Op woensdag 11 augustus 2004 16:35:06 UTC+2 schreef Bassam Karzeddin:   
   > DEAR NEWS GROUBSalam Every bodyThe 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)=0where,   
   a=-(u^4+3u^2+1)b=u^3+2uas 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   
   RegardsBassam KarzeddinP.O.Box 20Engineering & Project Circle AL-HUSSEIN BIN   
   TALAL UNIVERSITYMA,AN-   
   71111JORDAN *******************************Mobile : +962796666505   
      
   Is not right because with a= -(u^4+3u^2+1)  you calculate with the coefficient   
   a allready the u  and this will be a different u you will calculate    
   with b = u^3+2u  after that. So for a randum Quintic in Bring form you will   
   find two values for u and so the quintic will most times not factor.   
   There are special Quintic that do factor. But the most will not so this is no   
   general solution.   
      
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