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<record version="10" id="6851">
 <title>factoring all-one polynomials using the grouping method</title>
 <name>FactoringAllOnePolynomialsUsingTheGroupingMethod</name>
 <created>2005-03-06 22:00:58</created>
 <modified>2005-08-09 00:43:15</modified>
 <type>Example</type>
<parent id="6850">grouping method for factoring polynomials</parent>
 <creator id="6075" name="rspuzio"/>
 <author id="6075" name="rspuzio"/>
 <classification>
	<category scheme="msc" code="13P05"/>
 </classification>
 <related>
	<object name="AllOnePolynomial"/>
	<object name="CyclotomicPolynomial"/>
 </related>
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 <content>The method of grouping terms can be used to factor all-one polynomials, i.e. polynomials of the form
\[
\sum_{m=0}^{n-1} x^m
\]
when $n$ is composite. (When $n$ is prime, these polynomials are irreducible, so there is nothing to do in that case.)

Let us consider a few examples:

$n = 4$:
\begin{eqnarray*}
1 + x + x^2 + x^3 = \\
(1 + x) + (x^2 + x^3) = \\
(1 + x) + x^2 (1 + x) = \\
(1 + x) (1 + x^2)
\end{eqnarray*}

$n = 6$:
\begin{eqnarray*}
1 + x + x^2 + x^3 + x^4 + x^5 = \\
(1 + x + x^2) + (x^3 + x^4 + x^5) = \\
(1 + x + x^2) + x^3 (1 + x + x^2) = \\
(1 + x^3) (1 + x + x^2)
\end{eqnarray*}

$n = 8$:
\begin{eqnarray*}
1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 = \\
(1 + x + x^2 + x^3) + (x^4 + (x^5 + x^6 + x^7) = \\
(1 + x + x^2 + x^3) + x^4 (1 + x + x^2 + x^3) =\\
(1 + x^4) (1 + x + x^2 + x^3)
\end{eqnarray*}
Combining this result with the factorization we have for the case $n=4$, we obtain the following:
\begin{eqnarray*}
1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 = \\
(1 + x) (1 + x^2) (1 + x^4)
\end{eqnarray*}

$n = 9$:
\begin{eqnarray*}
1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8 = \\
(1 + x + x^2) + (x^3 + x^4 + x^5) + (x^6 + x^7  + x^8) = \\
(1 + x + x^2) + x^3 (1 + x + x^2) + x^6 (1 + x + x^2) = \\
(1 + x + x^2) (1 + x^3 + x^6)
\end{eqnarray*}

$n = 12$:
\begin{eqnarray*}
1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8 + x^9 + x^{10} + x^{11} = \\
(1 + x + x^2) + (x^3 + x^4 + x^5) + (x^6 + x^7 + x^8) + (x^9 + x^{10} + x^{11}) = \\
(1 + x + x^2) + x^3 (1 + x + x^2) + x^6 (1 + x + x^2) + x^9 (1 + x + x^2) = \\
(1 + x + x^2) (1 + x^3 + x^6 + x^9) = \\
(1 + x + x^2) ((1 + x^3) + (x^6 + x^9)) = \\
(1 + x + x^2) ((1 + x^3) + x^6 (1 + x^3)) = \\
(1 + x + x^2) (1 + x^3) (1 + x^6)
\end{eqnarray*}

It might be worth pointing out that the polynomials produced by this factorization are not all irreducible.  For instance,
\[ 1 + x^3 = (1 + x) (1 - x + x^2). \]
However, to obtain this factorization, one needs to use some techique other than the grouping method.  Likewise. the polynomial $1 + x^6$ is also reducible.</content>
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