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?
 A: Hints.


*

*The polynomial has real coefficients.  By looking at the two roots you are given, you can write down two more roots without calculation.

*What do you know about the product of the roots of a polynomial?

*What do you know about the sum of the roots of a polynomial?

A: You know two complex roots, hence together with their conjugates you know four of the five roots. You also know the product of all five roots, and you are looking for the negative sum of all five roots.
A: Let the roots be $a_1$, $a_2$, $a_3$, $a_4$, $a_5$. 
Now any polynomial, $(x+a_1)(x+a_2)(x+a_3)(x+a_4)(x+a_5)=0$ implies
$$(x^5)+(x^4)(a_1+a_2+a_3+a_4+a_5)+(x^3)(a_1a_2+a_2a_3+\cdots)+(x^2)(a_1a2a_3+a_2a_3a_4+\cdots)+(x^1)(a_1a_2a_3a_4+a_2a_3a_4a_5+\cdots)+(x^0)(a_1a_2a_3a_4a_5)=0$$
So, $a_1a_2a_3a_4a_5=4$, thus $(1+2i)(1-2i)(3-2i)(3+2i)a_5=4$ and so $a_5=\frac{4}{65}$.
Now $a_1+a_2+a_3+a_4+a_5=a$, meaning $(1+2i)+(1-2i)+(3+2i)+(3-2i)+\frac{4}{65}=a$, therefore $a=\frac{524}{65}$.
A: 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
