\newlec

\thm Rational Roots Theorem (RRT) If $f(x) = a_{n}x^{n} + ... + a_{1}x + a_{0} \in \mathbb{Z}[x]$ and $r = \tfrac{s}{t} \in \mathbb{Q}$ is a root of $f(x)$ over $\mathbb{Q}$ in lowest terms, then $s \mid a_{0}$ and $t \mid a_{n}$.
\vskip1em
\pf Plug $r$ into $f(x)$:
\[
0 =  a_{n}(\tfrac{s}{t})^{n} + ... + a_{1}(\tfrac{s}{t}) + a_{0}.
\]
Multiply by $t^{n}$
\[
0 = a_{n}s^{n} + a_{n-1}s^{n-1}t + ... + a_{1}st^{n-1} + a_{0}t^{n}.
\]
Rearranging gives
\[
a_{0}t^{n} = -s(a_{n}s^{n-1} + a_{n-1}s^{n-2}t + ... + a_{1}t^{n-1})
\]
and hence $s \mid a_{0}t^{n}$. Since $\gcd(s,t)=1$, we see that $\gcd(s,t^{n})=1$ (following from GCDPF) and hence $s \mid a_{0}$ by Coprimeness and Divisibility. Similarly, $t \mid a_{n}$. \qed
\vskip1em
\ex Find the roots of 
\[
2x^{3} + x^{2} - 6x - 3 \in \mathbb{R}[x]
\]
\soln By the Rational Roots Theorem, if $r$ is a root, then writing $r = \tfrac{s}{t}$, we have that $s \mid -3$ and $t \mid 2$. This gives the following possibilities for $r$:
\[
\pm1, \pm3, \pm\frac{3}{2}, \pm\frac{1}{2}
\]
Trying each of these possibilities one by one shows that $r = -\tfrac{1}{2}$ is a root since
\[
2(\tfrac{-1}{2})^{3} + (\tfrac{-1}{2})^{2} -6(\tfrac{-1}{2}) - 3 = \tfrac{-1}{4} + \tfrac{1}{4} + 3 - 3 = 0
\]
Hence $(x + \tfrac{1}{2})$ or $(2x+1)$ is a factor. By long division (or grouping and factoring), we see that
\[
2x^{3} + x^{2} - 6x - 3 =(2x+1)(x^{2}-3) = (2x+1)(x - \sqrt{3})(x + \sqrt{3})
\]
Hence all real roots are given by $\tfrac{-1}{2}, \pm \sqrt{3}$. \qed

\ic{This is the 15 minute mark.}

\newpage
\hdc

Factor $x^{3} - \tfrac{32}{15}x^{2} + \tfrac{1}{5}x + \tfrac{2}{15}$  as a product of irreducible polynomials over $\mathbb{R}$.
\vskip3em
\soln The above polynomial is equal to
\[
\tfrac{1}{15}(15x^{3} - 32x^{2} +3x + 2) = f(x)
\]
By the Rational Roots Theorem, possible roots are 
\[
\pm1, \pm\tfrac{1}{3}, \pm \tfrac{1}{5}, \pm\tfrac{1}{15}, \pm 2, \pm\tfrac{2}{3}, \pm \tfrac{2}{5}, \pm\tfrac{2}{15}, 
\]
Note that $x=2$ is a root. Hence by the Factor Theorem, $x-2$ is a factor. By long division:
\begin{center}
\includegraphics[scale=0.5]{LongDiv4.png}
\end{center}
we have that $f(x) = \tfrac{1}{15}(x-2)(15x^{2} -2x-1) = \tfrac{1}{15}(x-2)(5x+1)(3x-1)$ completing the question. \qed
\ic{This is the 30 minute mark}
\newpage

\ex Prove that $\sqrt{7}$ is irrational.
\vskip1em
\pf Assume towards a contradiction that $\sqrt{7} = x \in \mathbb{Q}$. Square both sides gives
\[
7 = x^{2} \quad\quad \implies \quad\quad 0 = x^{2} - 7
\]
Therefore, as a polynomial, $x^{2} - 7$ has a rational root. By the Rational Root Theorem, the only possible rational roots are given by $\pm1, \pm7$. By inspection, none of these are roots:
\[
(\pm1)^{2} - 7 = -6 \neq 0 \quad\quad\quad (\pm7)^{2} -7 = 42 \neq 0 
\]
Hence, $x$ cannot be rational. \qed

\ic{This is the 35 minute mark}
\newpage
\hdc

Prove that $\sqrt{5} + \sqrt{3}$ is irrational.
\vskip3em
\soln Assume towards a contradiction that $\sqrt{5} + \sqrt{3} = x \in \mathbb{Q}$. Squaring gives
\[
5 + 2\sqrt{15} + 3 = x^{2} \quad\quad \implies \quad\quad 2\sqrt{15} = x^{2}-8
\]
Squaring again gives
\[
60 = x^{4} - 16x^{2} + 64 \quad\quad \implies \quad\quad 0 = x^{4} - 16x^{2} + 4
\]
By the Rational Roots Theorem, the only possible roots are 
\[
\pm1, \pm2, \pm4
\]
A quick check shows that none of these work. \qed
\ic{This is the 45 minute mark}
\newpage
\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
\pf Write $p(x) = a_{n}x^{n}+... + a_{1}x + a_{0} \in \mathbb{R}[x]$ with $p(c) = 0$. Then:
\begin{align*}
p(\overline{c}) &= a_{n}(\overline{c})^{n}+... + a_{1}\overline{c} + a_{0}\\
&= \overline{a_{n}(c)^{n}}+... + \overline{a_{1}c} + \overline{a_{0}} && \text{ Since coefficients are real and PCJ.}\\
&= \overline{a_{n}(c)^{n}+... + a_{1}c + a_{0}} && \text{ By PCJ}\\
&= \overline{p(c)}\\
&= 0
\end{align*}

\ic{This is the 50 minute mark.}