MQ7 --- Factorization

f(x)=x^5+x+1, w=(-1+√3 i)/2,
f(w)=w^2+w+1=0, so x^2+x+1 is a factor of f(x)
then
f(x)=(x^2+x+1)(x^3-x^2+1)

2012-05-15 18:15:32 補充：
(1) x^2+x+1=0, x=(-1+√3 i)/2, (-1-√3 i)/2 = w, w^2
(2) x^3-x^2+1=0, x=u+1/(9u), uw+w^2/(9u), uw^2+w/(9u) , u=[(-25+3√69)/54]^(1/3)
let them be a, b, c
so, x^5+x+1=(x-w)(x-w^2)(x-a)(x-b)(x-c)

2012-05-15 18:54:32 補充：
Sorry! Amended as follows:
(1) x^2+x+1=0, x=(-1+√3 i)/2, (-1-√3 i)/2 = w, w^2
(2) x^3-x^2+1=0, x=u+1/(9u)+1/3, uw+w^2/(9u)+ 1/3, uw^2+w/(9u)+ 1/3
let them be a, b, c, where u=[(-25+3√69)/54]^(1/3)
so, x^5+x+1=(x-w)(x-w^2)(x-a)(x-b)(x-c)

x^5 + x + 1
= x^5 - x^2 + x^2 + x + 1
= x^2 (x^3 - 1) + (x^2 + x + 1)
= (x^2 + x + 1)[x^2 (x - 1) + 1]
= (x^2 + x + 1)(x^3 - x^2 + 1)