How to solve the following problem in Linear Algebra? Let $A\in M_3(\mathbb R$) which is not a diagonal matrix. 
Let $p$ be a polynomial  (in one variable), with real coefficients and of degree $3$ such that $p(A) = 0$.
Pick out the true statements:
(a) $p = cp_A$ where $c\in \mathbb R$ and $p_A$ is the characteristic polynomial of $A$
(b) If $p$ has a complex root (i.e. a root with non-zero imaginary part), then
    $p = cp_A$, with $c$ and $p_A$ as above
(c) If $p$ has a complex root, then $A$ is diagonalizable over $\mathbb C$.
I know if (b) holds then (c) will also hold.
Is there any relation between the characteristics polynomial and any other polynomial? I know that minimal polynomial will divide any other polynomial.
But how to think about these kind or problems?
 A: (a) is false: a nondiagonal matrix $A$ such that $A^2=0$ allows us to build a degree $3$ polynomial that is satisfied by $A$, but is not of the form $cx^3$; the characteristic polynomial of a nilpotent matrix is $x^3$ (or $-x^3$, depending on conventions).
(b) is true. If $p(A)=0$, then the minimal polynomial of $A$ divides $p(x)$; however, the minimal polynomial has real coefficients and, if it divides a (real) polynomial $p(x)$ with a complex root, then it has degree $1$, so $A$ would be diagonal.
(c) is true. Again, the minimal polynomial must divide $p(x)$ and, since $A$ is not diagonal, it must have degree $3$. So, in this case, $p(x)=cp_A(x)$ and so $A$ has distinct eigenvalues, so it is diagonalizable over $\mathbb{C}$.
A: I assume that you meant to say that $A$ is not diagonalizable, as opposed to simply not being diagonal.
(a) is false.  Consider $p(x) = x^3 - x^2$ and
$$
A = \pmatrix{0&1&0\\0&0&0\\0&0&0}
$$
(b) is true.  As soon as $A$ has a complex eigenvalue, we know that the minimal polynomial of $A$ must have degree $3$, since we must also have the conjugate eigenvalue and one real eigenvalue.
As you rightly deduced, (c) is true as well. 
