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[parent] factoring all-one polynomials using the grouping method (Example)

The method of grouping terms can be used to factor all-one polynomials, i.e. polynomials of the form

$\displaystyle \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$:

$\displaystyle 1 + x + x^2 + x^3 =$      
$\displaystyle (1 + x) + (x^2 + x^3) =$      
$\displaystyle (1 + x) + x^2 (1 + x) =$      
$\displaystyle (1 + x) (1 + x^2)$      

$ n = 6$:

$\displaystyle 1 + x + x^2 + x^3 + x^4 + x^5 =$      
$\displaystyle (1 + x + x^2) + (x^3 + x^4 + x^5) =$      
$\displaystyle (1 + x + x^2) + x^3 (1 + x + x^2) =$      
$\displaystyle (1 + x^3) (1 + x + x^2)$      

$ n = 8$:

$\displaystyle 1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 =$      
$\displaystyle (1 + x + x^2 + x^3) + (x^4 + (x^5 + x^6 + x^7) =$      
$\displaystyle (1 + x + x^2 + x^3) + x^4 (1 + x + x^2 + x^3) =$      
$\displaystyle (1 + x^4) (1 + x + x^2 + x^3)$      

Combining this result with the factorization we have for the case $ n=4$, we obtain the following:
$\displaystyle 1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 =$      
$\displaystyle (1 + x) (1 + x^2) (1 + x^4)$      

$ n = 9$:

$\displaystyle 1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8 =$      
$\displaystyle (1 + x + x^2) + (x^3 + x^4 + x^5) + (x^6 + x^7 + x^8) =$      
$\displaystyle (1 + x + x^2) + x^3 (1 + x + x^2) + x^6 (1 + x + x^2) =$      
$\displaystyle (1 + x + x^2) (1 + x^3 + x^6)$      

$ n = 12$:

$\displaystyle 1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8 + x^9 + x^{10} + x^{11} =$      
$\displaystyle (1 + x + x^2) + (x^3 + x^4 + x^5) + (x^6 + x^7 + x^8) + (x^9 + x^{10} + x^{11}) =$      
$\displaystyle (1 + x + x^2) + x^3 (1 + x + x^2) + x^6 (1 + x + x^2) + x^9 (1 + x + x^2) =$      
$\displaystyle (1 + x + x^2) (1 + x^3 + x^6 + x^9) =$      
$\displaystyle (1 + x + x^2) ((1 + x^3) + (x^6 + x^9)) =$      
$\displaystyle (1 + x + x^2) ((1 + x^3) + x^6 (1 + x^3)) =$      
$\displaystyle (1 + x + x^2) (1 + x^3) (1 + x^6)$      

It might be worth pointing out that the polynomials produced by this factorization are not all irreducible. For instance,

$\displaystyle 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.



"factoring all-one polynomials using the grouping method" is owned by rspuzio.
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See Also: all one polynomial, cyclotomic polynomial


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Cross-references: reducible, grouping method, irreducible, prime, composite, polynomials, all-one polynomials, factor, terms

This is version 10 of factoring all-one polynomials using the grouping method, born on 2005-03-06, modified 2005-08-09.
Object id is 6851, canonical name is FactoringAllOnePolynomialsUsingTheGroupingMethod.
Accessed 2857 times total.

Classification:
AMS MSC: 13P05 (Commutative rings and algebras :: Computational aspects of commutative algebra :: Polynomials, factorization)
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