Polynomial Roots

Let R be a ring, and let $P \in R[x]$. $a \in R$ is a root of P if P(a) = 0.

Proposition: Let F be a field, $P \in F[x]$ and $a \in F$, a is a root of P if and only if x-a divides P.

Proof: if P = (x-a)Q then P(a) = $(a-a)\cdot Q(a)$ = 0. Conversely, P = (x-a)Q+R (here deg(R) < 1), then $P(a) = 0 = (a-a) \cdot Q(a) + R(a)$ so (x-a) divides P.

Let F be a field, $P \in F[x], a \in F, n \in \mathbb{Z}_{>0}$. We say that a is a root of multiplicity n of P if $(x-a)^n \vert P$ and $(x-a)^{n+1} \not\vert P$.

Proposition: Let F be a field, and $a_i \in F$ be distinct (finitely many) roots of P of multiplicity $n_i$, then $deg(P) \geq \sum n_i$

Proof: by definition $(x-a_i)^{n_i} \vert P$. If $i \neq j$ then $a_i \neq a_j$ so $(x-a_i)$ is an irreducible not associated to $(x-a_j)$. Hence $\prod (x-a_i)^{n_i} \vert P$ and $p = \prod (x-a_i)^{n_i}Q$ which means $deg(P) = \sum deg(x-a_i)^{n_i}+deg(Q) \geq \sum n_i$.

Corollary: let F be a field, $P \in F[x]$. Then P has at most deg(P) roots, counting multiplicity.

Let $P \in R[x]$. Then $P'[x] \in R[x]$ (the derivative of P) is defined as follows $P = \sum a_ix^i \rightarrow P' = \sum(i-a_{i+1})x^i$.