Given some of the roots of the function $f(x) = x^3+bx^2+cx+d$, how do I find the coefficients of that function? Two of the roots of $f(x) = x^3+bx^2+cx+d$ are $3$ and $2+i$. How do I find b+c+d? The answer choices are -7, -5, 6, 9, and 25.

x^3+bx^2+cx+d = 0 has 3 solutions 3 and 2+i are given. third solution z is not.
this polynomial can decomposed as (x-3)(x-2-i)(x-z)=0 from here follows
1. 3*(2+i)*z = -d
2. 3z + 3(2+i) + z(2+i) = c
3. 3 + 2+i + z = -b
these admits a solution:
 b = -5 - i - z;
 c = 6 + 3 i + 5 z + i z, 
 d = -3 z(2 + i)
in order b to be real z = y-i
in order d to be real y = 2;
so b = -7;
c = 17;
d = -15; 
