+2y$ and

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

= $z^{2} - 4$ $z^2-4$

= ${(4 x + 2 y)}^{2} - 4$ $(4x+2y)^2-4$

= ${(4 x)}^{2} + 2 \cdot 4 x \cdot 2 y + {(2 y)}^{2} - 4$ $(4x)^2+2*4x*2y+(2y)^2-4$

= $16 x^{2} + 16 x y + 4 y^{2} - 4$ $16x^2+16xy+4y^2-4$

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