On Wednesday, August 25, 2004 at 4:50:33 PM UTC+3, Yogi wrote:
> On 25 Aug 2004, Bassam Karzeddin wrote:
> >On Wed, 11 Aug 2004 14:50:37 GMT, Richard Fateman wrote:
> >>What's the point? Something that is nice to the eye has no
> >>special algorithmic advantage. Consider
> >>X^5+(C+A)*X^4+(D+A*C+B)*X^3+(E+A*D+B*C)*X^2+(A*E+B*D)*X+B*E
> >>
> >>which is a quintic that always factors as..
> >>(X^2+A*X+B)*(X^3+C*X^2+D*X+E)
> >>
> >>So if you want to factor
> >>x^5+p*x^4+q*x^3+r*x^2+s*x+t
> >>
> >>just try to solve
> >>p=C+A
> >>q=D+A*C+B
> >> etc.
> >>for A,B,C,D,E.
> >>
> >>RJF
> >>
> >>
> >>Bassam Karzeddin wrote:
> >>> DEAR NEWS GROUB
> >>>
> >>> Salam Every body
> >>>
> >>> The reduced form of the Bring-Jerrard quintic equation is:
> >>>
> >>> x^5+a*x+b=(x^2+u*x-1)*(x^3-u*x^2+[u^2+1]*x-u^3-2*u)=0
> >>>
> >>> where,
> >>> a=-(u^4+3u^2+1)
> >>>
> >>> b=u^3+2u
> >>>
> >>> as you can see,it is very simple to prove,just by multiblication
> >>> you will get this algebraic factorization of the quintic.
> >>>
> >>> so,whatever value you assume for (u) ,you will always obtain fife
> >
> >>>
> >>> radical solutions to the quintic,
> >>>
> >>>
> >>>
> >>> Best Regards
> >>>
> >>> Bassam Karzeddin
> >>> P.O.Box 20
> >>> Engineering & Project Circle
> >>> AL-HUSSEIN BIN TALAL UNIVERSITY
> >>> MA,AN-71111
> >>> JORDAN
> >>> *******************************
> >>> Mobile : +962796666505
> >>>
> >
> >The point is it is good to start where others had finished.
> >consider the following reduced form of the quintic
> >
> > x^5+(3*b*a^2-b^2-a^4)*x+a*b*(2*b-a^2)=0
> >
> >This equation have always fife radical roots,for any assumed values
> >for (a & b),where (a,b) can be real or imaginary numbers.
> >
> >My question is (with out knowing much about Galios groub or Abel’s
> >impossible solution theorem)
> >
> >DOES THE ABOVE REDUCED FORM OF THE QUINTIC EQUATION CONTRADICTS THEM.
> >
> >Thanking YOU
> >
> >Best Regards
> >
> >Bassam Karzeddin
> >p.o.box 20
> >Engineering & projects circle
> >Al Hussein Bin Talal University
> >MA’AN - JORDAN
> No, it does not. Galois and Abel showed that THERE EXIST quintic equations that cannot be solved by radicals. That does not mean that there do not exist SOME that can be solved by radicals. By starting with a special form, you have found some of those that can be solved- there still exist others that cannot.

Where is that so lengthy thread published by myself since 2004?

Had it been stolen even in immoderated site as here? No woundrs !

BKK

On Friday, August 4, 2023 at 6:44:47 AM UTC+3, bassam karzeddin wrote:
> On Wednesday, August 25, 2004 at 4:50:33 PM UTC+3, Yogi wrote:
> > On 25 Aug 2004, Bassam Karzeddin wrote:
> > >On Wed, 11 Aug 2004 14:50:37 GMT, Richard Fateman wrote:
> > >>What's the point? Something that is nice to the eye has no
> > >>special algorithmic advantage. Consider
> > >>X^5+(C+A)*X^4+(D+A*C+B)*X^3+(E+A*D+B*C)*X^2+(A*E+B*D)*X+B*E
> > >>
> > >>which is a quintic that always factors as..
> > >>(X^2+A*X+B)*(X^3+C*X^2+D*X+E)
> > >>
> > >>So if you want to factor
> > >>x^5+p*x^4+q*x^3+r*x^2+s*x+t
> > >>
> > >>just try to solve
> > >>p=C+A
> > >>q=D+A*C+B
> > >> etc.
> > >>for A,B,C,D,E.
> > >>
> > >>RJF
> > >>
> > >>
> > >>Bassam Karzeddin wrote:
> > >>> DEAR NEWS GROUB
> > >>>
> > >>> Salam Every body
> > >>>
> > >>> The reduced form of the Bring-Jerrard quintic equation is:
> > >>>
> > >>> x^5+a*x+b=(x^2+u*x-1)*(x^3-u*x^2+[u^2+1]*x-u^3-2*u)=0
> > >>>
> > >>> where,
> > >>> a=-(u^4+3u^2+1)
> > >>>
> > >>> b=u^3+2u
> > >>>
> > >>> as you can see,it is very simple to prove,just by multiblication
> > >>> you will get this algebraic factorization of the quintic.
> > >>>
> > >>> so,whatever value you assume for (u) ,you will always obtain fife
> > >
> > >>>
> > >>> radical solutions to the quintic,
> > >>>
> > >>>
> > >>>
> > >>> Best Regards
> > >>>
> > >>> Bassam Karzeddin
> > >>> P.O.Box 20
> > >>> Engineering & Project Circle
> > >>> AL-HUSSEIN BIN TALAL UNIVERSITY
> > >>> MA,AN-71111
> > >>> JORDAN
> > >>> *******************************
> > >>> Mobile : +962796666505
> > >>>
> > >
> > >The point is it is good to start where others had finished.
> > >consider the following reduced form of the quintic
> > >
> > > x^5+(3*b*a^2-b^2-a^4)*x+a*b*(2*b-a^2)=0
> > >
> > >This equation have always fife radical roots,for any assumed values
> > >for (a & b),where (a,b) can be real or imaginary numbers.
> > >
> > >My question is (with out knowing much about Galios groub or Abel’s
> > >impossible solution theorem)
> > >
> > >DOES THE ABOVE REDUCED FORM OF THE QUINTIC EQUATION CONTRADICTS THEM.
> > >
> > >Thanking YOU
> > >
> > >Best Regards
> > >
> > >Bassam Karzeddin
> > >p.o.box 20
> > >Engineering & projects circle
> > >Al Hussein Bin Talal University
> > >MA’AN - JORDAN
> > No, it does not. Galois and Abel showed that THERE EXIST quintic equations that cannot be solved by radicals. That does not mean that there do not exist SOME that can be solved by radicals. By starting with a special form, you have found some of those that can be solved- there still exist others that cannot.
> Where is that so lengthy thread published by myself since 2004?
>
> Had it been stolen even in immoderated site as here? No woundrs !
>
> BKK

It seems that the whole comlete & quite lengthy issue was (removed, stolen, fabricated, ..., etc) by annymous acadimics for reasons they only know

However, people especially the academic proffessional mainstream mathematicians seem not wanting to understand the truth in this issue purely for physicological issues they do secretly understand nowadays

However, the truth would easily be grasped by talented students for sure

BKK