x³+2x²+2x+1
=(x³+1)+(2x²+2x)
=(x+1)(x²-x+1)+2x(x+1)
=(x+1)(x²+x+1)
=(x+1)[(x²+x+1/4)+3/4]
=(x+1)[(x+1/2)²+3/4]
[]内恒大于零,
所以方程有一个实数根x=-1,
一对复数根x=(1±i√3)/2
x³+2x²+2x+1
=(x³+1)+(2x²+2x)
=(x+1)(x²-x+1)+2x(x+1)
=(x+1)(x²+x+1)
=(x+1)[(x²+x+1/4)+3/4]
=(x+1)[(x+1/2)²+3/4]
[]内恒大于零,
所以方程有一个实数根x=-1,
一对复数根x=(1±i√3)/2