\newlec

\hdc
Compute the quotient and the remainder when $$x^{4} + 2x^{3} + 2x^{2} + 2x + 1$$ is divided by $g(x) = 2x^{2} + 3x + 4$ in $\mathbb{Z}_{5}[x]$.
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\soln

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\includegraphics[scale=0.5]{LongDiv2.png}
\end{center}

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\ic{This is the 10 minute mark}
\prop Let $f(x), g(x) \in \mathbb{F}[x]$ be nonzero polynomials. If $f(x) \mid g(x)$ and $g(x) \mid f(x)$, then $f(x) = cg(x)$ for some $c \in \mathbb{F}$.
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\pf By definition, there exists $q(x)$ and $\hat{q}(x)$ in $\mathbb{F}[x]$ such that 
\begin{align*}
f(x) &= g(x)q(x)\\
g(x) &= f(x)\hat{q}(x)
\end{align*}
Substituting the second equation into the first gives:
\[
f(x) = f(x)\hat{q}(x) q(x) \quad \implies \quad f(x)(1 - \hat{q}(x)q(x)) = 0
\]
As $f(x) \neq 0$, we see that $1 = \hat{q}(x)q(x)$. In fact, $\hat{q}(x)$ and $q(x)$ are nonzero. Now, note that $\deg(1) = 0$ and thus
\[
0 = \deg(\hat{q}(x)q(x)) = \deg(\hat{q}(x)) + \deg(q(x))
\]
(the last equality is an exercise - it holds in generality for nonzero polynomials). Therefore, $\deg(q(x)) = 0 = \deg(\hat{q}(x))$. Therefore, $q(x) = c \in \mathbb{F}$. Thus, substituting this into $f(x) = g(x)q(x)$ gives $f(x) = cg(x)$ completing the proof. \qed
\ic{This is the 25 minute mark}
\thm (Remainder Theorem (RT)) Suppose that $f(x) \in \mathbb{F}[x]$ and that $c \in \mathbb{F}$. Then, the remainder when $f(x)$ is divided by $x-c$ is $f(c)$.
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\pf By the Division Algorithm for Polynomials, there exists unique $q(x)$ and $r(x)$ in $\mathbb{F}[x]$ such that
\[
f(x) = (x-c)q(x) + r(x)
\]
with $r(x) = 0$ or $\deg(r(x)) < \deg(x-c) =1$. Therefore, $\deg(r(x)) = 0$. In either case, $r(x) = k$ for some $k \in \mathbb{F}$. Plug in $x=c$ into the above equation to see that $f(c) = r(c) = k$. Hence $r(x) = f(c)$. \qed
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\ex Find the remainder when $f(z) = z^{2} + 1$ is divided by 
\begin{enumerate}
\item $z-1$
\item $z+1$
\item $z + i+1$
\end{enumerate}
\soln 
\begin{enumerate}
\item By the Remainder Theorem, the remainder is $f(1) = (1)^{2} + 1 = 2$.
\item Note that $z+1 = z-(-1)$. By the Remainder Theorem, the remainder is $f(-1) = (-1)^{2} + 1 = 2$.
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\note $z^{2} + 1 = (z-1)(z+1) + 2$
\item Note that $z +i +1 = z-(-i-1)$. By the Remainder Theorem, the remainder is $f(-i-1) = (-i-1)^{2} + 1 = -1 + 2i + 1 + 1 = 2i+1$.
\end{enumerate}

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\hdc
In $\mathbb{Z}_{7}[x]$, what is the remainder when $4x^{3} + 2x + 5$ is divided by $x+6$?
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\soln Since $x+6 = x-1$ in $\mathbb{Z}_{7}$, we see by the Remainder Theorem that the remainder is
\[
4(1)^{3} + 2(1) + 5 = 11 \equiv 4 \mod{7}
\]
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\ic{Ideally this is the 40 minute mark.}
\thm (Factor Theorem (FT)) Suppose that $f(x) \in \mathbb{F}[x]$ and $c \in \mathbb{F}$. Then the polynomial $x-c$ is a factor of $f(x)$ if and only if $f(c) = 0$, that is, $c$ is a root of $f(x)$.
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\pf Note that $x-c$ is a factor of $f(x)$ if and only if $r(x) = 0$ via the Division Algorithm for Polynomials (DAP) which holds if and only if $r(x) = f(c) = 0$ via the Remainder Theorem (RT). \qed

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\hdc

Prove that there does not exist a real linear factor of $$f(x) = x^{8} + x^{3} + 1.$$
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\soln By the factor theorem, it suffices to show that $f(x)$ has no real roots. We will show that $f(x) > 0$ for all $x\in \mathbb{R}$.
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\textbf{Case 1:} Suppose that $|x| \geq 1$. Then $x^{8} + x^{3} \geq 0$ and hence $f(x) = x^{8} + x^{3} + 1 > 0$.
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\textbf{Case 2:} Suppose that $|x| < 1$. Then $|x^{3}| < 1$ and so  $x^{3}+1 > 0$ and hence $f(x) = x^{8} + x^{3} + 1 > 0$.
\ic{Note here that $-1 < x^{3} < 1$ and $x^{8} \geq 0$. This is the 50 minute mark.}