Editing Synthetic division (section)

=== For non-monic divisors ===

With a little prodding, the expanded technique may be generalised even further to work for any polynomial, not just [[Monic polynomial|monics]]. The usual way of doing this would be to divide the divisor <math>g(x)</math> with its leading coefficient (call it ''a''):
:<math>h(x) = \frac{g(x)}{a}</math>

then using synthetic division with <math>h(x)</math> as the divisor, and then dividing the quotient by ''a'' to get the quotient of the original division (the remainder stays the same). But this often produces unsightly fractions which get removed later and is thus more prone to error. It is possible to do it without first reducing the coefficients of <math>g(x)</math>.

As can be observed by first performing long division with such a non-monic divisor, the coefficients of <math>f(x)</math> are divided by the leading coefficient of <math>g(x)</math> after "dropping", and before multiplying.

Let's illustrate by performing the following division:

:<math>\frac{6x^3+5x^2-7}{3x^2-2x-1}</math>

A slightly modified table is used:

:<math>\begin{array}{cc}
    \begin{array}{rrr} \\ &1& \\ 2&& \\ \\&&/3 \\ \end{array}
    \begin{array}{|rrrr} 
        6 & 5 & 0 & -7 \\
          &     &  &     \\
          &    &   &     \\
        \hline
          &     &   &     \\ 
          &     &   &     \\   
    \end{array}
\end{array}</math>

Note the extra row at the bottom. This is used to write values found by dividing the "dropped" values by the leading coefficient of <math>g(x)</math> (in this case, indicated by the ''/3''; note that, unlike the rest of the coefficients of <math>g(x)</math>, the sign of this number is not changed).

Next, the first coefficient of <math>f(x)</math> is dropped as usual:

:<math>\begin{array}{cc}
    \begin{array}{rrr} \\ &1& \\ 2&& \\ \\&&/3 \\ \end{array}
    \begin{array}{|rrrr} 
        6 & 5 & 0 & -7 \\
          &     &  &     \\
          &    &   &     \\
        \hline
        6 &     &   &     \\ 
          &     &   &     \\   
    \end{array}
\end{array}</math>

and then the dropped value is divided by 3 and placed in the row below:

:<math>\begin{array}{cc}
    \begin{array}{rrr} \\ &1& \\ 2&& \\ \\&&/3 \\ \end{array}
    \begin{array}{|rrrr} 
        6 & 5 & 0 & -7 \\
          &     &  &     \\
          &    &   &     \\
        \hline
        6 &     &   &     \\ 
        2 &     &   &     \\   
    \end{array}
\end{array}</math>

Next, the '''new''' (divided) value is used to fill the top rows with multiples of 2 and 1, as in the expanded technique:

:<math>\begin{array}{cc}
    \begin{array}{rrr} \\ &1& \\ 2&& \\ \\&&/3 \\ \end{array}
    \begin{array}{|rrrr} 
        6 & 5 & 0 & -7 \\
          &   & 2 &     \\
          & 4 &   &     \\
        \hline
        6 &     &   &     \\ 
        2 &     &   &     \\   
    \end{array}
\end{array}</math>

The 5 is dropped next, with the obligatory adding of the 4 below it, and the answer is divided again:

:<math>\begin{array}{cc}
    \begin{array}{rrr} \\ &1& \\ 2&& \\ \\&&/3 \\ \end{array}
    \begin{array}{|rrrr} 
        6 & 5 & 0 & -7 \\
          &   & 2 &     \\
          & 4 &   &     \\
        \hline
        6 & 9   &   &     \\ 
        2 & 3   &   &     \\   
    \end{array}
\end{array}</math>

Then the 3 is used to fill the top rows:

:<math>\begin{array}{cc}
    \begin{array}{rrr} \\ &1& \\ 2&& \\ \\&&/3 \\ \end{array}
    \begin{array}{|rrrr} 
        6 & 5 & 0 & -7 \\
          &   & 2 &  3  \\
          & 4 & 6 &     \\
        \hline
        6 & 9 &   &     \\ 
        2 & 3 &   &     \\   
    \end{array}
\end{array}</math>

At this point, if, after getting the third sum, we were to try and use it to fill the top rows, we would "fall off" the right side, thus the third sum is the first coefficient of the remainder, as in regular synthetic division. But the values of the remainder are '''not''' divided by the leading coefficient of the divisor:

:<math>\begin{array}{cc}
    \begin{array}{rrr} \\ &1& \\ 2&& \\ \\&&/3 \\ \end{array}
    \begin{array}{|rrrr} 
        6 & 5 & 0 & -7 \\
          &   & 2 &  3  \\
          & 4 & 6 &     \\
        \hline
        6 & 9 & 8 & -4  \\ 
        2 & 3 &   &     \\   
    \end{array}
\end{array}</math>

Now we can read off the coefficients of the answer. As in expanded synthetic division, the last two values (2 is the degree of the divisor) are the coefficients of the remainder, and the remaining values are the coefficients of the quotient:

:<math>	\begin{array}{rr|rr} 
    2x &  +3 & 8x & -4 
\end{array}</math>

and the result is

:<math>\frac{6x^3+5x^2-7}{3x^2-2x-1} = 2x + 3 + \frac{8x - 4}{3x^2-2x-1}</math>