In the above question, I have mangaged to prove that $r_\varphi$ is a root of $f(x)$, namely, $f(\varphi(\overline{x}))=0$. So that can be used as a known fact. I am now trying to prove that the function mapping $\varphi$ to $r_\varphi$ is a bijection from the set of homomorphisms $\varphi$ to the set of roots of $f(x)$ in $R$. I have managed to prove that this function is injective but do not know how to prove that it is surjective. Can someone help me, please? Thanks so much.
Given a root $r \in R$ of $f$, let $\varphi$ be the homomorphism given by evaluation at $r$, that is, \begin{align*} \varphi: \mathbb{Z}[x] &\to R\\ g &\mapsto g(r) \, . \end{align*} Then $\varphi$ is a ring homomorphism and $\langle f \rangle \subseteq \ker(\varphi)$, so $\varphi$ descends to a well-defined homomorphism $\overline{\varphi}: \mathbb{Z}[x]/\langle f \rangle \to R$ on the quotient. Moreover, $$ \overline{\varphi}(\overline{x}) = \varphi(x) = r $$ so $r = r_\overline{\varphi}$, as desired.