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.
 A: 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 $x=\frac{y-1}{2}$
$\blacksquare$
A: When $2x=3$, then the expression becomes $$4x^2-2xy-4x+3y-3=9 - 3y -6 + 3y - 3 = 0$$
Hence the expression can be factored by $(2x-3)$, and you can see easily that
$$4x^2-2xy-4x+3y-3 = (2x-3)(2x-y+1)$$
A: Probably not the fastest, but I would like to suggest you use the general factorization method:
We have,
$$
\begin{align}
& 4 x^{2}-2 x y-4 x+3 y-3=4 x^{2}-x(2 y+4)+(3 y-3) \\
\implies & \Delta=(y+2)^{2}-4(3 y-3)=(y-4)^{2} \\
\implies & x_{1}=\frac{y+2+(y-4)}{4}=\frac{y-1}{2} \\
\implies & x_{2}=\frac{y+2-(y-4)}{4}=\frac{3}{2} \\
\implies & 4 x^{2}-2 x y-4 x+3 y-3=4\left(x-\frac{3}{2}\right)\left(x-\frac{y-1}{2}\right) \\
\implies & 4 x^{2}-2 x y-4 x+3 y-3=(2 x-3)(2 x-y+1).
\end{align}
$$
A: 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
$$4x^2 - 2xy - 4x + 3y - 3 = 4\Bigg(x - \dfrac{y - 1}{2}\Bigg)\Bigg(x - \dfrac{3}{2}\Bigg) = (2x - y + 1)(2x - 3).$$
A: 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
$$
(y-\alpha)(y-\beta)
$$
A: $$\begin{align}
4x^2-2xy-4x+3y-3
&=(4x^2-4x-3)-(2x-3)y\\
&=(2x-3)(2x+1)-(2x-3)y\\
&=(2x-3)(2x+1-y)
\end{align}$$
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.
A: 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,
$$\begin{align}\color{red}{4x^2}-2xy-4x+3y\color{blue}{-3}\\
=(\color{red}{2x}-y+\color{blue}{1})(\color{red}{2x}\color{blue}{-3}).\end{align}$$
A: $4x^2-2xy-4x+3y-3 = 4x^2+(-2y-4)x+3y-3=4x^2+[-2(y-1)-2(3)]x+3(y-1)=(2x-y+1)(2x-3).$
