求f(x)=2x^6-x^5-2x^4+3x^3-x^2-2x+1在实数域上的标准分解式

f(x)=2x^6-x^5-2x^4+3x^3-x^2-2x+1
f(1)=0
=>x-1 is a factor of f(x)
f(-1)=0
=>x+1 is a factor of f(x)
let
f(x) =(x-1)(x+1)(2x^4+k1x^3+k2x^2+k3x-1)
coef.of x,=> -k3=-2 =>k3 =2
coef.of x^2:-k2-1 =-1 =>k2=-2
coef.of x^3:-k1+k3=3 =>k1=-1
f(x)=(x-1)(x+1)(2x^4-x^3-2x^2+2x-1)
let
g(x) =2x^4-x^3-2x^2+2x-1
g(1)=0
g(-1)=0
let
g(x) = (x-1)(x+1)(2x^2+bx+1)
coef.of x
b=-2
g(x) = (x-1)(x+1)(2x^2-2x+1)
f(x)=2x^6-x^5-2x^4+3x^3-x^2-2x+1
=[(x-1)(x+1)]^2 .(2x^2-2x+1)

答：
f(x)=2x^6-x^5-2x^4+3x^3-x^2-2x+1
经观察x=1是其零点之一，分解为：
f(x)=(x-1)(2x^5+x^4-x^3+2x^2+x-1)，经观察x=-1是其零点之一
f(x)=(x-1)(x+1)(2x^4-x^3+2x-1)
f(x)=(x-1)(x+1)(x+1)(2x^3-3x^2+3x-1)
f(x)=2...

(x-1)(x+1)^2*(2x-1)(x^2-x+1)