\input{decls.tex}
\title{Polynomials}
\begin{document}
\maketitle
\begin{itemize}[leftmargin=10em]
	\item[Polynomial in $\C$] \quad $\forall n \in \N, a_{i \in [0, n]} \in \C, a_n \neq 0. \quad P(x) = \sum_{i=0}^{n} a_{i}x^i$
	\item[Degree] \quad $n$
	\item[Leading Coefficient] \quad $a_n$
	\item[Monic Polynomial] \quad Polynomial with $a_n = 1$
\end{itemize}
The Abel-Ruffini theorem states that there exists a degree-5 polynomial with roots that cannot be expressed with $+ - * / \surd$
\section*{Rational Root Theorem}
\[
	\forall \text{ polynomials } P(x) \text { with } \forall i. a_i \in \Z.\quad \forall \frac{p}{q} \in \Q.~P\paren{\frac{p}{q}} = 0 \implies p|a_0 \land q|a_n \\ % x|y x divides y
\]
This means that monic polynomials have no rational non-integral roots.
\section*{Polynomial Division}
\[\forall P(x), D(x).~ \exists Q(x), R(x).~ P(x) = D(x)Q(x) + R(x), \operatorname{degree}(R) < \operatorname{degree}(D)\]
\section*{Remainder Theorem}
\[\forall P(x), c. P(c) = 0 \iff \paren{x - c}|P(x)\]
Proof:
\begin{align*}
	\text{Given that }& P(c) = 0: \\
	& \exists Q(x), R(x).~P(x) = Q(x)(x - c) + R(x) & \text{Division of polynomials} \\
	& P(c) = Q(c)(c - c) + R(c) & \\
	& 0 = Q(c)(0) + R(c) & \\
	& R(c) = 0 & \\
	& \forall x.~R(x) = 0 & \text{As $D(x)$ has degree $1$, $R(x)$ must have degree $0$} \\
	& \therefore~(x - c)|P(x) \\
	\text{Given that }& (x - c)|P(x) \\
	& \exists Q(x).~P(x) = Q(x)(x - c) & \text{Division of polynomials, remainder $0$} \\
	& \therefore~P(c) = Q(c)(c - c) = 0\cdot Q(c) = 0
\end{align*}
\end{document}