I have the polynomial $$x^2(y+z) + y^2(x+z) + z^2(x+y)$$ that I want to factor.
I know how to factor $x^2(y-z) + y^2(z-x) + z^2(x-y)$. Because $(y - z) + (z - x) + (x - y) = 0$, we can use the substitution $(y-z) = -((z-x) + (x-y))$. We get $-x^2(z-x) - x^2(x-y) + y^2(z-x) + z^2(x-y)$. Grouping terms with $(z - x)$ together and terms with $(x - y)$ together, we get $$(y^2 - x^2)(z - x) + (z^2 - x^2)(x - y)$$ Factoring the difference of squares, $$(y - x)(y + x)(z - x) + (x - y)(z - x)(z + x)$$ Factoring out the $(x - y)(z - x)$, $$(x - y)(-y - x)(z - x) + (x - y)(x + z)(z - x) = (x - y)(z - y)(z - x) = -(x - y)(y - z)(z - x)$$
However in the case $x^2(y+z) + y^2(x+z) + z^2(x+y)$, I don't see an easy substitution similar to the one above.
We can use the multi-variable factor theorem, but it takes guessing and I want to avoid guessing. I would like a solid way to factor this.
it does not factor at all, as this matrix has full rank:
$$ \left( \begin{array}{rrrrrrrrrr} 0 & 0 & 0 & 2 & 2 & 0 & 1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 2 & 2 & 0 & 1 & 0 & 1 \\ 1 & 2 & 0 & 0 & 2 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 2 & 0 & 0 & 0 & 2 & 1 \\ 1 & 0 & 2 & 1 & 2 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 2 & 0 & 1 & 2 & 0 & 0 \\ 0 & 0 & 1 & -1 & 0 & 1 & 0 & -1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & -1 & 0 & 0 & -1 & 0 \\ \end{array} \right) $$
This is Corollary 3 on the top of page 213 in Schinzel, Polynomials with Special Regard to Reducibility published 200, Cambridge University Press. Chapter 3, section 2 is called Theorems of Ruppert.
Also see Brookfield, although I don't believe he includes this particular result.