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.

Treat $4x^2 - 2xy - 4x + 3y - 3$ as a quadratic in $x$.
Set $4x^2 - 2x(y + 2) + 3(y - 1) = 0$, then solve for $x$ (using the Quadratic Formula), to factor the LHS:
$$x = \dfrac{2(y+2) \pm \sqrt{(2(y+2))^2 - 4(4)(3)(y-1)}}{8} = \dfrac{y + 2 \pm \sqrt{y^2 + 4y + 4 - 12y + 12}}{4} = \dfrac{y + 2 \pm \sqrt{y^2 - 8y + 16}}{4} = \dfrac{y + 2 \pm \sqrt{(y - 4)^2}}{4} = \dfrac{(y + 2) \pm (y - 4)}{4}$$
so that the LHS factors as
$$4x^2 - 2xy - 4x + 3y - 3 = 4\Bigg(x - \dfrac{y - 1}{2}\Bigg)\Bigg(x - \dfrac{3}{2}\Bigg) = (2x - y + 1)(2x - 3).$$
