How to factorize $2x^2-3xy-2y^2+7x+6y-4=0$ without software? Based on the previous question, I tried to find the roots of x with variable y includes in the equation like this:
$2x^2-x(3y-7)-(2y^2-6y+4)=0$
Then, I can apply the Quadratic formula like here: https://en.wikipedia.org/wiki/Quadratic_formula
$x= \frac {(3y-7)\pm \sqrt{(3y-7)^2-4(-2(2y^2-6y+4))}}{4}$
$x=\frac{(3y-7)\pm \sqrt{(25y^2-90y+81)}}{4}$
Until then, I cannot go further because the equation inside the square root is cannot factorized.
Then, I'm trying to use the other way like this:
$2x^2-x(3x-7)-(2y^2-6y+4)= (ax+by+c)(dx+ey+f)$
$2x^2-x(3x-7)-(2y^2-6y+4)=(adx^2+aexy+afx+bdxy+bey^2+bfy+cdx+cey+cf)$
$2x^2-x(3x-7)-(2y^2-6y+4)=(adx^2+x(aey+bdy+af+cd)+bey^2+bfy+cey+cf)$
But, I think this equation still hard to solve and need some effort to find the answer.
How can I get the easiest way to factorize this equation and also without software too?
 A: Since the polynomial is quadratic in both $x$ and $y$, you can write it in the form:
$$f(x,y) =2 x^2 - x (3 y - 7) - (2 y^2 - 6 y + 4) $$
$$= p^i F_{ij} p^j=(x-x_0,y-y_0)^i F_{ij}(x-x_0,y-y_0)^j$$
where $p^i=(x-x_0,y-y_0)$, while $(x_0,y_0)$ is the value at the minimum/maximum and $F_{ij}$ is the Fisher matrix defined as $F_{ij} =\frac12 \partial_i \partial_j f(x,y)|_{(x,y)->(x_0,y_0)}$.
Simple algebra gives:
$$
F_{ij}=\left(
    \begin{matrix}
    2 & -3/2 \\
    -3/2 & -2 \\
    \end{matrix}
\right), 
$$
and $(x_0,y_0)=(-2/5,9/5)$.
A: Your method is good and correct.
If your all polynomial factors are also polynomials, then $\Delta\left[\Delta_x\right]=0$ or $\Delta\left[\Delta_y\right]=0$.
Indeed, we have
$\Delta\left[\Delta_x\right]=45^2-25\times 81=0$.
This implies,
$$y_1=y_2=\frac 95$$
You can conclude that
$$\begin{align}25y^2-90y+81&=25\left(y-\frac 95\right)^2\\
&=(5y-9)^2\end{align}$$
Thus you get,
$$x_{1,2}=\frac{3y-7±5y-9}{4}$$

The advantage of this method is:
If factoring isn't possible, you'll see it right away.
For example, if $\Delta=25y^2-90y+85$, then we could quickly conclude that polynomial factorization of $f(x,y)$ is impossible.
A: If you factorise the quadratic terms first, leaving unknowns for the constant terms:
$$(2x+y+a)(x-2y+b)$$
Now compare coefficients of $x$ and $y$, and solve $$2b+a=7, b-2a=6\implies a=-1,b=4$$
and this agrees with $ab=-4$
A: Let graph the quadratic form below : https://www.desmos.com/calculator/ezwhow2rmf
$$f(x,y)=2x^2-3xy-2y^2+7x+6y-4$$
Notice that with conics (red curve) you can always find a translation to center it at the origin (green curve), which is equivalent to eliminate terms in $x$ and $y$ and reduce to terms in $x^2,y^2,xy$ only
Just set $\begin{cases}X=x-a\\Y=y-b\end{cases}$ and substitute $f(x,y)=f(X+a,Y+b)$ and try to remove terms in $X^1$ and $Y^1$.
In our case this gives:
$$\underbrace{(2a^2-3ab-2b^2+7a+6b-4)}_{f(a,b)}+\underbrace{(4a-3b+7)}_{f_x(a,b)}X+\underbrace{(-3a-4b+6)}_{f_y(a,b)}Y+2X^2-3XY-2Y^2$$
Annulating the partial derivatives $f_x,f_y$ gives you a system to find $a,b$ directly.
This leads to $(a,b)=(-\frac 25,\frac 95)$.
The simplified translated conic becomes: $$2X^2-3XY-2Y^2$$

And this equation is way easier to factorize into $(2X+Y)(X-2Y)$

