I have a polynomial $$F(x_1,x_2,x_3,x_4)=k(x_1+x_2)(x_3+x_4)$$
The polynomial can be described by $$F(x_1,x_2,x_3,x_4)=0\iff (x_1+x_2)=0\mbox{ or }(x_3+x_4)=0$$
Is there a way to formally derive minimum total degree polynomial through interpolation from vanishing conditions?
I am looking at simple derivation of multivariat polynomials from conditions.
This particular polynomial has degree $2$. The easiest way to note this is simply to apply two (true) statements:
Using this, and the fact that there are two solutions desired to $$F(t,t,t,t-2)=0$$ which are $t=0$ and $t=1$, it follows that $F$ is at least quadratic.
Analogously to the one dimensional case, you can construct such a polynomial by fixing $a_1,\ldots,a_n$ - that is, a direction for the parametrization - and computing $F(x_1,\ldots,x_n)$ as the product of the $t$ such that $(a_1t + x_1,\ldots,a_nt+x_n)$ is desired as a root.
To give a less abstract answer, consider the polynomial $F(x_1,x_2,x_3,x_4)$ with roots exactly when $x_1+x_2=0$ or $x_3+x_4=0$. To find a form for this, given the roots, notice the following:
Notice that this represents restricting the domain of $F$ to a single line, parameterized in $t$. However, given that $P(t)$ is univariate and has roots at $\frac{x_1+x_2}{2}$ and $\frac{x_3+x_4}2$, we can find a form for $P(t)$. Assuming the leading coefficient is $1$, this implies that $$P(t)=\left(t-\frac{x_1+x_2}{2}\right)\left(t-\frac{x_3+x_4}2\right)$$ However, since we defined $P(t)=F(x_1-t,x_2-t,x_3-t,x_4-t)$, if we set $t=0$ we get: $$F(x_1,x_2,x_3,x_4)=\frac{x_1+x_2}2\cdot \frac{x_3+x_4}2$$ which is, just a constant coefficient off of what was desired, and which most certainly satisfies the desired conditions.