If 2i is a root of x^3+2x^2+cx+8=0, find c

Help! I do not understand. I want to know how to do it, so if you could try and show me some steps that would be greta. thank you so much i can't tell you how much it means to me!

f(x) = x^3+2x^2+cx+8
f(2i) = 0
(2i)^3 +2(2i)^2 +c(2i) + 8 =0
-8i - 8 + 2ic +8 =0
2ic = 8i
c = 8i/2i
= 4

ok. So, I'd do this by long division. If you know 2i is a root, you know that (x-2i) is part of the factored form. You also know with an imaginary root, there has to be another imaginary root opposite of the first one. So you know (x+2i) exists. Multiply these two together, and you get x^2-4i^2, which equals x^2+4.
now, use long division to divide x^3+2x^2+cx+8 by x^2+4. You will be forced, in the end of your long division, to calculate what c is, but it will be really easy then. Good luck.

Just plug 2i into x^3+2x^2+cx+8=0, & isolate c

Remember i = sqrt(-1) & i^2 = -1

2i is a root implies that x = 2i solves the equation.

Plug in 2i for every x and solve for c:
8(-i) - 8 + 2i*c + 8 = 0
c = 4