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A polynomial $p\in \Fb[x]$ of degree $2$ or $3$ is reducible if and only
if it has at least a root in $\Fb$.
\\
Or, equivalently, \emph{Or, equivalently,}
\\
A polynomial $p\in \Fb[x]$ of degree $2$ or $3$ is irreducible if and only
if it has no roots in $\Fb$.
...
\begin{Exercise}\label{Factorize}
Factorize the following polynomials in $\Fb[x]$:
\begin{itemize}
\item
$p_1(x)=x^5+x^4+x+1$ $p_1=x^5+x^4+x+1$
\item
$p_2(x)=x^4+x^2+1$ $p_2=x^4+x^2+1$
\end{itemize}
Hint: try all possible irreducible
factors. factors of degree $1$ and $2$.
\end{Exercise}
%
\begin{Exercise}\label{IrredDeg4}
Find all the irreducible polynomials of degree $2$ in $\Fb[x]$.\\
Use this result to prove that
$p(x)=x^4+x+1$ $f=x^4+x+1$ is irreducible in $\Fb[x]$.
\end{Exercise}
...
to define one more operation: \emph{the division of polynomials}.
\\
A consequence of more advanced theory for polynomials with coefficients in a field is the
following. If we consider two polynomials
$f(x)$ $f$ and
$g(x)$ $g$ in $\Fb[x]$,
we can find two other polynomials, say
$q(x), r(x)$, $q, r$, such that
$f(x)=g(x)\cdot q(x) +r(x)$ $f=g\cdot q+r$ and
$\deg(r(x))< \deg(g(x))$. $\deg(r)< \deg(g)$. \\
The polynomial
$q(x)$ $q$ is called \emph{quotient}, whereas
$r(x)$ $r$ is the
\textbf{remainder}. If
$r(x)=0$ $r=0$ then we say that
$g(x)$ $g$ \emph{divides}
$f(x)$, $f$, or that
we have performed an \emph{exact division} between
$f(x)$ $f$ and
$g(x)$. $g$.
%\\
%This is a general property and, in particular, it holds for polynomials in bits.
\begin{Example}\label{QuotRem}
Let us consider $f=x^2+1, \, g=x$ in $\Fb[x]$; we clearly have $q=x$
and $r=1$:
$$x^2+1=x\cdot x+1.$$
If we consider
$f=x^2+1, $f=1+x^2, \, g=x+1$ in $\Fb[x]$,
it holds we note that
$x^2+1=(x+1)\cdot (x+1)$, so $q=g $ and $r=0.$
\end{Example}
Let us now see how to compute the quotient and the reminder; we will use
...
\end{center}
If in $f(x)$ a term of degree smaller than $\deg(f)$ does not appear, we put
it in the table with coefficient
zero. \textbf{zero}.
\\
Now we divide the term of maximal degree of $f$ by that of $g$,
getting a new monomial $m$; in our particular case $m=x$
...
\end{tabular}
\end{center}
Then, we multiply $m\cdot g$ and we
pose insert the result in the table under
$f(x)$, $f$, so that the terms of same degree are lined up.
\begin{center}
\begin{tabular}{ccc|cc}
...
& & & &
\end{tabular}
\end{center}
Sum We sum the
lined up lined-up monomials (remember that the coefficients belong to $\Fb$, so
$1+1 = 0$!!!!); the obtained sum is the \emph{partial reminder} of our division.
\begin{center}
...
\end{tabular}
\end{center}
If the partial
reminder's remainder's degree is greater or equal than the degree of
$g$ $g$, \emph{then}
repeat the procedure;
if not, otherwise, stop
and: and observe that
\begin{itemize}
\item the polynomial under $g$ is the quotient $q$
\item the partial
reminder becomes remainder is the
reminder (final) remainder $r$
\end{itemize}
In our example $q=x$ and $r=1$.
...
\begin{Exercise}\label{QuoRemEx}
Perform the division between the following polynomials in $\Fb[x]$
\begin{itemize}
\item
$f(x)= $f= x^3+x^2+1$,
$g(x)=x^3+1$; $g=x^3+1$;
\item
$f(x)= $f= x^4+x $,
$g(x)=x+1$; $g=x+1$;
\item
$f(x)=x^5+x^3+1 $f=x^5+x^3+1 $,
$g(x)=x^2+x+1$; $g=x^2+x+1$;
\item
$f(x)=x^6+x^5+x^4+x^3+x^2+x+1 $f=x^6+x^5+x^4+x^3+x^2+x+1 $,
$g(x)=x^3+x^2+1$. $g=x^3+x^2+1$.
\end{itemize}
Which of them are exact divisions?
\end{Exercise}
