Question

x^4-2x^3-x^2+2x+1=0因式分解

给一个过程。答案是完全平方数。
要过程。

Answer 1

x^4-2x^3-x^2+2x+1
=(x^4-2x^3+x^2)-2x^2+2x+1
=x^2(x-1)^2-2x(x-1)+1
=[x(x-1)-1]^2
=(x^2-x-1)^2

故x^4-2x^3-x^2+2x+1=0
(x^2-x-1)^2=0
x^2-x-1=0
x=(1±√5)/2

Answer 2

2x^4+x^3-2x-1=0
x^2(2x+1)-(2x+1)=0
(2x+1)(x^2-1)=0
(2x+1)(x+1)(x-1)
=0
2x+1=0,x+1=0,x-1=0
x=-1/2,x=-1,x=1

Answer 3

x^4-2x^3-x^2+2x+1=0
x^2(x^2-2x+1)-2x^2+2x+1=0
x^2(x-1)^2-2x(x-1)+1=0
[x(x-1)-1]^2=0
x(x-1)-1=0
x^2-x-1=0
x^2-x+1/4-1/4-1=0
(x-1/2)^2-5/4=0
(x-1/2-√5/2)(x-1/2+√5/2)=0
x=1/2+√5/2或x=1/2-√5/2