<s>[INST] 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.
 [/INST] $$\begin{align}
4x^2-2xy-4x+3y-3
&=(4x^2-4x-3)-(2x-3)y\\
&=(2x-3)(2x+1)-(2x-3)y\\
&=(2x-3)(2x+1-y)
\end{align}$$
If the quadratic $4x^2-4x-3$ hadn't had $2x-3$ as a factor, the polynomial in $x$ and $y$ would not have factored.
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