Answer:
(x+1+i)(x+1-i) goes with x^2+2x+2
(x+2i)(x-2i) goes with x^2+4
(x-2+2i)(x-2-2i) goes with x^2-4x+8
Explanation:
(x+1+i)(x+1-i)
(x+[1+i])(x+[1-i])
Use foil.
First: x(x)=x^2
Outer: x(1+i)=x+ix
Inner: x(1-i)=x-ix
Last: (1+i)(1-i)=1-i^2 since 1+i and 1-i are conjugates
__Add together to get: x^2+2x+1-i^2
We can actually simplify this because i^2=-1
So x^2+2x+1-i^2=x^2+2x+1-(-1)=x^2+2x+2
(x+2i)(x-2i)
These are conjugates so just do first and last of foil.
First: x(x)=x^2
Last: 2i(-2i)=-4i^2=-4(-1)=4
==Adding together gives x^2+4
(x-2+2i)(x-2-2i)
(x+[-2+2i])(x+[-2-2i])
This is similar to first.
Foil time!
First: x(x)=x^2
Outer: x(-2-2i)=-2x-2ix
Inner: x(-2+2i)=-2x+2ix
Last: (-2-2i)(-2+2i)=4-4i^2 (multiplying conjugates again)
==Add together giving us x^2-4x+4-4i^2
This can be simplified since i^2=-1.
So applying this gives us x^2-4x+4-4(-1)
=x^2-4x+4+4
=x^2-4x+8