\newlec

\hdc
How many of the following statements are true?
 
 \begin{itemize}
\item Every complex cubic polynomial has a complex root.
\item When $x^3+6x-7$ is divided by a quadratic polynomial $ax^2+bx+c$ in $\mathbb{R}[x]$, \\
then the remainder has degree 1.
\item If $f(x), g(x) \in \mathbb{Q}[x]$, then $f(x)g(x) \in \mathbb{Q}[x]$.
\item Every non-constant polynomial in $\mathbb{Z}_5[x]$ has a root in $\mathbb{Z}_5$.
\end{itemize}

\begin{enumerate}[A)]
\item 0
\item 1
\item 2
\item 3
\item 4
\end{enumerate}
\vskip3em
\soln The first statement is true by the Fundamental Theorem of Algebra. The second is false since $x-1$ is a factor of the cubic polynomial and so there must be a quadratic factor as well. The third is true since $\mathbb{Q}[x]$ forms a ring. The last is false since say $f(x) = x(x-1)(x-2)(x-3)(x-4)+1$ has no roots over $\mathbb{Z}_{5}[x]$. Hence the answer is $2$.

\newpage

Recall:
\vskip1em
\thm (Conjugate Roots Theorem (CJRT)) If $c \in \mathbb{C}$ is a root of a polynomial $p(x) \in \mathbb{R}[x]$ (over $\mathbb{C}$) then $\overline{c}$ is a root of $p(x)$. 
\vskip1em
\note This is not true if the coefficients are not real, for example $(x+i)^{2} = x^{2} + 2ix - 1$.
\vskip1em
\ex Factor
\[
f(z) = z^{5} - z^{4} -z^{3} + z^{2} - 2z +2
\]
over $\mathbb{C}$ as a product of irreducible elements of $\mathbb{C}[x]$ given that $i$ is a root.
\vskip1em
\pf Note by CJRT that $\pm i$ are roots. By the Factor Theorem, we see that $(z-i)(z+i) = z^{2}+1$ is a factor. Note that $z-1$ is also a factor since the sum of the coefficients is $0$.Hence, $(z^{2}+ 1)(z-1) = z^{3}-z^{2}+z-1$ is a factor. By long division,
\begin{center}
\includegraphics[scale=0.25]{LongDiv5.png}
\end{center}
we see that $f(z) = (z^{3}-z^{2}+z-1)(z^{2}-2) = (z-i)(z+i)(z-1)(z-\sqrt{2})(z + \sqrt{2})$ is a full factorization. \qed

\newpage

Factor $f(z) = z^{4} -5z^{3} + 16z^{2} -9z -13$ over $\mathbb{C}$ into a product of irreducible polynomials given that $2-3i$ is a root.
\vskip1em
Factors are (using the Factor Theorem and CJRT)
\[
(z-(2-3i))(z-(2+3i)) = z^{2} - 4z + 13
\]
After long division,
\[
f(z) = (z^{2} - 4z + 13)(z^{2}-z-1)
\]
By the quadratic formula on the last quadratic,
\begin{align*}
z &= \frac{-(-1) \pm \sqrt{(-1)^{2} -4(1)(-1)}}{2(1)}\\
&= \frac{1 \pm \sqrt{5}}{2}
\end{align*}
Hence, $f(z) = (z-(2-3i))(z-(2+3i))(z - (1 + \sqrt{5})/2)(z - (1-\sqrt{5})/2)$. \qed

\newpage

\thm (Real Quadratic Factors (RQF)) Let $f(x) \in \mathbb{R}[x]$. If $c \in \mathbb{C} -\mathbb{R}$ and $f(c) = 0$, then there exists a $g(x) \in \mathbb{R}[x]$ such that $g(x)$ is a real quadratic factor of $f(x)$.
\vskip1em
\pf Take 
\begin{align*}
g(x) &= (x-c)(x-\overline{c})\\
&= x^{2} - (c + \overline{c})x + c\overline{c}\\
&= x^{2} -2\Re(c)x + |c|^{2} \in \mathbb{R}[x]
\end{align*}
It suffices to show that $g(x)$ is a factor of $f(x)$. By the Division Algorithm for Polynomials, there exists a unique $q(x)$ and $r(x)$ in $\mathbb{R}[x]$ such that
\[
f(x) = g(x)q(x) + r(x)
\]
with $r(x)= 0$ or $\deg(r(x)) < \deg(g(x)) = 2$, that is, $r(x)$ is either constant or linear. Substituting $x=c$ into the above gives
\[
0 = f(c) = g(c)q(c) + r(c)=r(c)
\]
and hence $r(c) = 0$. Assume towards a contradiction that $r(x)$ is linear. By definition, $r(x) = ax+b \in \mathbb{R}[x]$ with $a \neq 0$. Then 
\[
r(c) = ac + b = 0 \quad\quad \implies \quad\quad c = \tfrac{-b}{a} \in \mathbb{R}
\]
and this is a contradiction. Therefore, $r(x)$ is a constant polynomial and since $r(c) = 0$, we have that $r(x) = 0$ and thus $g(x) \mid f(x)$. \qed
\vskip1em
\thm (Real Factors of Real Polynomials (RFRP)) 
Let $f(x) = a_{n}x^{n} + ... + a_{1}x + a_{0} \in \mathbb{R}[x]$. Then $f(x)$ can be written as a product of real linear and real quadratic factors,
\vskip1em
\pf By CPN, $f(x)$ has $n$ roots over $\mathbb{C}$. Let $r_{1}, r_{2},... ,r_{k}$ be the real roots and let $c_{1}, c_{2}, ... ,c_{\ell}$ be the strictly complex roots. By CJRT, complex roots come in pairs, say $c_{2} = \overline{c_{1}}$, $c_{4} = \overline{c_{3}}$, ..., $c_{\ell} = \overline{c_{\ell-1}}$ (hence also $\ell$ is even). For each pair, by RQF, we have an associated quadratic factor, say $q_{1}(x), q_{2}(x), ... ,q_{\ell/2}(x)$. By the Factor Theorem, each real root corresponds to a linear factor, say $g_{1}(x), ..., g_{k}(x)$. Hence
\[
f(x) = cg_{1}(x)...g_{k}(x)q_{1}(x)...q_{\ell/2}(x)
\]
where $c$ is the coefficient of the leading term completing the proof. \qed

\newpage
\hdc

Prove that a real polynomial of odd degree has a real root.
\vskip3em
\soln Assume towards a contradiction that $p(x)$ is a real polynomial of odd degree without a root. By the Factor Theorem, we know that if $p(x)$ cannot have a real linear factor. By Real Factors of Real Polynomials, we see that
\[
p(x) = q_{1}(x) ... q_{k}(x)
\]
for some quadratic factors $q_{i}(x)$. Now, taking degrees shows that 
\[
\deg(p(x)) = 2k
\]
contradicting the fact that the degree was of $p(x)$ is odd. Hence, the polynomial must have a real root. \qed