Another proof that $A[x]\otimes M\cong M[x]$

Question

Let $A$ be a ring and $M$ be an $A$-module. If $x$ is an indeterminate, I know that $A[x]\otimes_AM\cong M[x]$ ($A[x],M[x]$ are clearly $A$-modules). However my question is if one can prove this equality using the exact sequences, plus the facts that $M\otimes_A-$ is right-exact and that $A[x]\otimes _A-$ is exact.