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Question Number 68422 by ajfour last updated on 10/Sep/19

Answered by ajfour last updated on 10/Sep/19

C=10m^3 p^2 +b(m^3 +6m^2 p+3mp^2 )   +c(3m^2 +6mp+p^2 )+d(6m+4p)   +10e  D=10m^2 p^3 +b(3m^2 p+6mp^2 +p^3 )   +c(m^2 +6mp+3p^2 )+d(4m+6p)   +10e  C−D=0 gives_(−)   10m^2 p^2 (m−p)  +b[(m^3 −p^3 )+6mp(m−p)−3mp(m−p)]  +c[3(m^2 −p^2 )−(m^2 −p^2 )]  +d[6(m−p)−4(m−p)]+0 = 0  If m would be p then z=m=p  otherwise dividing by m−p gives    10m^2 p^2 +b(m^2 +4mp+p^2 )  +c[2(m+p)]+2d=0    .....(I)  (if m≠−p   even)  eq.(i)×(m+p) gives  10m^2 p^2 (m+p)  +b[m^3 +4m^2 p+mp^2 +m^2 p+4mp^2      +p^3 ]+c[2m^2 +4mp+2p^2 ]     +d(2m+2p)=0       ......(i)  Now C+D=0 gives_(−)   10m^2 p^2 (m+p)+  b[m^3 +p^3 +6mp(m+p)+3mp(m+p)]  +c(4m^2 +4p^2 +12mp)  +d(10m+10p)+20e =0   ....(II)    (II)−(i) gives_(−)     b[4mp(m+p)]+   c[2m^2 +8mp+2p^2 ]+   d[8(m+p)]+20e=0  ⇒   2bmp(m+p)+c[(m+p)^2 +2mp]  +4d(m+p)+10e=0    ....(III)    Now lets transform eq. (I) &(III)  let  m+p=s , and mp=g  Eq. (I)_(−)    10g^2 +b(s^2 +2g)+2cs+2d=0    Eq.(III)_(−)     2bgs+c(s^2 +2g)+4ds+10e=0  ⇒  g=−((cs^2 +4ds+10e)/(2(bs+c)))  substituting for g in eq.(I)_(−)   10(cs^2 +4ds+10e)^2 −  4b(cs^2 +4ds+10e)(bs+c)  +4(bs^2 +2cs+2d)(bs+c)^2 =0  ⇒  5(cs^2 +4ds+10e)^2      −2b(bs+c)(cs^2 +4ds+10e)     +2(bs+c)^2 (bs^2 +2cs+2d)=0  This is a degree 4 eq. in s.  So it seems i shall obtain s  and g=−((cs^2 +4ds+10e)/(2(bs+c))) .  m,p = (s/2)±(√((s^2 /4)−g))  Now we then have     At^5 +Bt^4 +Et+F=0  or  t^5 +((B/A))t^4 +((E/A))t+(F/A)=0  but if m=p  clearly z=m  hence all coefficients get zero  and we need  to change   to z=((mt+p)/(t−1)) ; this case i shall take  up in another post..

C=10m3p2+b(m3+6m2p+3mp2)+c(3m2+6mp+p2)+d(6m+4p)+10eD=10m2p3+b(3m2p+6mp2+p3)+c(m2+6mp+3p2)+d(4m+6p)+10eCD=0gives10m2p2(mp)+b[(m3p3)+6mp(mp)3mp(mp)]+c[3(m2p2)(m2p2)]+d[6(mp)4(mp)]+0=0Ifmwouldbepthenz=m=potherwisedividingbympgives10m2p2+b(m2+4mp+p2)+c[2(m+p)]+2d=0.....(I)(ifmpeven)eq.(i)×(m+p)gives10m2p2(m+p)+b[m3+4m2p+mp2+m2p+4mp2+p3]+c[2m2+4mp+2p2]+d(2m+2p)=0......(i)NowC+D=0gives10m2p2(m+p)+b[m3+p3+6mp(m+p)+3mp(m+p)]+c(4m2+4p2+12mp)+d(10m+10p)+20e=0....(II)(II)(i)givesb[4mp(m+p)]+c[2m2+8mp+2p2]+d[8(m+p)]+20e=02bmp(m+p)+c[(m+p)2+2mp]+4d(m+p)+10e=0....(III)Nowletstransformeq.(I)&(III)letm+p=s,andmp=gEq.(I)10g2+b(s2+2g)+2cs+2d=0Eq.(III)2bgs+c(s2+2g)+4ds+10e=0g=cs2+4ds+10e2(bs+c)substitutingforgineq.(I)10(cs2+4ds+10e)24b(cs2+4ds+10e)(bs+c)+4(bs2+2cs+2d)(bs+c)2=05(cs2+4ds+10e)22b(bs+c)(cs2+4ds+10e)+2(bs+c)2(bs2+2cs+2d)=0Thisisadegree4eq.ins.Soitseemsishallobtainsandg=cs2+4ds+10e2(bs+c).m,p=s2±s24gNowwethenhaveAt5+Bt4+Et+F=0ort5+(BA)t4+(EA)t+FA=0butifm=pclearlyz=mhenceallcoefficientsgetzeroandweneedtochangetoz=mt+pt1;thiscaseishalltakeupinanotherpost..

Commented by TawaTawa last updated on 10/Sep/19

wow, God bless you sir and increase your knowledge. I will learn  this with an example.

wow,Godblessyousirandincreaseyourknowledge.Iwilllearnthiswithanexample.

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