Q(x) is a quartic polynomial with roots 2-i and 3+2i
Find the equation for Q(x) if Q(0)=13
Answer is .2(x^4-10x^3+42x^2-82x+65)
Assuming that the coefficients of the quartic are all real, you can say that 2 + i and 3 - 2i are also roots.
Therefore linear factors are (x - [2 - i]), (x - [2 + i]), (x - [3 + 2i]) and (x - [3 + 2i]).
So quadratic factors are (x - [2 - i])(x - [2 + i]) and (x - [3 + 2i])(x - [3 + 2i]).
Now note that
(x - [2 - i])(x - [2 + i] = ([x - 2] + i)([x - 2] - i] = [x - 2]^2 - (i)^2
using the difference of two squares formula
= (x - 2)^2 + 1 = x^2 - 4x + 5.
Do (x - [3 + 2i])(x - [3 + 2i]) the same way to get x^2 - 6x + 13 (check this).
So the quartic is Q(x) = a(x^2 - 4x + 5)(x^2 - 6x + 13).
Now sub Q(0) = 13 and solve for a:
13 = a(5)(13) => a = 1/5.