We know that every operator on a finite-dimensional, non-zero, complex vector space has an eigenvalue. The proof of this theorem, as done in Axler's Linear Algebra Done Right book, depends on the fundamental theorem of algebra, where a complex polynomial $p(z)$ is constructed and applied over the operator, i.e., $p(T)$.
Suppose $T \in L(V)$ where $V$ is a real vector space of dimension $n>0$. If the corresponding (real) polynomial $p(x)$ happens to have $n$ roots (including multiplicities), then we can use the proof above to show that $T$ has at least one eigenvalue, which is one of the (real) roots. Here $p(x)$ is constructed the same way as in Axler's proof.
If $p(x)$ has no root, can we conclude that $T$ has no eigenvalue?
If $p(x)$ has less than $n$ roots, can we say anything about the number of eigenvalues of $T$?