<s>[INST] 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?
 [/INST] 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 $a=\frac{524}{65}$.
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