快 明天交 解一元三次方程x3-2x-1=0

x^3-2x-1=0
很明显有一根是-1
所哟
x^3-2x-1可以分解为
(x+1)(x^2+bx-1)
=x^3+bx^2-x+x^2+bx-1
=x^3+(b+1)x^2+(b-1)x-1
对比有:
b-1=-2
b+1=0

b=-1

所以
x3-2x-1=0
(x+1)(x^2-x-1)=0
x+1=0,
x=-1
x^2-x-1=0
x=[1+根号(5)]/2

x^3-2x-1
=x^3+x^2-x^2-2x-1
=x^2(x+1)-(x+1)^2
=(x+1)(x^2-x-1)=0
x1=-1
x2,3=(1±√5)/2

x³-2x-1=0
(x²+1)-(2x+2)=0
(x+1)(x²-x+1)-2(x+1)=0
(x+1)(x²-x+1-2)=0
(x+1)(x²-x-1)=0
x+1=0或x²-x-1=0
x=-1或x=(1±√5)/2

X3-X-(X+1)=0
X(X2-1)-(X+1)=0
X(X+1)(X-1)-(X+1)=0
(X+1)(X2-X-1)=0
x1=-1 X2=(1+根号5)/2 X3=(1-根号5)/2