<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] A better approach would be to treat the given expression as a quadratic in $x$ and, so we can re-write the expression as $ 4x^2 -2x(y+2) +3(y-1)$. Set it equal to zero, and solve for $x$ using the quadratic formula.
$x = \frac{ 2(y+2) \pm \sqrt{4(y+2)^2-48(y-1)}} {8}$
This gives us the solutions $x=\frac{3}{2} $ and $x=\frac{y-1}{2}$
$\blacksquare$
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