If $(1 + 2i)$ and $(3 - 2i)$ are two roots of $x^5 + ax^4 + bx^3 + cx^2 + dx + 4$, then $a$ =? Consider the polynomial $x^5 + ax^4 + bx^3 + cx^2 + dx + 4$ where $a, b, c, d$ are real
numbers. If $(1 + 2i)$ and $(3 - 2i)$ are two roots of this polynomial then what is the
value of a?
Well, I know only the 4 roots, which are obvious from what are given (the conjugates serve as the other two), but what next?

i think the solution will come from vieta's theorem .
after this multiple of all the roots will give a 4 . let the fifth root be r then ,
Acc to vieta theorem: product of roots = (-1)^n (a0/an) = -4
-4 = (1+ 2i)(1-2i)(3-2i) (3+2i) r
find the value of r from this ..
r= -65/4
 now after this again from vietas theorem 
a = (-1)((1+2i) + (1-2i) + (3+2i) +(3-2i) + r ( r is the value u got above ) 
a= 33/4
