First, when you say "it is symmetric", you probably mean that the polynomial $P(t) = a_n t^n + ... + a_1 t + a_0$ satisfies $a_{n-i} = a_i$ for all $0 \leq i \leq n$, not that the matrix is symmetric, since it is the former condition which implies that the set of roots is invariant under taking reciprocals (and also that $0$ is not a root). Such a polynomial is more commonly called palindromic.

The connection with the matrix seems unhelpful, because every polynomial is the characteristic polynomial of some matrix, e.g. its companion matrix. (It could possibly become helpful if you had some additional information about the matrix.)

You ask whether one can tell which primes divide the discriminant from the coefficients of the polynomial. The answer is a resounding yes, although perhaps not in a way which will be satisfying to you: you can compute the discriminant directly from the coefficients of the polynomial and then you can factor it! The formula you gave is actually not very good for *computing* the discriminant: for that it is better to use

$\operatorname{disc}(P) = (-1)^{\frac{(n)(n-1)}{2}} \frac{\operatorname{Res}(P,P')}{a_n}$,

where $P'(t)$ is the derivative and $\operatorname{Res}$ is the resultant, computed using its interpretation as the determinant of the Sylvester matrix.

[Thanks to Michel Coste for pointing out that the discriminant is not quite equal to the resultant of $P$ and $P'$ when $P$ is not monic.]

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