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Question Number 171835 by Mikenice last updated on 21/Jun/22
solve for x: x^5 +x^4 +1=0
solveforx:x5+x4+1=0
Answered by floor(10²Eta[1]) last updated on 21/Jun/22
x^5 +x^4 +1=(x^2 +ax+1)(x^3 +bx^2 +cx+1) x^5 +(a+b)x^4 +(ab+c+1)x^3 +(1+ac+b)x^2 +(a+c)x+1 a+b=1⇒b=1−a ab+c=−1⇒ab−a=−1 ac+b=−1⇒−a^2 +b=−1⇒a^2 +a−2=0⇒a=−2 or 1 a+c=0⇒a=−c only works a=1⇒b=0, c=−1 ⇒x^5 +x^4 +1=(x^2 +x+1)(x^3 −x+1)=0 x^2 +x+1=0⇒x=((−1±i(√3))/2) x^3 −x+1=0 note that (a+b)^3 −3ab(a+b)−(a^3 +b^3 )=0 let x=a+b x^3 −3abx−(a^3 +b^3 )=0 ⇒3ab=1⇒a^3 b^3 =(1/(27))=P a^3 +b^3 =−1=S ⇒a^3 , b^3 are the roots of z^2 +z+(1/(27))=0⇒z=((−1±(5/(3(√3))))/2)=a^3 and b^3 ⇒x=a+b=(((−1+(5/(3(√3))))/2))^(1/3) +(((−1−(5/(3(√3))))/2))^(1/3)
x5+x4+1=(x2+ax+1)(x3+bx2+cx+1)x5+(a+b)x4+(ab+c+1)x3+(1+ac+b)x2+(a+c)x+1a+b=1⇒b=1−aab+c=−1⇒ab−a=−1ac+b=−1⇒−a2+b=−1⇒a2+a−2=0⇒a=−2or1a+c=0⇒a=−conlyworksa=1⇒b=0,c=−1⇒x5+x4+1=(x2+x+1)(x3−x+1)=0x2+x+1=0⇒x=−1±i32x3−x+1=0notethat(a+b)3−3ab(a+b)−(a3+b3)=0letx=a+bx3−3abx−(a3+b3)=0⇒3ab=1⇒a3b3=127=Pa3+b3=−1=S⇒a3,b3aretherootsofz2+z+127=0⇒z=−1±5332=a3andb3⇒x=a+b=−1+53323+−1−53323

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