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.

  • $\begingroup$ There are infinitely many other polynomials that satisfy this condition, e.g., $(x_1^2 - x_2^2)(x_3^3 + x_4^3)$ or $(x_1^2 - x_2^2)(x_3^3 + x_4^3)$ or their sum $(x_1^2 - x_2^2)(x_3^3 + x_4^3) + (x_1 + x_2)(x_4^4 - x_3^4)$ $\endgroup$
    – Simon S
    Dec 31, 2014 at 1:35
  • $\begingroup$ These are quartic or above. $\endgroup$
    – Turbo
    Dec 31, 2014 at 1:42
  • $\begingroup$ Ah, ok so your conditions include being the minimum possible degree? Then $a(x_1 + x_2)$ or $b(x_3 + x_4)$ also works, as well of course as the trivial polynomial of zero. $\endgroup$
    – Simon S
    Dec 31, 2014 at 1:44

1 Answer 1


This particular polynomial has degree $2$. The easiest way to note this is simply to apply two (true) statements:

  • If $F(x_1,\ldots,x_n)$ is a polynomial of degree $d$, then $F(a_1 t + b_1,\ldots,a_n t + b_n)$ for constant $a_i$ and $b_i$ is a polynomial of degree at most $d$ - that is, it is a polynomial alone any line.
  • A polynomial of degree $d$ has precisely $d$ roots in the complex plane.

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:

The polynomial $P(t)=F(x_1-t,x_2-t,x_3-t,x_4-t)$ should have a root exactly when $x_1+x_2=2t$ or $x_3+x_4=2t$.

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.

  • $\begingroup$ Does your $P$ relate to my $F$? $\endgroup$
    – Turbo
    Dec 31, 2014 at 3:16
  • $\begingroup$ @Turbo Yes; I changed the $P$'s in my answer to $F$'s in an edit, because it is my intention that I am talking exactly about the function you bring up (and polynomials in general) $\endgroup$ Dec 31, 2014 at 3:20
  • $\begingroup$ I do not understand very well. Could you make the calculations explicit? $\endgroup$
    – Turbo
    Dec 31, 2014 at 3:30
  • $\begingroup$ @Turbo I wrote down a calculation. I hope it helps; I admit this method may be a little confusing (especially since it uses a dummy variable $t$ which appears in the middle of calculation, and disappears by the end), but it works. $\endgroup$ Dec 31, 2014 at 3:41

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