Here is the problem... I will forever be indebted to anyone who can help me, since nobody else seems to be able to. I'm sure it's probably fairly simple and I am going to kick myself later for not being able to figure it out on my own.
Find the cubic equation that has -1 and 2i as roots.
A. y = X^3 - X^2 + 4x - 4
B. y = X^3 + X^2 - 4x - 4
C. y = X^3 + X^2 + 4x + 1
D. y = X^3 + X^2 + 4x + 4
E. y = X^3 - X^2 + 4x - 1
I get the answer of "A" and "D". Here's how I worked through it, and I hope it is right (and that someone with more math-fu can correct me if I am in error):
If 2i is a root, then so is -2i. Similarly, if -1 is a root, then 1 is too. So I take those two roots of 2i and multiply the binomials:
(x+2i)*(x-2i) -> x^2-2ix+2ix-4i (FOIL method) -> x^2+4 (because i=sqrt(-1) and a sqrt*sqrt undoes the root)
Now take that polynomial (x^2+4) and multiply it by a binomial containing your other root (-1):
(x^2+4)*(x-1) -> x^3+4x-x^2-4 (FOIL method) -> x^3-x^2+4x-4 -> choice "A"
(x^2+4)*(x+1) ->x^3+4x+x^2+4 (FOIL method) -> x^3+x^2+4x+4 -> choice "D"