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.

Since I saw it working, I wanted to write this method for readers as well.

We see that, the polynomial $4x^2-2xy-4x+3y-3$ contains the term of $y$, but not the term of $y^2$.
Thus, we can come to the following conclusion:

If factorization is possible, then only one multiplier polynomial contains the term of $«y»$, but the other multiplier polynomial does not.

Suppose that, the polynomial $4x^2-2xy-4x+3y-3$ is  a factorable polynomial.

Let $ax+by+c=0,\thinspace b≠0$ be the multiplier polynomial, where $-\frac ab=m$ and $-\frac cb=n$.
Then there exist $m,n\in\mathbb R$ such that, if $y=mx+n$, then $4x^2-2xy-4x+3y-3=0$.

This implies,
$$\begin{align}x^2(4-2m)+x(3m-2n-4)+(3n-3)=0\end{align}$$
Hence,
$$\begin{align}&\begin{cases}4-2m=0\\ 3m-2n-4=0\\ 3n-3=0\end{cases}\\
\implies &(m,n)=(2,1).\end{align}$$
This immediately tells us, $y-(2x+1)$ or $2x-y+1$ is a factor.
This means,
$$\begin{align}\color{red}{4x^2}-2xy-4x+3y\color{blue}{-3}\\
=(\color{red}{2x}-y+\color{blue}{1})(\color{red}{2x}\color{blue}{-3}).\end{align}$$
