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.
On
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 $x=\frac{y-1}{2}$
$\blacksquare$
On
Probably not the fastest, but I would like to suggest you use the general factorization method:
We have,
$$ \begin{align} & 4 x^{2}-2 x y-4 x+3 y-3=4 x^{2}-x(2 y+4)+(3 y-3) \\ \implies & \Delta=(y+2)^{2}-4(3 y-3)=(y-4)^{2} \\ \implies & x_{1}=\frac{y+2+(y-4)}{4}=\frac{y-1}{2} \\ \implies & x_{2}=\frac{y+2-(y-4)}{4}=\frac{3}{2} \\ \implies & 4 x^{2}-2 x y-4 x+3 y-3=4\left(x-\frac{3}{2}\right)\left(x-\frac{y-1}{2}\right) \\ \implies & 4 x^{2}-2 x y-4 x+3 y-3=(2 x-3)(2 x-y+1). \end{align} $$
On
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 $$4x^2 - 2xy - 4x + 3y - 3 = 4\Bigg(x - \dfrac{y - 1}{2}\Bigg)\Bigg(x - \dfrac{3}{2}\Bigg) = (2x - y + 1)(2x - 3).$$
On
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 $$ (y-\alpha)(y-\beta) $$
On
$$\begin{align} 4x^2-2xy-4x+3y-3 &=(4x^2-4x-3)-(2x-3)y\\ &=(2x-3)(2x+1)-(2x-3)y\\ &=(2x-3)(2x+1-y) \end{align}$$
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.
On
Thinking it might be helpful to readers, I used another alternative method in this answer that I've seen works.
First, let's write the given polynomial in general factored form:
$$\begin{align}&\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace 4x^2-2xy-4x+3y-3\\ =&(a_1x+a_2y+a_3)(a_4x+a_5y+a_6)\end{align} \tag 1$$
where $a_1,a_2,a_3,a_4,a_5,a_6\in\mathbb R.$
Then we have,
$$\begin{align}4x^2-2xy-4x+3y-3=&\;\,\,a_1a_4x^2+a_2a_5y^2\\ &+(a_1a_5+a_2a_4)xy\\ &+(a_1a_6+a_3a_4)x\\ &+(a_2a_6+a_3a_5)y\\ &+\,\,a_3a_6\end{align}$$
This gives us,
$$\begin{cases}a_1a_4=4\\a_2a_5=0\\ a_1a_5+a_2a_4=-2\\a_1a_6+a_3a_4=-4\\ a_2a_6+a_3a_5=3\\a_3a_6=-3 \end{cases}$$
Since $a_2=0$ (without loss of generality), then we have
$$\begin{cases}a_1a_4=4\\ a_1a_5=-2\\a_1a_6+a_3a_4=-4\\ a_3a_5=3\\a_3a_6=-3 \end{cases}$$
Since $a_3≠0$, then $a_6=-a_5$. This implies,
$$\begin{align}\begin{cases}a_1a_4=4\\ a_1a_5=-2\\a_3a_4-a_1a_5=-4\\ a_3a_5=3 \end{cases}&\implies \begin{cases}a_1a_4=4\\ a_1a_5=-2\\a_3a_4=-6\\ a_3a_5=3 \end{cases}\end{align} $$
Since $a_1,a_3≠0$, then $a_5=-\frac 12a_4$. This implies,
$$\begin{align}\begin{cases}a_1a_4=4\\a_1a_5=-2\\a_3a_4=-6\\ a_3a_5=3 \end{cases}&\implies \begin{cases}a_1a_4=4 \\a_3a_4=-6\end{cases}\\ &\implies a_1=-\frac 23a_3\end{align} $$
Let $a_3=k,\thinspace k≠0$, then we get
$$\begin{cases}a_1=-\frac {2k}{3}\\ a_2=0\\a_3=k\\a_4=-\frac {6}{k}\\ a_5=\frac 3k\\ a_6=-\frac 3k\end{cases}$$
Finally, we conclude that
$$\begin{align}\left(a_1x+a_2y+a_3\right)\left(a_4x+a_5y+a_6\right)&=\left(-\frac {2k}{3} x+k\right)\left(-\frac 6kx+\frac 3ky-\frac 3k\right)\\ &=\left(-\frac 23x +1\right)\left(-6x+3y-3\right)\\ &=\left(2 x-3\right)\left(2 x-y+1\right).\end{align}$$
Explanation: $(1)$
We see that, the degree of the polynomial $4x^2-2xy-4x+3y-3$ is $2$.
Then, if the polynomial $4x^2-2xy-4x+3y-3$ is a factorable polynomial, we claim that its possible multiplier should be $ax+by+c$, such that if $a=0$ then $b≠0$ or if $b=0$, then $a≠0$ (We accept the $ax+by+c$ as a non-constant polynomial).
Because, if the multiplier polynomial had a degree of $2$ (e.g. $ax+by+cxy+d$ ), then the second multiplier polynomial would have to be a constant polynomial. Otherwise, the polynomial obtained by multiplying two polynomials would have a degree of at least $3$.
Thus, factorization would not be possible. Because the constant polynomial is not the multiplier polynomial we are looking for. This tells us that, one of the multiplier polynomial must be of the form
$$ax+by+c.$$
Knowing that the degree of the polynomial $4x^2-2xy-4x+3y-3$ is $2$, then we can easily write that,
$$\begin{align}&\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace 4x^2-2xy-4x+3y-3\\ =&(a_1x+a_2y+a_3)(a_4x+a_5y+a_6).\end{align}$$
On
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,
$$\begin{align}\color{red}{4x^2}-2xy-4x+3y\color{blue}{-3}\\ =(\color{red}{2x}-y+\color{blue}{1})(\color{red}{2x}\color{blue}{-3}).\end{align}$$
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 $$4x^2-2xy-4x+3y-3 = (2x-3)(2x-y+1)$$