What is (x+2i) x (x-2i)?

(x+2i)(x-2i)

= x(x-2i) + 2i(x-2i)

= x^2 - 2ix + 2ix -4i^2

= x^2 - 4(-1)

= x^2 + 4

X 2i

in case you distribute from the beginning up, the equation could in basic terms be x+2 • x-2 • x+2i • x -2i. So, it fairly is 4 x's, and in case you upload togeher each and all the +2s and -2s, youd have 0. (+2+-2=0. And +2i+-2i=0. they cancel eachother out) so which you basically have the 4 x. So the equation simplified is 4x. in basic terms look at each and all the 2s. They cancel eachother out, and your left with basically the x's.

F.O.I.L. = First Outer Inner Last

First = x * x = x^2
Outer = x * -2i = -2ix
Inner = 2i * x = 2ix
Last = 2i * -2i = -4i^2 = -4(-1) = 4

Add them all up:
x^2 - 2ix + 2ix + 4 = x^2 + 4

Use the formula
( a + b )( a - b ) = a² - b²
= ( x + 2i )( x - 2i )
= ( x )² - ( 2i )²
= x² - 4i²
= x² + 4

hope this helps
Bye !!

(x + 2i)(x - 2i)
= x(x) - x(2i) + 2i(x) - 2i(2i)
= x^2 - 2ix + 2ix - 4i^2
= x^2 - 4(-1)
= x^2 - (-4)
= x^2 + 4

See my answer to your other question!

( x + 2 i ) ( x - 2 i ) = x ² - 4 i ² = x ² + 4