 Solution

The prime ideals of $\mathbb{R}[x]$ are $(0)$ and the maximal ideals generated by the monic irreduciblesFor a nonzero prime $I=(p)$, use the residue classes of $1,x,\dots ,x^{N-1}$, where $N=n\mathrm{deg}p$. Multiplication by $x$ on $\mathbb{R}[x]/(p^n)$ is represented by the companion matrix of the monic polynomial $p^n$. In particular, for $p=x-a$ this may equivalently be written as one size-$n$ Jordan block with eigenvalue $a$. The ideal $(0)$ does not give a finite-dimensional quotient.