## maxima-lang-fr

 [Maxima-lang-fr] factorisation avec radicaux From: Valere Bonnet - 2007-12-09 19:14:46 Attachments: Message as HTML Bonjour, j'aimerais savoir s'il est possible d'utiliser maxima pour factoriser des expressions avec des radicaux; par exemple: $x^2-2\sqrt3+3$ ; deviendrait: $(x-\sqrt3)^2$. Merci d'avance. 
 Re: [Maxima-lang-fr] factorisation avec radicaux From: Stavros Macrakis - 2007-12-12 22:10:32 Pas directement. Mais, en supposant que Maxima arrive =E0 trouver les racines du polynomial, il est facile =E0 faire. Attention! La m=E9thode d=E9montr=E9e ne prend pas compte des eventuels polynomiaux o=F9 Maxima ne trouve pas les racines explicites. poly: x^4-x-1$sols: solve(poly,x); [x =3D -sqrt((3*(3^-(3/2)*sqrt(283)/2+1/2)^(2/3)-4)^(3/2)+6*sqrt(3)*sqrt(3= ^-(3/2)*sqrt(283)/2+1/2))*%i /(2*sqrt(3)*(3^-(3/2)*sqrt(283)/2+1/2)^(1/6)*(3*(3^-(3/2)*sqrt(283)/2+1= /2)^(2/3)-4)^(1/4)) -sqrt(3*(3^-(3/2)*sqrt(283)/2+1/2)^(2/3)-4)/(2*sqrt(3)*(3^-(3/2)*sqrt(2= 83)/2+1/2)^(1/6)), x =3D sqrt((3*(3^-(3/2)*sqrt(283)/2+1/2)^(2/3)-4)^(3/2)+6*sqrt(3)*sqrt(3^= -(3/2)*sqrt(283)/2+1/2))*%i /(2*sqrt(3)*(3^-(3/2)*sqrt(283)/2+1/2)^(1/6)*(3*(3^-(3/2)*sqrt(283)/2+1= /2)^(2/3)-4)^(1/4)) -sqrt(3*(3^-(3/2)*sqrt(283)/2+1/2)^(2/3)-4)/(2*sqrt(3)*(3^-(3/2)*sqrt(2= 83)/2+1/2)^(1/6)), x =3D sqrt(3*(3^-(3/2)*sqrt(283)/2+1/2)^(2/3)-4)/(2*sqrt(3)*(3^-(3/2)*sqr= t(283)/2+1/2)^(1/6)) -sqrt(6*sqrt(3)*sqrt(3^-(3/2)*sqrt(283)/2+1/2)-(3*(3^-(3/2)*sqrt(283)/2= +1/2)^(2/3)-4)^(3/2)) /(2*sqrt(3)*(3^-(3/2)*sqrt(283)/2+1/2)^(1/6)*(3*(3^-(3/2)*sqrt(283)/2+= 1/2)^(2/3)-4)^(1/4)), x =3D sqrt(6*sqrt(3)*sqrt(3^-(3/2)*sqrt(283)/2+1/2)-(3*(3^-(3/2)*sqrt(283= )/2+1/2)^(2/3)-4)^(3/2)) /(2*sqrt(3)*(3^-(3/2)*sqrt(283)/2+1/2)^(1/6)*(3*(3^-(3/2)*sqrt(283)/2+1= /2)^(2/3)-4)^(1/4)) +sqrt(3*(3^-(3/2)*sqrt(283)/2+1/2)^(2/3)-4)/(2*sqrt(3)*(3^-(3/2)*sqrt(2= 83)/2+1/2)^(1/6))] fact: xreduce("*",makelist(x-rhs(q),q,sols)); (x-sqrt(6*sqrt(3)*sqrt(3^-(3/2)*sqrt(283)/2+1/2)-(3*(3^-(3/2)*sqrt(283)/2= +1/2)^(2/3)-4)^(3/2)) /(2*sqrt(3)*(3^-(3/2)*sqrt(283)/2+1/2)^(1/6)*(3*(3^-(3/2)*sqrt(283)/2+= 1/2)^(2/3)-4)^(1/4)) -sqrt(3*(3^-(3/2)*sqrt(283)/2+1/2)^(2/3)-4)/(2*sqrt(3)*(3^-(3/2)*sqrt(2= 83)/2+1/2)^(1/6))) *(x+sqrt(6*sqrt(3)*sqrt(3^-(3/2)*sqrt(283)/2+1/2)-(3*(3^-(3/2)*sqrt(283)= /2+1/2)^(2/3)-4)^(3/2)) /(2*sqrt(3)*(3^-(3/2)*sqrt(283)/2+1/2)^(1/6)*(3*(3^-(3/2)*sqrt(283)/= 2+1/2)^(2/3)-4)^(1/4)) -sqrt(3*(3^-(3/2)*sqrt(283)/2+1/2)^(2/3)-4)/(2*sqrt(3)*(3^-(3/2)*sqrt= (283)/2+1/2)^(1/6))) *(x-sqrt((3*(3^-(3/2)*sqrt(283)/2+1/2)^(2/3)-4)^(3/2)+6*sqrt(3)*sqrt(3^-= (3/2)*sqrt(283)/2+1/2))*%i /(2*sqrt(3)*(3^-(3/2)*sqrt(283)/2+1/2)^(1/6)*(3*(3^-(3/2)*sqrt(283)/= 2+1/2)^(2/3)-4)^(1/4)) +sqrt(3*(3^-(3/2)*sqrt(283)/2+1/2)^(2/3)-4)/(2*sqrt(3)*(3^-(3/2)*sqrt= (283)/2+1/2)^(1/6))) *(x+sqrt((3*(3^-(3/2)*sqrt(283)/2+1/2)^(2/3)-4)^(3/2)+6*sqrt(3)*sqrt(3^-= (3/2)*sqrt(283)/2+1/2))*%i /(2*sqrt(3)*(3^-(3/2)*sqrt(283)/2+1/2)^(1/6)*(3*(3^-(3/2)*sqrt(283)/= 2+1/2)^(2/3)-4)^(1/4)) +sqrt(3*(3^-(3/2)*sqrt(283)/2+1/2)^(2/3)-4)/(2*sqrt(3)*(3^-(3/2)*sqrt= (283)/2+1/2)^(1/6))) expand(%); x^4+(3*(sqrt(283)/(6*sqrt(3))+1/2)^(2/3)-4)^(3/4)*sqrt((3*(sqrt(283)/(6*sqr= t(3))+1/2)^(2/3)-4)^(3/2) +6*sqrt(3)*sqrt(sqrt(283)/(6*sqrt(3))+1/2))*%i*x /(12*sqrt(3)*sqrt(sqrt(283)/(6*sqrt(3))+1/2)) -(sqrt(283)/(6*sqrt(3))+1/2)^(1/6)*sqrt((3*(sqrt(283)/(6*sqrt(3))+= 1/2)^(2/3)-4)^(3/2) +6*sqrt(3)*sqrt(sqrt(283)/(6*sqrt(3))+1/2))*%i*x etc. etc. radcan(%),algebraic:true; x^4-x-1 On Dec 9, 2007 2:14 PM, Valere Bonnet wrote: > j'aimerais savoir s'il est possible d'utiliser maxima pour factoriser des > expressions avec > des radicaux; par exemple:$x^2-2\sqrt3+3$; deviendrait:$(x-\sqrt3)^2\$.