How to factorize $x^2y^2+a(x^2y+xy^2)+b(x^2+y^2)+cxy+d(x+y)+e$?

Question

What is the condition on complex constants $a,b,c,d,e$ that allows us to factorize $$x^2y^2+a(x^2y+xy^2)+b(x^2+y^2)+cxy+d(x+y)+e$$ in to linear, quadratic or cubic factors?

Answer 1

One possibility:

$\displaystyle \qquad \, \left[ xy+ \frac{a+\sqrt{a^2-4b}}{2} (x+y)- b+\frac{c}{2}+ \frac{ac-2ab-2d}{2\sqrt{a^2-4b}} \right] \\ \quad \times \displaystyle \left[ xy+ \frac{a-\sqrt{a^2-4b}}{2} (x+y)- b+\frac{c}{2}- \frac{ac-2ab-2d}{2\sqrt{a^2-4b}} \right] \\ \displaystyle = x^2y^2+axy(x+y)+b(x^2+y^2)+cxy+d(x+y)-\frac{b(2b-c)^2+ad(2b-c)+d^2}{a^2-4b}$

$$e=-\frac{b(2b-c)^2+ad(2b-c)+d^2}{a^2-4b}$$

In particular $b=\dfrac{a^2}{4}$,

$\displaystyle \qquad \, \left[ xy+ \frac{a(x+y)}{2}+ \frac{c}{2}-\frac{a^2}{4}+ \sqrt{\left( \frac{c}{2}-\frac{a^2}{4} \right)^2-e} \; \right] \\ \quad \times \displaystyle \left[ xy+ \frac{a(x+y)}{2}+ \frac{c}{2}-\frac{a^2}{4}- \sqrt{\left( \frac{c}{2}-\frac{a^2}{4} \right)^2-e} \; \right] \\ \displaystyle = x^2y^2+axy(x+y)+\frac{a^2(x^2+y^2)}{4}+ cxy+ \left( \frac{ac}{2}-\frac{a^3}{4} \right)(x+y)+e$

$$d=\frac{ac}{2}-\frac{a^3}{4}$$