If I have simple polynomial like this: $12 - 8 x - x^2 + x^3$ and I know that it has zero at x=2, I can get Taylor series near x=2, which is $5(x-1)^2$ and I get this picture:
So in 1D case I can naturally divide $(12 - 8 x - x^2 + x^3) / (x-1)^2$ = (x+3)/5 . And this division allows me to find second zero (x=-3) easily.
Now lets consider this 2D function $f = (x - y^2)^2 + (y - x^2)^2$:
Its clear that $f>=0$ and has two zeroes $x=y=0$ and $x=y=1$. Taylor expansion near $x=y=0$ is $t = x^2 (1 - 2 y) + y^2 - 2 x y^2$. But it doesn't seem to be possible to divide $f$ by $t$ without getting some crazy fractionals.
My question is: is it possible somehow to transform f, so it no longer has root $x=y=0$, but still has only root $x=y=1$?