Synthetic division via the greedy strategy I was looking at the expanded synthetic division within Wikipedia. I was stumped by how to come up with and perform the 'compactified' version of synthetic division.
Does anyone know how to do it?

I came up with three shorthand methods from noticing certain patterns, but the one I describe below is the simplest:


*

*Write the coefficients of the dividend on a bar


$\begin{array}{cc}
    \begin{array}{|rrrrrrrr} 
        a & b & c & d & e & f & g & h \\
        \hline 
    \end{array}
\end{array}$


*

*Negate the coefficients of the divisor. Write in every coefficient of the divisor but the first (leading coefficient) one on the left.


$\begin{array}{cc}
    \begin{array}{rrrr} i &j & k & l \\ \end{array}
    &
    \begin{array}{|rrrrrrrr} 
        a & b & c & d & e & f & g & h \\
        \hline 
    \end{array}
\end{array}$


*

*From the number of coefficients placed on the left side, count the number of dividend coefficients above the bar and then place a vertical bar on the row below. This vertical bar marks the separation between the quotient and the remainder.


$\begin{array}{cc}
    \begin{array}{rrrr} i &j & k & l \\ \\ \end{array}
    &
    \begin{array}{|rrrr|rrrr} 
        a & b & c & d & e & f & g & h \\
        \hline 
         &  &  &  & &  &  &  \\
    \end{array}
\end{array}$


*

*Drop the first coefficient of the dividend below the bar.


$\begin{array}{cc}
    \begin{array}{rrrr} i &j & k & l \\ \\ \end{array}
    &
    \begin{array}{|rrrr|rrrr} 
        a & b & c & d & e & f & g & h \\
        \hline 
         m &  &  &  & &  &  &  \\
    \end{array}
\end{array}$


*

*Multiply the last dropped/summed number to each negated coefficients on the left (starting with the left most); skip this step if the summed number is zero. Place each product on top of the subsequent columns.


$\begin{array}{cc}
    \begin{array}{rrrr}  \\ i &j & k & l \\ \\ \end{array}
    &
    \begin{array}{|rrrr|rrrr} 
         & mi & mj & mk & ml &  &  &  \\
        a & b & c & d & e & f & g & h \\
        \hline 
         m &  &  &  & &  &  &  \\
    \end{array}
\end{array}$


*

*Perform an column-wise addition on the next column.


$\begin{array}{cc}
    \begin{array}{rrrr}  \\ i &j & k & l \\ \\ \end{array}
    &
    \begin{array}{|rrrr|rrrr} 
         & mi & mj & mk & ml &  &  &  \\
        a & b & c & d & e & f & g & h \\
        \hline 
         m &  n &  &  & &  &  &  \\
    \end{array}
\end{array}$


*

*Repeat the previous two steps. Stop when you performed the previous two steps on the number just before the vertical bar.


$\begin{array}{cc}
    \begin{array}{rrrr}  \\  \\ i &j & k & l \\ \\ \end{array}
    &
    \begin{array}{|rrrr|rrrr} 
         &    & ni & nj & nk &  &  &  \\
         & mi & mj & mk & ml & nl  &  &  \\
        a & b & c & d & e & f & g & h \\
        \hline 
         m &  n & o &  & &  &  &  \\
    \end{array}
\end{array}$
$\begin{array}{cc}
    \begin{array}{rrrr} \\ \\  \\ i &j & k & l \\ \\ \end{array}
    &
    \begin{array}{|rrrr|rrrr} 
         &    &    & oi & oj &  &  &  \\
         &    & ni & nj & nk & ok &  &  \\
         & mi & mj & mk & ml & nl  & ol  &  \\
        a & b & c & d & e & f & g & h \\
        \hline 
         m &  n &  o & p & &  &  &  \\
    \end{array}
\end{array}$
$\begin{array}{cc}
    \begin{array}{rrrr} \\ \\ \\  \\ i &j & k & l \\ \\ \end{array}
    &
    \begin{array}{|rrrr|rrrr} 
         &    &    &    & pi &  &  &  \\
         &    &    & oi & oj & pj &  &  \\
         &    & ni & nj & nk & ok & pk &  \\
         & mi & mj & mk & ml & nl  & ol  & pl  \\
        a & b & c & d & e & f & g & h \\
        \hline 
         m &  n &  o & p & q &  &  &  \\
    \end{array}
\end{array}$


*

*Perform the remaining column-wise additions on the subsequent columns (getting the remainder).


$\begin{array}{cc}
    \begin{array}{rrrr} \\ \\ \\  \\ i &j & k & l \\ \\ \end{array}
    &
    \begin{array}{|rrrr|rrrr} 
         &    &    &    & pi &  &  &  \\
         &    &    & oi & oj & pj &  &  \\
         &    & ni & nj & nk & ok & pk &  \\
         & mi & mj & mk & ml & nl  & ol  & pl  \\
        a & b & c & d & e & f & g & h \\
        \hline 
         m &  n &  o & p & q & r & s &  t \\
    \end{array}
\end{array}$


*

*The results below the horizontal bar would be interpreted with increasing degree from right to left beginning with degree zero for both the remainder and the result.


EDIT: I went ahead and edited the wiki.
