Solution
= Solution
Divide $f$ by $m$ in the <polynomial ring> over the field: $f=qm+r$, where $r=0$ or $\deg r<\deg m$. If $f(a)=0$, then $r(a)=0$, and minimality forces $r=0$. Conversely a multiple of $m$ vanishes at $a$. Therefore
$$
\boxed{f(a)=0\iff m\mid f.}
$$
This includes the zero <polynomial>.