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Question Number 176458 by mathlove last updated on 19/Sep/22

x^2 +x=1 ((x^5 +8)/(x+1))=?

x2+x=1x5+8x+1=?

Answered by Rasheed.Sindhi last updated on 19/Sep/22

x^2 +x=1; ((x^5 +8)/(x+1))=? x^2 =1−x⇒x^3 =x−x^2 =x−(1−x)=2x−1 ⇒x^5 =2x^3 −x^2 =2(2x−1)−(1−x) ⇒x^5 =5x−3 ((x^5 +8)/(x+1))=((5x−3+8)/(x+1))=((5x+5)/(x+1))=5(((x+1)/(x+1)))=5

x2+x=1;x5+8x+1=?x2=1−x⇒x3=x−x2=x−(1−x)=2x−1⇒x5=2x3−x2=2(2x−1)−(1−x)⇒x5=5x−3x5+8x+1=5x−3+8x+1=5x+5x+1=5(x+1x+1)=5

Commented by mathlove last updated on 20/Sep/22

thanks

thanks

Answered by Rasheed.Sindhi last updated on 20/Sep/22

AnOther way..._((Replacing 1′s )) determinant (((x^2 +x=1; ((x^5 +8)/(x+1))=?))) ((x^5 +8)/(x+1))=((x^5 +8(x^2 +x))/(x+x^2 +x))=((x^5 +8x^2 +8x)/(x^2 +2x)) =((x^4 +8x+8)/(x+2))=((x^4 +8x+8(x^2 +x))/(x+2(x^2 +x))) =((x^4 +8x^2 +16x)/(2x^2 +3x)) =((x^3 +8x+16)/(2x+3))=((x^3 +8x+16(x^2 +x))/(2x+3(x^2 +x)))=((x^3 +16x^2 +24x)/(3x^2 +5x)) =((x^2 +16x+24)/(3x+5))=((x^2 +16x+24(x^2 +x))/(3x+5(x^2 +x))) =((25x^2 +40x)/(5x^2 +8x))=((25x+40)/(5x+8))=((5(5x+8))/((5x+8))) =5

AnOtherway...(Replacing1′s)x2+x=1;x5+8x+1=?x5+8x+1=x5+8(x2+x)x+x2+x=x5+8x2+8xx2+2x=x4+8x+8x+2=x4+8x+8(x2+x)x+2(x2+x)=x4+8x2+16x2x2+3x=x3+8x+162x+3=x3+8x+16(x2+x)2x+3(x2+x)=x3+16x2+24x3x2+5x=x2+16x+243x+5=x2+16x+24(x2+x)3x+5(x2+x)=25x2+40x5x2+8x=25x+405x+8=5(5x+8)(5x+8)=5

Answered by Rasheed.Sindhi last updated on 20/Sep/22

AnOther way..._((Replacing x^(2 ) ′s)) x^2 +x=1; ((x^5 +8)/(x+1))=? x^2 =1−x ((x^5 +8)/(x+1))=((x(x^2 )^2 +8)/(x+1))=((x(1−x)^2 +8)/(x+1)) =((x−2x^2 +x^3 +8)/(x+1))=((x−x^2 (2−x)+8)/(x+1)) =((x−(1−x)(2−x)+8)/(x+1)) =((x−2+x+2x−x^2 +8)/(x+1))=((−x^2 +4x+6)/(x+1)) =((−(1−x)+4x+6)/(x+1))=((−1+x+4x+6)/(x+1)) =((5x+5)/(x+1))=((5(x+1))/((x+1)))=5

Prime causes double exponent: use braces to clarifyx2+x=1;x5+8x+1=?x2=1−xx5+8x+1=x(x2)2+8x+1=x(1−x)2+8x+1=x−2x2+x3+8x+1=x−x2(2−x)+8x+1=x−(1−x)(2−x)+8x+1=x−2+x+2x−x2+8x+1=−x2+4x+6x+1=−(1−x)+4x+6x+1=−1+x+4x+6x+1=5x+5x+1=5(x+1)(x+1)=5

Answered by Rasheed.Sindhi last updated on 20/Sep/22

A_(pproach) ^(nOther) ... x^2 +x=1; ((x^5 +8)/(x+1))=? determinant (((x(x+1)=1))) ((x^5 +8)/(x+1)) =((x(x^5 +8))/(x(x+1))) [x(x+1)=1] =x^6 +x^5 −x^5 +8x =x^5 (x+1)−x^5 +8x =x^4 ∙x(x+1)−x^5 +8x =x^4 +x^4 −x^5 −x^4 +8x =2x^4 −x^3 .x(x+1)+8x =2x^4 +2x^3 −3x^3 +8x =2x^2 .x(x+1)−3x^3 +8x =5x^2 −3x^3 −3x^2 +8x =5x^2 −3x.x(x+1)+8x =5x^2 −3x+8x =5x^2 +5x =5x(x+1) =5

ApproachnOther...x2+x=1;x5+8x+1=?x(x+1)=1x5+8x+1=x(x5+8)x(x+1)[x(x+1)=1]=x6+x5−x5+8x=x5(x+1)−x5+8x=x4⋅x(x+1)−x5+8x=x4+x4−x5−x4+8x=2x4−x3.x(x+1)+8x=2x4+2x3−3x3+8x=2x2.x(x+1)−3x3+8x=5x2−3x3−3x2+8x=5x2−3x.x(x+1)+8x=5x2−3x+8x=5x2+5x=5x(x+1)=5

Commented by Tawa11 last updated on 20/Sep/22

Great sir

Greatsir

Commented by Rasheed.Sindhi last updated on 20/Sep/22

Thanks miss

Thanksmiss

Commented by mathlove last updated on 02/Oct/22

thanks a lot

thanksalot

Answered by Peace last updated on 20/Sep/22

x^2 =1−x,x^3 =x−x^2 =2x−1 x^4 =2x^2 −x=2−3x x^5 =2x−3x^2 =5x−3 ((x^5 +8)/(x+1))=((5x−3+8)/(x+1))=5

x2=1−x,x3=x−x2=2x−1x4=2x2−x=2−3xx5=2x−3x2=5x−3x5+8x+1=5x−3+8x+1=5

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