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 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:
$\begin{array}{cc} \begin{array}{|rrrrrrrr} a & b & c & d & e & f & g & h \\ \hline \end{array} \end{array}$
$\begin{array}{cc} \begin{array}{rrrr} i &j & k & l \\ \end{array} & \begin{array}{|rrrrrrrr} a & b & c & d & e & f & g & h \\ \hline \end{array} \end{array}$
$\begin{array}{cc} \begin{array}{rrrr} i &j & k & l \\ \\ \end{array} & \begin{array}{|rrrr|rrrr} a & b & c & d & e & f & g & h \\ \hline & & & & & & & \\ \end{array} \end{array}$
$\begin{array}{cc} \begin{array}{rrrr} i &j & k & l \\ \\ \end{array} & \begin{array}{|rrrr|rrrr} a & b & c & d & e & f & g & h \\ \hline m & & & & & & & \\ \end{array} \end{array}$
$\begin{array}{cc} \begin{array}{rrrr} \\ i &j & k & l \\ \\ \end{array} & \begin{array}{|rrrr|rrrr} & mi & mj & mk & ml & & & \\ a & b & c & d & e & f & g & h \\ \hline m & & & & & & & \\ \end{array} \end{array}$
$\begin{array}{cc} \begin{array}{rrrr} \\ i &j & k & l \\ \\ \end{array} & \begin{array}{|rrrr|rrrr} & mi & mj & mk & ml & & & \\ a & b & c & d & e & f & g & h \\ \hline m & n & & & & & & \\ \end{array} \end{array}$
$\begin{array}{cc} \begin{array}{rrrr} \\ \\ i &j & k & l \\ \\ \end{array} & \begin{array}{|rrrr|rrrr} & & ni & nj & nk & & & \\ & mi & mj & mk & ml & nl & & \\ a & b & c & d & e & f & g & h \\ \hline m & n & o & & & & & \\ \end{array} \end{array}$
$\begin{array}{cc} \begin{array}{rrrr} \\ \\ \\ i &j & k & l \\ \\ \end{array} & \begin{array}{|rrrr|rrrr} & & & oi & oj & & & \\ & & ni & nj & nk & ok & & \\ & mi & mj & mk & ml & nl & ol & \\ a & b & c & d & e & f & g & h \\ \hline m & n & o & p & & & & \\ \end{array} \end{array}$
$\begin{array}{cc} \begin{array}{rrrr} \\ \\ \\ \\ i &j & k & l \\ \\ \end{array} & \begin{array}{|rrrr|rrrr} & & & & pi & & & \\ & & & oi & oj & pj & & \\ & & ni & nj & nk & ok & pk & \\ & mi & mj & mk & ml & nl & ol & pl \\ a & b & c & d & e & f & g & h \\ \hline m & n & o & p & q & & & \\ \end{array} \end{array}$
$\begin{array}{cc} \begin{array}{rrrr} \\ \\ \\ \\ i &j & k & l \\ \\ \end{array} & \begin{array}{|rrrr|rrrr} & & & & pi & & & \\ & & & oi & oj & pj & & \\ & & ni & nj & nk & ok & pk & \\ & mi & mj & mk & ml & nl & ol & pl \\ a & b & c & d & e & f & g & h \\ \hline m & n & o & p & q & r & s & t \\ \end{array} \end{array}$
EDIT: I went ahead and edited the wiki.
This is showing a compact (notional shortcut) for doing synthetic division as the non-compactified variant is just a lot of writing.
So, I have worked an example using a seventh-order equation so we can follow along with the Wiki site you reference for expanded synthetic division.
One VERY important note in this approach is that you are jumping between filling in a row value and filling in a column value. Just as soon as you can fill in a value, you do so. I am going to provide you all of the values, but do it in order and see how to fill in the triangle (if you will) of values on your own.
To actually understand why it works, you will have to use the SAME example with the previous method on the site, so you can see that this is just a very compactified variant of the previous method. Please do this or you won't understand the compactified method!
Using compactified synthetic division, find the quotient and remainder of the following
$\frac{x^{7} + 2x^{6} + 3x^{5} + 4x^{4} +5x^{3} + 6x^{2} + 7x +8}{x^{4} -2x^{3} -3x^{2}-4x -5}$
Note, when the items below says calculate, recall that you are bouncing between row and column calculations and doing each one just as soon as having the necessary values permit.
I am going to write these in the order as shown on the web page (you are effectively starting at Row 1 and working up to Row 5) - so,
Row 5: Calculate: $p i = 96$
Row 4: Calculate: $o i = 28$ $o j = 42$ $p j = 144$
Row 3: Calculate: $n i = 8$ $n j = 12$ $n k = 16$ $o k = 56$ $p k = 192$
Row 2: Calculate: $m i = 2$ $m j = 3$ $m k = 4$ $m l = 5$ $n l = 20$ $o l = 70$ $p l = 240$
Row 1: Write out the entire bottom row, that is:
$i = 2$, $j = 3$, $k = 4$, $l = 5$ $\vert$ $a=1$ $b=2$ $c=3$ $d=4$ $e=5$ $f=6$ $g=7$ $h=8$
Row 0: this is the resultant quotient and remainder, and it is being filled out just as soon as we have all the necessary column values by just summing all of the values in the respective column.
$m=1$ $n=4$ $o=14$ $p=48$ $\vert$ $q=164$ $r=226$ $s=269$ $t=248$
Therefore, using the compactified variant of synthetic division yields the following quotient and remainder
$x^{3} + 4 x^{2} + 14 x + 48$ + $\frac{164 x^{3} + 226 x^{2} + 269 x +248}{x^{4} -2x^{3} -3x^{2}-4x -5}$
Lets verify the result using WolframAlpha.
So, now try this approach with the previous example they show on the web site and you'll see how it works. I also recommend you do several more examples of varying sizes to make sure you get it! You can check your work using your favorite CAS or WolframAlpha.
Update
The compactified - synthetic division variant is just re-writing synthetic division is a sort of shorthand. Synthetic division is just writing long polynomial division in shorthand.
Regards