Vieta's Formula - Cubic Polynomial

Question

Suppose $p(x)$ is a monic cubic polynomial with real coefficients such that $p(3-2i)=0$ and $p(0)=-52$.

Determine $p(x)$ (in expanded form).

I know that the roots are $3-2i$, $3+2i$, and something else, so it is $(x-(3-2i)) \cdot (x-(3+2i))$ then multiplied by the other root. Do I try to plug in $x=0$ and try to find something?

Answer 1

Your intentions are correct. As your polinomial is monic, it will be $$p(x)=[x-(3-2i)][x-(3+2i)](x-\alpha)=(x^2-6x+13)(x-\alpha)$$ I've multiplied $[x-(3-2i)][x-(3+2i)]$ because the substitution $x=0$ will be easier.

Now, plug $x=0$ and find $\alpha$.