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Find-the-sum-of-the-fifth-powers-of-the-roots-of-x-3-2x-2-x-1-0-by-applying-synthetic-division-




Question Number 200465 by faysal last updated on 19/Nov/23
Find the sum of the fifth powers of  the roots of x^3 −2x^2 +x−1=0 by  applying synthetic division
Findthesumofthefifthpowersoftherootsofx32x2+x1=0byapplyingsyntheticdivision
Commented by mr W last updated on 19/Nov/23
you posted your first question on  15/09/2020. since then you have never  given even a single one feedback.  even when people are talking to a  stone wall, it gives an echo back...
youpostedyourfirstquestionon15/09/2020.sincethenyouhavenevergivenevenasingleonefeedback.evenwhenpeoplearetalkingtoastonewall,itgivesanechoback
Answered by mr W last updated on 19/Nov/23
say the roots are a,b,c.  a+b+c=2  ab+bc+ca=1  abc=1  (a+b+c)^2 =a^2 +b^2 +c^2 +2(ab+bc+ca)  ⇒a^2 +b^2 +c^2 =2^2 −2×1=2    (ab+bc+ca)^2 =a^2 b^2 +b^2 c^2 +c^2 a^2 +2abc(a+b+c)  ⇒a^2 b^2 +b^2 c^2 +c^2 a^2 =1^2 −2×1×2=−3    (a+b+c)^3 =a^3 +b^3 +c^3 −3abc+3(a+b+c)(ab+bc+ca)  ⇒a^3 +b^3 +c^3 =2^3 +3×1−3×2×1=5    (a^2 +b^2 +c^2 )(a^3 +b^3 +c^3 )=a^5 +b^5 +c^5 +(a+b+c)(a^2 b^2 +b^2 c^2 +c^2 a^2 )−abc(ab+bc+ca)  ⇒a^5 +b^5 +c^5 =2×5−2×(−3)+1×1=17 ✓
saytherootsarea,b,c.a+b+c=2ab+bc+ca=1abc=1(a+b+c)2=a2+b2+c2+2(ab+bc+ca)a2+b2+c2=222×1=2(ab+bc+ca)2=a2b2+b2c2+c2a2+2abc(a+b+c)a2b2+b2c2+c2a2=122×1×2=3(a+b+c)3=a3+b3+c33abc+3(a+b+c)(ab+bc+ca)a3+b3+c3=23+3×13×2×1=5(a2+b2+c2)(a3+b3+c3)=a5+b5+c5+(a+b+c)(a2b2+b2c2+c2a2)abc(ab+bc+ca)a5+b5+c5=2×52×(3)+1×1=17
Answered by ajfour last updated on 19/Nov/23
let   x^5 =t  ⇒ x^2 (2x^2 −x+1)=t  2x(2x^2 −x+1)−(2x^2 −x+1)+x^2 =t  4(2x^2 −x+1)−3x^2 +3x−1=t  5x^2 −x+3=t     5(2x^2 −x+1)−x^2 =(t−3)x  9(x+t−3)+25=5(t+2)x  (5t+1)x=9t−2  x=((9t−2)/(5t+1))  (9t−2)^3 −2(5t+1)(9t−2)^2      +(5t+1)^2 (9t−2)−(5t+1)^3 =0  Σ_(i=1) ^3 x_i ^5 =Σ_(i=1) ^3 t_i =−((coeff of t^2 )/(coeff of t^3 ))  =−(((−6×81+360−216−50+90−75))/((729−810+225−125)))  =−(((−323))/(19)) = 17
letx5=tx2(2x2x+1)=t2x(2x2x+1)(2x2x+1)+x2=t4(2x2x+1)3x2+3x1=t5x2x+3=t5(2x2x+1)x2=(t3)x9(x+t3)+25=5(t+2)x(5t+1)x=9t2x=9t25t+1(9t2)32(5t+1)(9t2)2+(5t+1)2(9t2)(5t+1)3=03i=1xi5=3i=1ti=coeffoft2coeffoft3=(6×81+36021650+9075)(729810+225125)=(323)19=17
Commented by mr W last updated on 20/Nov/23
great!
great!

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