The following question on polynomials arose as a potentially helpful intermediate step on a proof of a Theorem that I want to demonstrate. Its statement is quite elementary, and I can think of a number of applications it might have to the study of Hilbert polynomials in commutative algebra and certain combinatorial sequences arising in discrete geometry.



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> **Conjecture 1**: Let $(c\_0,\ldots,c\_n)$ be non-negative integers and consider the polynomial $P(x) = \sum\_{i=0}^n c\_i \binom{x+n-i}{n}$. Assume that $P(x)$ has at least one negative real-root and denote by $-r$ the largest (in absolute value) negative real-root of $P(x)$. Then either $\lfloor r \rfloor = 0$ or $-\lfloor r\rfloor$ is a root of $P(x)$ too.
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A weaker statement that could also be useful:



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> **Conjecture 1'**: Let $(c\_0,\ldots,c\_n)$ be non-negative integers and consider the polynomial $P(x) = \sum\_{i=0}^n c\_i \binom{x+n-i}{n}$. If $P(-1)\neq 0$, then the negative real roots of $P(x)$ (if any) lie in the interval $(-1,0)$.
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Location of the negative real roots of certain integer-valued polynomials
Conjecture 1' looked a bit easier to test. Unless I've misinterpreted the question (not impossible), I believe the conjecture is false. If we look at the case $n=2$, the polynomial is
$$\begin{align}
P(x)&=c\_0\binom{x+2}{2}+c\_1\binom{x+1}{2}+c\_2\binom{x}{2} \\
&=\frac12\left(c\_0(x+2)(x+1)+c\_1 (x+1)x + c\_2 x(x-1)\right) \\
&=\frac12\left(\left(c\_0+c\_1+c\_2 \right)x^2+\left(3c\_0+c\_1-c\_2 \right)x+2c\_0 \right)
\end{align}$$


Taking $\left(c\_0,c\_1,c\_2\right)=\left(18,1,1\right)$ gives
$$P(x)=10x^2+27x+18=(2x+3)(5x+6)$$


with two negative roots, each less than $-1$.


Is there a reformulation that would be worth checking?


