# Minimal polynomial of an $n\times n$ matrix $A$ is $x^3+2x+2$; then $3$ divides $n$

Let $A$ be an $n × n$ matrix with rational entries such that the minimal polynomial of $A$ is $x^3 + 2x+2$.

Prove that $3$ divides $n$.

I think there is no rational root of this polynomial but how I use this. Please give me hint.

That is a monic polynomial with integer coefficients, so any rational roots must be integer. Also none of the divisors $-2,-1,1,2$ of the constant term $2$ are roots, so there are no rational roots at all. Being of degree$~3$ this means the polynomial is irreducible over$~\Bbb Q$.
Now use the theorem (kind of converse to Cayley-Hamilton) that the characteristic polynomial$~\chi$ divides a power of the minimal polynomial. (This is so because the characteristic polynomial of $A$ is the product of the invariant factors associated to $IX-A$, all of which divide to minimal polynomial.) The minimal polynomial being irreducible this means that $\chi$ itself is a power of the minimal polynomial, whence its degree$~n$ is a multiple of the degree $~3$ of the minimal polynomial.
If $A$ is any $n\times n$ matrix with rational entries, the vector space $\def\Q{\Bbb Q}\Q^n$ can be made into a module over the polynomial ring $\Q[X]$ by having $X$ act as multiplication by$~A$ (and constant polynomials by scalar multiplication). If $P\in\Q[X]$ is any polynomial annihilating$~A$ (that is $P[A]=0$) then the action passes to the quotient to define a $\Q[X]/(P)$-module structure on $\Q^n$. Now here one can take in particular $P=X^3+2X+3$ which is irreducible over$~\Q$, so that $K=\Q[X]/(P)$ is a field, over which $\Q^n$ becomes a vector space. One then has $$n=\dim_\Q(\Q^n)=[K:\Q]\dim_K(\Q^n)=3\dim_K(\Q^n),$$ which proves that $3\mid n$.