How to factorize $2x^2-3xy-2y^2+7x+6y-4=0$ without software?

Question

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?

Answer 1

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 $(x_0,y_0)=(-2/5,9/5)$.

Answer 2

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,

$$x_{1,2}=\frac{3y-7±5y-9}{4}$$

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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.

Answer 3

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 $ab=-4$

Answer 4

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 $(2X+Y)(X-2Y)$