For every $n\geq 3$ ,exists $g_n(x) \in \mathbb{Z}[x]$ of degree 2 : $A^n=g_n(A)$

Question

Let $V=\mathbb{R}_2[x]$ and $f: V \to V $ is a linear map ,such as : $f(\phi(x))=\phi(x)'+\phi(1)x\;,\;\forall \;\phi(x) \in V$. $B=\{1,x,x^2\}$ a standard basis of V.
a)Prove that there exists a basis of $V$ constructed by eigenvalues of $f$ and find one.
b)If $A$ is the matrix of $f$ constructed by the basis b ,then $\forall\; n \geq 3 \;\; \exists\; g_n(x) \in \mathbb{Z}[x]$ of degree 2 for which $g_n(A)=A^n$.


For a) i worked this way : I found that $A=\begin{pmatrix} 0 & 1 & 0 \\ 1 & 1 & 3 \\ 0 & 0 & 0 \\ \end{pmatrix}$, with a characteristic polynomial $p(x)=x(x^2-x-1)$
So we get 3 distinct eigenvalues $x_1=0 \; \;,x_{2,3}=\frac{1 \pm \sqrt{5}}{2}$ and thus $A$ is diagonalizable and of course we can get 3 linear independent eigenvectors and constract a basis of V. The basis I found consists of the basis of eigenspaces $V_A(0),V_A(\frac{1 +\sqrt{5}}{2}),V_A(\frac{1 - \sqrt{5}}{2})$ respectively , $v=\{(-3,0,1)^t\;,\;(-1+\sqrt{5},2,0)\;,\;(-1-\sqrt{5},2,0) \}$.

b) is the part that kind of fries my brain .If anyone can help or provide a hint it would be appreciated.