Factoring $4x^2-2xy-4x+3y-3$. Why isn't it working? I want to factorise the following expression.
$$4x^2-2xy-4x+3y-3$$
Here are the ways I tried
$$4x^2-2xy-4x+3y-3=\left(2x-\frac y2\right)^2+3y-3-\frac{y^2}{4}-4x=\left(2x-\frac y2\right)^2+\frac{12y-12-y^2}{4}-4x=\left(2x-\frac y2\right)^2-\frac 14(y^2-12y+12)-4x$$
Now I need to factor the quadratic $y^2-12y+12$.
So, I calculated discriminant
$$D=12^2-4\times 12=96\implies \sqrt D=4\sqrt 6.$$
This means that the multipliers of quadratic are not rational.  So I don't know how to proceed anymore.

Probably not the fastest, but I would like to suggest you use the general factorization method:
We have,
$$
\begin{align}
& 4 x^{2}-2 x y-4 x+3 y-3=4 x^{2}-x(2 y+4)+(3 y-3) \\
\implies & \Delta=(y+2)^{2}-4(3 y-3)=(y-4)^{2} \\
\implies & x_{1}=\frac{y+2+(y-4)}{4}=\frac{y-1}{2} \\
\implies & x_{2}=\frac{y+2-(y-4)}{4}=\frac{3}{2} \\
\implies & 4 x^{2}-2 x y-4 x+3 y-3=4\left(x-\frac{3}{2}\right)\left(x-\frac{y-1}{2}\right) \\
\implies & 4 x^{2}-2 x y-4 x+3 y-3=(2 x-3)(2 x-y+1).
\end{align}
$$
