Question

Resolva a equação 2x^4 -x^3-6x^2-x+2=0
(considere:x+1/x=t)​

Answer 1

solução:

2x⁴ -x³-6x²-x+2=0

2x³*(x-1/2)-6x²-x+2=0

####-6x²-x+2=0

####x'=1/2  e x''=-2/3

####ax²+bx+c=a*(x-x')*(x-x'')

####-6x²-x+2 =-6*(x-1/2)*(x+2/3)

2x³*(x-1/2)-6*(x-1/2)*(x+2/3)=0

(x-1/2)*[2x³-6*(x+2/3)]=0

(2x-1)*[x³-3*(x+2/3)]=0

(2x-1)*(x³-3x-2)=0

(2x-1)*(x³-x -2x-2)=0

(2x-1)*[x*(x²-1)-2*(x+1)]=0

(2x-1)*[x*(x-1)*(x+1)-2*(x+1)]=0

(2x-1)*[(x+1)*(x*(x-1)-2)]=0

(2x-1)*(x+1)*(x²-x-2)=0

2x-1=0 ==>x'=1/2

x+1=0 ==>x''=-1

x²-x-2= 0   ==>x'''=2 e x''''=-1 (raiz dupla)

Raízes={-1,  1/2 e 2}

2x⁴-x³-6x²-x+2=(x+1)²*(x-1/2)*(x-2)