Can we factor multivariate polynomial $x+xy+y$?

Question

It seems to me we can only get $x(1+y)+y$ or $x+(x+1)y$, but maybe something better is possible with complex numbers?

It has zeros at $y=-\frac{x}{x+1}$, but how to use that fact?

Answer 1

One way to use your identification of the zeros is to note that any factor of the polynomial would have to vanish only at those zeros, so nothing of the form $ax+by+c$ will do. You could try dividing out a factor of $y+\frac{x}{x+1}$ (if you don't mind non-polynomial factor)s; but what would happen if you did? The other factor would be $x+1$; and now you can see what's gone wrong. When $x=-1$, $x+xy+y=-1\ne 0$, but the factors are $0$ and undefined. So no, there's no sensible way to factorise $x+xy+y$. (I say "sensible" because of course we can take a constant factor out, but that's uninteresting.)

Answer 2

A possible factorization is :

$\begin{align}x+xy+y &=x+xy+y+1-1\\ &=x(y+1)+(y+1)-1 \\ &=(x+1)(y+1)-1 \\ &=\left(\sqrt{(x+1)(y+1)}\right)^2-1^2 ,\text{assuming $x,y \gt -1$} \\ &=\left(\sqrt{(x+1)(y+1)}+1\right)\cdot\left(\sqrt{(x+1)(y+1)}-1\right)\end{align}$

Answer 3

Depending on the choice of allowable coefficients, some multivariate polynomials factor as a product of two non-constant polynomials, some do not.

For example:

• With rational coefficicients, $x^2-y^2$ factors as $x^2-y^2=(x+y)(x-y)$.$\\[6pt]$
• Allowing real coefficients, $x^2-2y^2$ factors as $x^2-2y^2=\left(x+y\sqrt{2}\right)\left(x-y\sqrt{2}\right)$.$\\[6pt]$
• Allowing arbitrary complex coefficients, $x^2+y^2$ factors as $x^2+y^2=(x+yi)(x-yi)$.$\\[6pt]$

However, even with arbitrary complex coefficients, $x+xy+y$ doesn't factor as a product of two non-constant polynomials.