Third degree polynomial with zeros $6$ and $-5+2i$ that passes through $(2,-63)$? 
Find a third degree polynomial with real coefficients which has $6$ and $-5+2i$ as zeros and $f(2)=-63$

I tried to substitute in the equation
$$ f(x)=ax^3 +bx^2 + cx +d = 0$$
by the two given zeros and use $f(2)= -63$. But I think we need another equation to determine all the coefficients.

$$\begin{align}f(x)&=a(x-\alpha)(x-\beta)(x-\gamma)\end{align}$$
Using complex conjugate theorem
$$x=-5+2i\text{ is a root}\implies x=-5-2i\text{ is a root}\\(x+5)^2=-4\\x^2+10x+29=0$$
The other factor is $(x-6)$
So far, we have
$$\begin{align}a(x-6)(x^2+10x+29)&=0\end{align}$$
Use the fact that $f(2)=-63$
$$\begin{align}a(-4)(4+20+29)&=-63\\212a&=63\\a&=\dfrac{63}{212}\end{align}$$
$$f(x)=\dfrac{63}{212}(x-6)(x^2+10x+29)$$
is the required polynomial.
