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'factoring all-one polynomials using the grouping method'
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| Title of object: |
factoring all-one polynomials using the grouping method |
| Canonical Name: |
FactoringAllOnePolynomialsUsingTheGroupingMethod |
| Type: |
Example |
| Created on: |
2005-03-06 22:00:58 |
| Modified on: |
2005-03-07 16:15:39 |
| Classification: |
msc:13P05 |
Revision comment (for changes between this and next version):
| Changes for correction #6278 ('polynimial'). |
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Content:
The method of grouping terms can be used to factor all-one polynimials, 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 ireeducible. 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. |
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