What primes divide the discriminant of a polynomial? Given a monic polynomial $p(t) = t^n + ... + c_1 t + c_0$ with integer (or rational) coefficients and with roots $a_1, \dots a_n$, we can compute its discriminant, which is defined to be $\prod_{i< j}(a_i - a_j)^2$.
In my case, I have a polynomial which is the characteristic polynomial of some invertible matrix $T$.  It is palindromic -- i.e., $c_{n-i} = c_i$ for all $0 \leq i \leq n$ -- so the roots come in inverse pairs $a$ and $\frac{1}{a}$.  There are no repeated roots, so the discriminant is non-zero.
My question is: is there any way of knowing which primes divide this discriminant, i.e. from the coefficients of the polynomial or from the matrix $T$?
 A: I disagree with the definition of the discriminant as the resultant of $P$ and $P'$.
When $P$ is a polynomial with integer coefficients, then a prime $q$ should divide the discriminant of $P$ if and only if the reduction of $P$ modulo $q$ has a multiple root (possibly at infinity, when the degree decreases by at least 2 under reduction). But now consider $P=2X^2+ 3X+1$. The resultant of $P$ and $P'$ is $-2$, and the reduction of $P$ modulo 2 has no multiple root. In this case, the well known discriminant $b^2-4ac$ is actually 1. The correct relation between the discriminant and the resultant for a polynomial $P(t)=a_nt^n+\cdots+a_1t+a_0$ is $\mathrm{disc}(P)= (-1)^{n(n-1)/2}\mathrm{res}(P,P')/a_n$. 
A: 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.]
A: I see that Pete beat me to the Resultant response, so I'll give a slightly different answer. For more on this, see page 21 of Ribbenboim's Classical Theory of Algebraic Numbers.
Let $p(x)=x^n+a_1x^{n-1}+\cdots + a_n$. We'd like to find the discriminant $D(p)$ of $p(x)$ using the coefficients only (i.e. without knowing the roots).
Set $p_k=\alpha_1^k + \cdots + \alpha_n^k$, where the $\alpha_i$ are the roots of $p(x)$ and $k=0,1,2,...$. Now for the amazing part. We can find all of the $p_i$ without actually computing any of the $\alpha_i$!
Explicitly, $p_0=n$, $p_1=-a_1$ and $p_i$ for $i>1$ can be computed recursively using the Newton Formulas.
Then 
$D(p)=\displaystyle\det\begin{bmatrix}
  p_0 & p_1 & \cdots & p_{n-1} \newline
  p_1 & p_2 & \cdots & p_n \newline
  \vdots &\vdots & &\vdots\newline
  p_{n-1}&p_n &\cdots &p_{2n-2}
\end{bmatrix}$
