If $R$ is a commutative ring and $a\in R$, then I want to show that $$I=(f_1(x),\dots, f_r(x),x-a)=(f_1(a),\dots,f_r(a),x-a)$$ where $f_i(x)\in R[x]$, and use this fact to show that $$\frac{R[x]}{I}\simeq\frac{R}{(f_1(a),\dots,f_r(a))}.$$
For proving the second statement, I think I have to use the fact that if $\varphi:R\rightarrow S$ is a homomorphism of rings with kernel $I$ and $\varphi(J)=\overline{J}$ where $J$ is some ideal of $R$, then $$\frac{R/I}{\overline{J}}\cong\frac{R}{I+J}$$ but I am stuck. Any help is appreciated.
For the first fact, you can reason in the quotient ring $R[x]/(x-a)$. There, $x^n\equiv a^n$ so $f_i(x)=c_nx^n+\cdots+c_1x+c_0\equiv c_na^n+\cdots+c_1a+c_0$. This implies that in $R[x]/I$, $f_i(a)\equiv f_i(x)\equiv 0$, so $f_i(a)\in I$. Conversely, for $I'=(f_1(a),\cdots,f_r(a),x-a)$, in $R[x]/I'$, $f_i(x)\equiv f_i(a)\equiv 0$, so $f_i(x)\in I'$. Thus, $I=I'$.
For the second, first consider the evaluation map $R[x]\to R$ defined by $f(x)\mapsto f(a)$. This is surjective, and the kernel is $(x-a)$ (by the division algorithm), so $R[x]/(x-a)$ is isomorphic to $R$. For $\overline{I}$ being the image of $I$ in $R[x]/(x-a)$, we then have $(R[x]/(x-a))/\overline{I}=R[x]/((x-a)+I)=R[x]/I$. For $J$ being the image of $I$ in $R$, $R/J=R/(f_1(a),\dots,f_r(a),a-a)=R/(f_1(a),\dots,f_r(a))$.
This is using the fact that for an isomorphism, we may quotient by corresponding ideals to get isomorphic quotient rings.