Two Polynomials with a Common Quadratic Factor Let $f(x)=x^3-ax^2-bx-3a$ and $g(x)=x^3+(a-2)x^2-bx-3b$. If they have a common quadratic factor, then find the value of $a$ and $b$.
My Attempt
Let $h(x)$ be the common quadratic factor. Then $h(x)$ also the factor of $g(x)-f(x)$, that is
$$(2a-2)x^2+(3a-3b)$$
Since $h(x)$ a quadratic factor, then
$$(2a-2)x^2+(3a-3b)=k \cdot h(x)$$
Where $k$ is a constant.
But, i don't know how to continue, because there are many possible values for $k$.
Any advice?

Since $(2a-2)x^2+(3a-3b)$ is a common factor and the common factor is said to be a quadratic, the (monic) common factor must look like $h(x) = x^2+\dfrac 32 \cdot \dfrac{a-b}{a-1}$.
This is ugly. So let's try and avoid the direction that this is taking us. We can assume that $h(x) = x^2 - \alpha$ where $\alpha = -\dfrac 32 \cdot \dfrac{a-b}{a-1}$ and, for some $u$ and for some $v$
\begin{align}
   f(x) &= (x-u)(x^2-\alpha) \\
   x^3-ax^2-bx-3a &= x^3 - ux^2 - \alpha x +\alpha u \\
   u &= a \\
   \alpha &= b \\
   \alpha u &= -3a
\end{align}
\begin{align}
   g(x) &= (x-v)(x^2-\alpha) \\
   x^3+(a-2)x^2-bx-3b &= x^3 - vx^2 - \alpha x +\alpha v \\
   v &= 2-a \\
   \alpha &= b \\
  \alpha v &= -3b
\end{align}
Since $\alpha = b$, then $\alpha v = -3b 
       \implies b v = -3b
       \implies b(v+3)=0$.
So, either $b=0$ or $v=-3$
If $b=0$, then we must also have $a=0$.
In which case, 
\begin{align}
   f(x) &=x^3 \\
   g(x) &=x^3-2x^2 \\
   h(x) &=x^2
\end{align}
If $v=-3$, then $a=5$, $u=5$, $\alpha = -3$, and $b=-3$.
In which case
\begin{align}
   f(x) &= x^3-5x^2+3x-15 \\
   g(x) &= x^3+3x^2+3x+9 \\
   h(x) &= x^2+3
\end{align}
