I'm trying to show that $M_n(R[x]) \cong (M_n(R)[x]$ so I consider the mapping that sends an element $A \in M_n(R[x])$ to the polynomial whose coefficients are matrices in which the entries of those matrices are the coefficients of the polynomials of in $A$. So for example:
$$\begin{bmatrix} x^2 & x \\ 3 & 2x+1 \end{bmatrix} \mapsto \begin{bmatrix} 1 & 0 \\ 0 & 0 \\ \end{bmatrix} x^2 + \begin{bmatrix} 0 &1 \\ 0 & 2 \end{bmatrix} x + \begin{bmatrix} 0 & 0 \\ 3 & 1 \\ \end{bmatrix} $$ Clearly $\phi(A+B) = \phi(A) + \phi(B)$ however it doesn't seem to hold for multiplication. Is this the correct mapping I'm suppose to be considering or is there another that would work better, thanks in advance.
Note that $M_n(R[x])$ is a free $R[x]$ module, and has basis the elementary matrices $E_{ij}$ which are all $0$ except for the $ij$th entry, which is 1. Thus a basis for $M_n(R[x])$ as a free $R$ module is $x^kE_{ij}$. Similarly, a basis for $M_n(R)[x]$ is $E_{ij}x^k$, where I'm using the order of multiplication to distinguish these two rings.
You're trying to show that these $R$-algebras are isomorphic as (presumably) $R$-algebras, so the map should be $R$-linear. Thus such a map is defined by what it does on an $R$-generating set, or in this case since the $R$-algebras are free modules, an $R$-basis. Your map $\phi$ is the map defined by sending $x^kE_{ij}$ to $E_{ij}x^k$. Since it is $R$-linear, it certainly preserves addition, so we just need to check that this map does indeed preserve multiplication. For this, it suffices to check on the basis. $$\phi\newcommand\of[1]{\left({#1}\right)}\of{\of{x^kE_{ij}}\of{x^{k'}E_{i'j'}}} =\phi\of{x^{k+k'}E_{ij}E_{i'j'}}=\phi\of{\delta_{ji'}x^{k+k'}E_{ij'}}=\delta_{ji'}E_{ij'}x^{k+k'},$$ and $$\phi\of{x^kE_{ij}}\phi\of{x^{k'}E_{i'j'}}=\of{E_{ij}x^k}\of{E_{i'j'}x^{k'}}=E_{ij}E_{i'j'}x^{k+k'}=\delta_{ji'}E_{ij'}x^{k+k'}.$$ Thus $\phi$ preserves the multiplication as well.
If you think I've made an error, do let me know.