Polynomial division in 3 variables I'm trying to compute $$\frac{x^3-2x^2-xyz+2xy+yz}{x-2}$$
I have written this in the form $$\begin{align}x^3-2x^2-xyz+2xy+yz&=(x-2)(x^2+bx+cy+dz)\\
x^3-2x^2-(yz+2y)x+yz&= x^3 + (b-2)x^2 + (cy+dz-2b)x+(-2cy-2dz)
\end{align}$$
This leaves me with the following equations
$$\begin{align}b-2&=-2\\
cy+dz-2b&=yz+2y \\
-2cy-2dz &= yz
\end{align}$$
From this we can see that $b=0$
Therefore we are left with $$\begin{align}cy+dz&=yz+2y \\
cy +dz&=-\frac{yz}{2}
\end{align}$$
I'm now stuck, how can I find values for $c$ and $d$ from this? If I have hit a dead-end, feel free to suggest a different method for solving this equation

Let $F(x)={x}^{3}-xyz-2\,{x}^{2}+2\,xy+yz$. Then by Bézout's theorem we have 
$F(x)=(x-2) Q(x)+F(2).$
Now $F(2)=4y-yz$.
$$
F(x)-F(2)={x}^{3}-xyz-2\,{x}^{2}+2\,xy+2\,yz-4\,y=\left( x-2 \right)  \left( {x}^{2}-yz+2\,y \right).
$$
Thus 
$$
\frac{{x}^{3}-xyz-2\,{x}^{2}+2\,xy+yz}{x-2}={x}^{2}-yz+2\,y+\frac{4y-yz}{x-2}
$$
