Lemma 1 (cf. factor theorem)   Let $R$ be a commutative ring with identity and let $p(x)\in R[x]$ be a polynomial with coefficients in $R$ . The element $a\in R$ is a root of $p(x)$ if and only if $(x-a)$ divides $p(x)$ .

Proof. See proof of factor theorem using division.

Theorem 1   Let $F$ be a field and let $p(x)$ be a non-zero polynomial in $F[x]$ of degree $n\geq 0$ . Then $p(x)$ has at most $n$ roots in $F$ (counted with multiplicity).

Proof. We proceed by induction. The case $n=0$ is trivial since $p(x)$ is a non-zero constant, thus $p(x)$ cannot have any roots.

Suppose that any polynomial in $F[x]$ of degree $n$ has at most $n$ roots and let $p(x)\in F[x]$ be a polynomial of degree $n+1$ . If $p(x)$ has no roots then the result is trivial, so let us assume that $p(x)$ has at least one root $a\in F$ . Then, by the lemma above, there exist a polynomial $q(x)$ such that: $$p(x)=(x-a)\cdot q(x).$$ Hence, $q(x)\in F[x]$ is a polynomial of degree $n$ . By the induction hypothesis, the polynomial $q(x)$ has at most $n$ roots. It is clear that any root of $q(x)$ is a root of $p(x)$ and if $b\neq a$ is a root of $p(x)$ then $b$ is also a root of $q(x)$ . Thus, $p(x)$ has at most $n+1$ roots, which concludes the proof of the theorem.