How does number of signs of pivots same as number of signs of eigenvalues help in finding eigenvalues? Knowing that the number of the signs of pivots of a matrix is equal to the signs of eigenvalues of a matrix, that is the number of positive/negative pivots is the same the as the number of positive/negative eigenvalues, how does this help or simplify the computation to find the eigenvalues of a large matrix? It is said that it would and is a helpful fact to know but I don't understand how is this going to help.
I still have to go through the determinant and polynomial method to find the roots, wouldn't I?
 A: I'm assuming you're talking about the use of the $\mathbf L\mathbf D\mathbf L^T$ decomposition of a symmetric matrix, since it only makes sense to speak of "signs of eigenvalues" when all the eigenvalues are real like in the symmetric case.
The reason why this works out is due to the inertia theorem of Sylvester: briefly put, for a symmetric matrix $\mathbf A$ and some nonsingular matrix $\mathbf W$, $\mathbf A$ and $\mathbf W\mathbf A\mathbf W^T$ have the same number of positive, negative, and zero eigenvalues. ($\mathbf A$ and $\mathbf W\mathbf A\mathbf W^T$ are then said to be congruent.)
Since the relation $\mathbf A=\mathbf L\mathbf D\mathbf L^T$ is a congruence relation, one can then inspect the signs of the diagonal entries (pivots) of $\mathbf D$ to determine the inertia of $\mathbf A$. If the count of signs you obtained for $\mathbf D$ and for the eigenvalues of $\mathbf A$ don't match, then you've made an error somewhere.


I still have to go through the determinant and polynomial method to find the roots, wouldn't I?

For hand computations, yes. One efficient method used by computers involves performing a similarity transformation of the original symmetric matrix into a tridiagonal matrix, a matrix whose nonzero entries are on the diagonal, subdiagonal, and superdiagonal. There is then a simple recursion that can be used for computing the characteristic polynomial (and this recursion can in fact be shown to be equivalent to performing an $\mathbf L\mathbf D\mathbf L^T$ decomposition of the tridiagonal matrix), and any number of methods for finding the roots of a polynomial are then available.
A: Places where I found this fact useful is testing whether a matrix is positive definite or not. A symmetric positive definite matrix is a square matrix such that all eigenvalues are positive, and the matrix itself is symmetric. There are few useful algorithms we can use if we know a matrix is positive definite, such as Cholesky decomposition, but finding eigenvalues can be a pain.
In this particular case, if we find that all the pivots of our square matrix are positive through something like elimination, then we can also know that all of our eigenvalues are positive.
As far as I know, it doesn't make our life simpler when it comes to finding the eigenvalues, but it does make it simpler in finding other computations that require positive eigenvalues.
If you are needed to compute eigenvalues anyways, it can be a good check to see if you got the right eigenvalues. Compare signs, if they match then you can feel a little at ease that you have the right answer.
