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.

Or
$$
\begin{align}
4x^2-2xy-4x+3y-3 &= \\
4x^2-4x-3-y(2x-3) &= \\
(2x-3)(2x+1)-y(2x-3) &= \\
(2x-3)(2x+1-y)
\end{align}
$$
As far as factoring quadratics is concerned (your question about factoring $y^2-12y+12$) if the roots are $\alpha$ and $\beta$ the quadratic factors as
$$
(y-\alpha)(y-\beta)
$$
