Re: a fault in the Factor[] function for polynomials?

• To: mathgroup at smc.vnet.net
• Subject: [mg70479] Re: a fault in the Factor[] function for polynomials?
• From: Roger Bagula <rlbagula at sbcglobal.net>
• Date: Tue, 17 Oct 2006 02:59:15 -0400 (EDT)
• References: <egse02\$cpp\$1@smc.vnet.net> <egvbhd\$rue\$1@smc.vnet.net>

```Peter Pein wrote:

>
>>Yes and there are many more "errors" in Factor:
>>
>>try
>>Factor[%, Extension -> (Sqrt[3]{-1, 1})]
>>or
>>Factor[1 - 14*x^4 + x^8, Extension -> Sqrt[2 + Sqrt[3]*{-1, 1}]]
>>but the "worst" is definitely
>>Factor[1 - 14*x^4 + x^8,
>>   Extension -> Evaluate[x /. Solve[1 - 14*x^4 + x^8 == 0]]]
>>
>>did you assume uniqueness of factorization into non-linear terms?
>>
>>Peter
>>
>>
>>
I was basing my argument on the actual root factors as found by Mathematica:
Solve[1 - 14x^4 + x^8=0,x]

\!\({{x \[Rule] \(-\@\(2 - \@3\)\)}, {x \[Rule] \(-\[ImaginaryI]\)\
\@\(2 - \
\@3\)}, {x \[Rule] \[ImaginaryI]\ \@\(2 - \@3\)}, {x \[Rule] \@\(2 -
\@3\)}, \
{x \[Rule] \(-\@\(2 + \@3\)\)}, {x \[Rule] \(-\[ImaginaryI]\)\ \@\(2 + \
\@3\)}, {x \[Rule] \[ImaginaryI]\ \@\(2 + \@3\)}, {x \[Rule] \@\(2 +
\@3\)}}\)

ExpandAll[Product[(z-x/. Solve[1 - 14x^4 + x^8=0,x][[n]]),{n,1,8}]]

1 - 14 z^4 + z^8

```

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