Does the Cayley-Hamilton theorem go both ways

If $p$ is a polynomial and $p(A) = 0$, then it can be deduced that the eigenvalues of $A$ are roots of $p$. (As noted in other answers and comments, it does not necessarily follow that every root of $p$ is an eigenvalue of $A$.)

Proof: Let $\lambda$ be an eigenvalue of $A$. By definition, there is a nonzero vector $v$ with $Av = \lambda v$. From $Av = \lambda v$ it follows that $A^2 v = A(Av) = A(\lambda v) = \lambda(Av) = \lambda^2 v$, that $A^3 v = A(A^2 v) = A(\lambda^2 v) = \lambda^2 Av = \lambda^3 v$, and more generally $A^n v = \lambda^n v$ for any positive integer $n$, which implies that $q(A) v = q(\lambda) v$ for every polynomial $q$. Taking $q = p$ to be the polynomial for which $p(A) = 0$, we deduce that $0 v = p(\lambda) v$, which implies that $p(\lambda) v = 0$, which (since $v$ is a nonzero vector) implies that $p(\lambda) = 0$, i.e., that $\lambda$ is a root of $p$.

Note that this argument requires only the definition of "eigenvector" and not the theory of polynomials. In particular, we do not need to know that there exists a "minimal" polynomial of $A$, or identify how other polynomials relate to it. But that broader theory is certainly crucial to a general understanding of operators and the polynomial relations that they satisfy.

The answser to this question is no. This is due to the fact that if a matrix is a root of some some arbitrary polynomial f(n), that implies that f(n) is a multiple of the minimal polynomial of that matrix.