\documentclass[english]{article}
\usepackage{setspace}
\usepackage{babel}
\usepackage{amsmath}

\setlength{\oddsidemargin}{0in}
\setlength{\evensidemargin}{0in}
\setlength{\topmargin}{0in}
\setlength{\textheight}{8.5in}
\setlength{\textwidth}{6.5in}
\setlength{\parskip}{0.1in}
\setlength{\parindent}{0in}

\title{Notebook Problem 5.6.34}
\author{Jason Wojciechowski}

\begin{document}
\maketitle

{\bf Theorem:} Let $f(x) \in F[x]$ where $F$ is a field, and let
$\alpha \in F$. Then the remainder $r(x)$ when $f(x)$ is divided by
$x-\alpha$ is $f(\alpha)$.

{\bf Proof:} When we use long division to divide $x-\alpha$ into the
polynomial $a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0$, the first
term in the quotient is $a_nx^{n-1}$. When this term is multiplied by
$\alpha$ and subtracted from $a_{n-1}x^{n-1}$, we're left with
$(a_{n-1} + \alpha )x^{n-1}$, so that the next term in the quotient is
$(a_{n-1} + \alpha )x^{n-2}$. We multiply this by $\alpha$ and
subtract the result from the next term in the numerator, leaving us
with $(a_{n-2}+\alpha a_{n-1} + \alpha ^2)x^{n-2}$.

Thus, every time we multiply a term in the quotient by $\alpha$ and
subtract from the next term in the numerator, we multiply the previous
remainder by $\alpha$ and then add a constant $a_k$ term. This whole
expression is, of course, multiplied by some $x^k$, but the exponent
in that term is decreasing, so it eventually reaches $x^0 = 1$. At
that point, we will just have multiplied $a_1 + a_2\alpha + a_3\alpha
^2 + \ldots + a_n\alpha ^{n-1}$ by $\alpha$ and added an $a_0$,
leaving us with $a_0 + a_1\alpha + a_2\alpha ^2 + \ldots +
a_{n-1}\alpha^{n-1} + a_n\alpha ^n$, which is $f(\alpha)$.

\end{document}