Prove that if $n\, |\, (x-a)$ and $n \, | \, f(x)$ then $n \, | \, f(a)$ for a polynomial $f$.
It is true if $\operatorname{deg}(f)=1$ or $2$, but what for greater degree of $f(x)$?
I don't know how to continue or if it's any other solution. If possible I would like to prove this without induction.
In general, for polynomial division of $f(x)$ by $g(x)$ we have unique polynomials $q(x),r(x)$ such that $$f(x) = g(x)q(x) + r(x)$$
Where $\deg(r(x)) < \deg(g(x))$. Letting $g(x)=x-a$ we get
$$f(x) = (x-a)q(x) + c$$
for some constant $c$. Since only a constant has lesser degree than $x-a$. The Remainder Theorem tells us this constant must be $f(a)$. So
$$\color{blue}{f(x)} = \color{red}{(x-a)}q(x) + f(a)$$
So if $n$ divides $\color{blue}{f(x)}$ and $\color{red}{(x-a)}$, it must divide $f(a)$.