Closed-form expressions for the zeros of $\text{Li}_{-n}(x)$? Consider the first few polylogarithm functions $\text{Li}_{-n}(x)$, where $-n$ is a negative integer and $x\in\mathbb R$ (plotted below). Observation suggests that $\text{Li}_{-1}(x)$ has one zero (at $x=0$), $\text{Li}_{-2}(x)$ has two zeros (one at $x=0$, and one at $x=-1$), and more generally, $\text{Li}_{-n}(x)$ has $n$ zeroes (one at $x=0$, and the rest satisfying $x<0$).
Are there closed-form expressions for these zeroes? In other words, is there a general formula giving the $m$th zero of $\text{Li}_{-n}(x)$?


 A: Extending a bit off of Claude Leibovici's answer, the factorization of palindromic polynomials can be done algebraically in general up to the 9th degree. In particular, if the degree is odd ($n$ is even) then $x+1$ can be factored out, giving an even degree palindromic polynomial. Then the degree is even and the substitution $t=x+\frac1x$ may be used to halve the degree of the polynomial.
For example with $n=7$ we compute $t^2$ and $t^3$ to get
$$\begin{cases}t^2=x^2+2+\frac1{x^2}\\t^3=x^3+3x+\frac3x+\frac1{x^3}\end{cases}\implies\begin{cases}x^2+\frac1{x^2}=t^2-2\\x^3+\frac1{x^3}=t^3-3t\end{cases}$$
and thus see that
\begin{align}&x^6+120x^5+1191x^4+2416x^3+1191x^2+120x+1\\&=x^3\left(x^3+120x^2+1191x+2416+\frac{1191}x+\frac{120}{x^2}+\frac1{x^3}\right)\\&=x^3[(t^3-3t)+120(t^2-2)+1191t+2416]\\&=x^3(t^3+120t^2+1188t+2176)\end{align}
which may then be algebraically solved.
Interestingly, it is shown in this paper that these Eulerian polynomials of degree $k$ have at least one irreducible factor over $\mathbb Q$ of degree at greater than $p_L(k)$, the largest prime less than $k$, and for $k=10$ ($n=11$) it is irreducible over $\mathbb Q$.
