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