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?

Question

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.

Answer 1

A cubic polynomial will have three roots. We were given all three roots, since $f(2+i)=0 \space \implies \space f(2-i)=0$. Now we can factor the cubic and write $$x^3+bx^2+cx+d = (x-3)(x-(2+i))(x-(2-i))$$ Once you expand the RHS you can easily find $b,c,d$.

Answer 2

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;