http://blogfiles.wfmu.org/LG/CUT/Church_Universal_and_Triumphant_-_06_-_Decree_10_05.mp37 (175x^2-15x+2)=(5a_1(x)+7)(5a_2(x)+7)wherethea'sarerootsofa^2-(7x-1)a+(49x^2-14x)=0.(Theabovewasderived.Butthatderivationisnotnecessarytounderstandtheimplications.)Thea'scanbefoundusingthequadraticformula:a_1(x)=((7x-1)+/-sqrt((7x-1)^2-4(49x^2-14x)))/2a_2(x)=((7x-1)-/+sqrt((7x-1)^2-4(49x^2-14x)))/2orviceversa.Lettingb_1(x)=a_1(x)+1,assuminga_2(0)=0,Ihavea_1(x)=b_1(x)-1,andsubstitutinggives:7(175x^2-15x+2)=(5b_1(x)+2)(5a_2(x)+7)useaspecialconstruction:7(175x^2-15x+2)=(5a_1(x)+7)(5a_2(x)+7)wherethea'sarerootsofa^2-(7x-1)a+(49x^2-14x)0.Thea'scanbefoundusingthequadraticformula,butaresouglyIusuallydon'tbothertogivethem,butmaybeit'llhelpforthosewhowonderwhattheylooklike:a_1(x)=((7x-1)+/-sqrt((7x-1)^2-4(49x^2-14x)))/2a_2(x)=((7x-1)-/+sqrt((7x-1)^2-4(49x^2-14x)))/2orviceversa.Atx=0,oneofthea'sis0,whiletheotheris-1,asa^2+a=0,hasthosesolutions(youcancheckwiththeexplicitequationsaswellifyouwish).Soitpaystogetfunctionswherebothare0atx=0,asthattriviallyfinishestheexercise.Lettingb_1(x)=a_1(x)+1,assuminga_2(0)=0,Ihavea_1(x)=b_1(x)-1,andsubstitutinggives:7(175x^2-15x+2)=(5b_1(x)+2)(5a_2(x)+7)Sonowinthefieldofcomplexnumbersyouhavetheinterestingresultthatthe7wasmultipliedinonlyoneway,whichisthewayIpickedasfundamentallytheexampleisnodifferentfrom7(x^2+3x+2)=(x+2)(7x+7).IfI'dpickedanotherway,youmighthave:7(175x^2 -15x+2)=(5b_1(x)+14)(5a_2(x)+1)IfI'dwantedtosplitthe7withsquareroots,youcouldhave:7(175x^2-15x+2)=(5b_1(x)+2sqrt(7))(5a_2(x)+sqrt(7))ooverthecomplexplaneyouhavetheanswerastowhatIwantedandtheonlywaythe7couldhavemultiplied.Thatisamathematicalabsolute.It'smathematicallyimpossibleforthe7tohavemultipliedanydifferentwaywith:7(175x^2 -15x+2)=(5b_1(x)+2)(5a_2(x)+7)whereb_1(0)=a_2(0)=0.>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>[[JSHsPUBLISHEDPAPER,]]http://dl.free.fr/gMojEGqSw/jsh.pdfElectronicJournal:SouthwestJournalofPureandAppliedMathematicsInternet:http://rattler.cameron.edu/swjpam.htmlISSN1083-0464Issue2,December2003,pp.6 -8.Submitted:July25,2003.Published:December312003.ADVANCEDPOLYNOMIALFACTORIZATIONJamesHarrisAbstract.Algebraicmethodfordeterminingdistributionoffactorswithinapolynomialfactorization,whichbreaksthroughwhatwasseenasabarrierfromoverinterpretationsofGaloisTheory.A.M.S.(MOS)SubjectClassificationCodes.11R04,11R09KeyWordsandPhrases.Polynomialfactorization,Galoistheory,Factorizationlemma,RingofalgebraicintegersAdvancedPolynomialFactorizationApproached.DeterminingthedistributionoffactorswithinirrationalalgebraicintegershaslongbeenconsideredimpossibleasitisnotpossibletodousingGaloisTheory.Howeverasimpletechniquethroughtheintroductionofmorevariablesmakesitpossible.Tohighlightthestandardbeliefconsiderthealgebraicintegerrootsofx2+x?5.Whileyouknowthatthealgebraicintegerfactorsarethemselvesfactorsof5,caneithernothavenonunitfactorsof5?Howdoyouknow?InlookingtoconsiderdistributionofalgebraicintegerfactorswithinafactorizationI'llbeusingamorecomplicatedexamplethanx2+x?5.Thispaperwillshow,usingbasicalgebraicmethods,thatgiventhefactorization,intheringofalgebraicintegers,65x3?12x+1=(a1x+1)(a2x+1)(a3x+1)oneofthea'siscoprimeto5.FirstI'llneedasimplelemmatogeneralizebeyondfactorsofapolynomialthatarethemselvespolynomials.E-mailAddress:jstevh@msn.comc2003CameronUniversityTypesetbyAMS-TEXADVANCEDPOLYNOMIALFACTORIZATION7FactorizationLemma:.GivenafactorgofapolynomialP(x),furtherdefinedasafactorforallx,whichmeansthatthevalueofgforavalue'a'ofxisafactorofP(a),withintheringofalgebraicintegers,thereexistsrandcsuchthatg=r+cwherer=0,orvariesasxvaries,andcisafactoroftheconstanttermP(0)andisitselfconstant.Letx=0,thengmustbeafactorofP(0),soatthatpointc=g.Ifwhenxdoesnotequal0,g=c,r=0.Ifwhenxdoesnotequal0,g6=ctheremustexistrwhichvarieswithx.Thatis,r=g-c.Asanexampleconsider?x+1whichisanonpolynomialfactorofx+1,andwhilethereareaninfinityofirrationalsolutionsconsidertherationalsolutionatx=35.ThenIhave?35+1=6=5+1;thereforewhenx=35,g=6,r=5,andc=1.Butfordifferentvaluesofx,gandrwillvary,whilecwillnot.PrimaryArgument.Given65x3?12x+1=(a1x+1)(a2x+1)(a3x+1)intheringofalgebraicintegers.LetP(m)=f2((m3f4?3m2f2+3m)x3?3(?1+mf2)xu2+u3f)Herefisanonunit,nonzeroalgebraicintegercoprimeto3andx,anduanonunit,nonzeroalgebraicintegercoprimetof.NoteP(m)hasafactorthatisf2.Thatexpressioncomesfromexpanding(v3+1)x3?3vxy2+y3,usingthesubstitutionsv=?1+mf2,andy=uf,whereadditionalvariablesprovideanadditionaldegreeoffreedom.NowconsiderthefactorizationP(m)=(a1x+uf)(a2x+uf)(a3x+uf)wheremultiplyingoutshowsthata1a2a3=m3f6?3m2f4+3mf2=f2(m3f4?3m2f2+3m)soa1a2a3=mf2(m2f4?3mf2+3).Therefore,atleastoneofthea'scannotbecoprimetom,andatleastoneoftheasmustequal0whenm=0.(Note:Thea'sarerootsofamonicpolynomialwithalgebraicintegercoefficientssotheyarealgebraicintegers.)NoticethattheconstanttermP(0)isgivenbyP(0)=f2(3xu2+u3f)andalsothatP(0)/f2=3xu2+u3f,whichiscoprimetof.8SOUTHWESTJOURNALOFPUREANDAPPLIEDMATHEMATICSThenIhavethefactorofP(m),g1,whereg1=a1x+uf,wherehereIalsohavethata1isnotcoprimetom.¿Frommyfactorizationlemma,Ihavethat,whenm=0g1=c=ufmeaningfisafactoroftheconstantterm.Therefore,exactlytwoofthea'sequal0,whenm=0,togetthefactorf2intheconstanttermP(0),whileonemustnotequal0,orf3wouldbethefactor.NowasnotedbeforeingeneralP(m)hasafactorthatisf2,andseparatingthatfactoroff,givesaconstanttermcoprimetof;therefore,giveng1=a1x+ufwherewithm=0,g1givesafactoroffitmusthavethatsamefactoringeneral,provingthattwoofthea'shaveafactorthatisf.Therefore,onefactoriscoprimetof.Nowlettingm=1,f=?5,whereIcanletu=1asitsvaluedoesn'tchangethea's,Ihave(m3f6?3m2f4+3m)x3?3(?1+mf2)xu2+u3=65x3?12x+1whichmaybemoreeasilyseenfromusingv=?1+mf2=4,y=1with(v3+1)x3?3vxy2+y3.Therefore,withthefactorization65x3?12x+1=(a1x+1)(a2x+1)(a3x+1)oneofthea'siscoprimeto5,whichshowswheresomeofthealgebraicintegerfactorsdistributedespitethefactorsbeingirrational.Butthatsaysthatwitha^2-(7x-1)a+(49x^2-14x)=0onlyoneoftherootsisaproductof7,andIsayitthatweirdwayasI'monthecomplexplaneand"factor"isn'treallymeaningful.Afterall,ifthe7isbeingmultipliedtimesonefactor,sothatonlyonerootisaproductof7andsomethingelse,thenhowinthehellcoulditsfactorsendupintheotherroot?ButmathematiciansteachthatitDOESifwithintegerxa^2-(7x-1)a+(49x^2-14x)=0http://dl.free.fr/gMojEGqSw/jsh.pdf>>>>>>>>>>>>>>>>SOJSHisstillf*ckingaroundwiththesamewrongBSforYEARSandYEARS. 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