x^4+x^3-8x^2+x+1=0
用待定系數(shù)法
令(x^2+ax+1)(x^2+bx+1)=0
x^4+(a+b)x^3+(ab+2)x^2+(a+b)x+1=0
a+b=1
ab=-10
a,b是方程y^2-y-10=0的兩根.
(y-1/2)^2=41/4
a=(1+√41)/2 b=(1-√41)/2
x^4+x^3-8x^2+x+1=0
[x^2+(1+√41)x/2+1][x^2+(1-√41)x/2+1]=0
x^2+(1+√41)x/2+1=0 x=[-(1+√41)±√(26+2√41)]/4
x^2+(1-√41)x/2+1=0 x=[-(1-√41)±√(26-2√41)]/4
x=[-(1+√41)±√(26+2√41)]/4或x=[-(1-√41)±√(26-2√41)]/4
一共4個解.