As Fernando Muro points out in the comments, Sylvester's law of inertia is probably the easiest way to determine the signature. You diagonalize the symmetric matrix by orthogonal transformations, by the Gram-Schmidt process. This is essentially as easy as Gaussian elimination.
A symmetric matrix is equivalent to a bilinear form, with basis $\{v_1, \ldots, v_n\}$, and matrix entries inner products $\langle v_i,v_j\rangle, i,j\leq n$. If $\langle v_1,v_1\rangle \neq 0$, then replace $v_i$ with $v_i-v_1 \langle v_1,v_i\rangle/\langle v_1,v_1\rangle$, and proceed by induction (this has the effect of performing the same row and column operations, so as to keep the matrix symmetric). The only variation on the standard Gram-Schmidt is what to do when $\langle v_1,v_1\rangle=0$, which clearly means the form is indefinite. If $\langle v_1, v_i\rangle=0$ for all $i\leq n$, then you can proceed by induction. Otherwise, by permuting indices, you may assume that $\langle v_1,v_2\rangle \neq 0$. Replace $v_1$ with $v_2+tv_1$, such that $\langle v_2,v_2\rangle+2t\langle v_1,v_2\rangle\neq 0$, and proceed as with the usual Gram-Schmidt. Once you have diagonalized in this way, Sylvester's law implies that the signature of the diagonal matrix is the same as that of the original one.