一元三次方程x^3-3x^2+2求解!要过程

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一元三次方程x^3-3x^2+2求解!要过程
一元三次方程x^3-3x^2+2求解!要过程

一元三次方程x^3-3x^2+2求解!要过程
答:
x^3-3x^2+2
=(x^3-x^2)-2(x^2-1)
=(x-1)x^2-2(x-1)(x+1)
=(x-1)(x^2-2x-2)
所以:
x^3-3x^2+2=0
(x-1)(x^2-2x-2)=0
解得:x=1或者x^2-2x+1=3
解得:x=1或者x=1+√3或者x=1-√3

(x³-x²)-(2x³-2)=0
x²(x-1)-2(x+1)(x-1)=0
(x-1)(x²-2x-2)=0
x=1,x=1-√3,x=1+√3

x^3-3x^2+2=0
(x^3-1)-3x^2+3=0
(x-1)(x^2+x+1)-3(x^2-1)=0
(x-1)(x^2+x+1)-3(x+1)(x-1)=0
(x-1)(x^2+x+1-3(x+1))=0
(x-1)(x^2-2x-2)=0
x-1=0
x1=1
x^2-2x-2=0
(x-1)^2=3
x-1=±√3
x=1±√3
x2=1-√3
x3=1+√3

原=x^3-x^2-2x^2+2=x^2(x-1)-2(x+1)(x-1)=(x-1)(x^2-2x-2)=(x-1)(x-1-√3)(x-1+√3)