Let $F$ be a field, $f(x)$ be an irreducible polynomial in $F[x]$ and $I =(f(x))$. Let $f(x)= a_nx^n+···+a_1x+a_0, a_i \in F$ for $i=0,...,n$. And, $\bar{x} = x + I ∈ F[x]/I$ and $\bar{a_i} = a_i + I \in F[x]/I, \forall a_i \in F \subseteq F[x].$
Show that $\bar{a}_{n}(\bar{x})^n+...+\bar{a}_{1}\bar{x}+\bar{a}_{0}=0_{F[x]/I}$.
Here, all I know is that $I$ is a maximal ideal and $F[x]/I$ is a field. But, I don't know how to prove $\bar{a}_{n}(\bar{x})^n+...+\bar{a}_{1}\bar{x}+\bar{a}_{0}=0_{F[x]/I}$. Any help would be much appreciated.
The ring hom $\rm\:g\mapsto \bar g = g + \left<f\right> \:$ preserves $\rm\:\color{#a0f}{sums}\,\ \&\,\ \color{#0a0}{products}\:$ so it preserves $\rm\color{#c00}{polynomials}$ by induction, since polynomials are compositions of sums and products. $ $ More explicitly
$$ \begin{eqnarray} \rm 0\, =\ \overline{\color{#c00}{f(x)}}\: &=&\rm\ \ \overline{a_n x^n +\,\cdots + a_1 x + a_0}\\ &=&\rm\,\ \overline{a_n x^n}\, +\,\cdots + \overline{a_1 x} + \overline a_0\quad by\ \ \ \color{#a0f}{\overline{g+h}\ =\, \overline g + \overline h}\ \ \ \,\forall\ g,h \in F[x]\\ &=&\rm\,\ \overline a_n\, \overline x^n+\,\cdots + \overline a_1\overline x + \overline a_0\quad by\ \ \ \color{#0a0}{\overline{g\, *\, h}\, =\, \overline g\, *\, \overline h}\ \ \ \forall\ g,h\in F[x] \\ &=&\rm\ \bar{\color{#c00}f}(\overline{\color{#c00}x})\\ \end{eqnarray}$$