2、设f(x)=(1+a)x4+x3-(3a+2)x2-4a,试证明对任意实数a：

1:f(x)=ax4-3ax2-4a+x4+x3-2x2
=a(x4-3x2-4)+x4+x3-2x2
=a(x2+1)(x2-4)+x2(x2+x-2)
=a(x-2)(x+2)(x2+1)+x2(x+2)(x-1)
=(x+2)(ax3+ax-2ax2-2a+x3-x2)
可知当f(x)=0时无论a取何值时总有相同实根
x=-2
2:f(x)=a(x4-3x2-4)+x4+x3-2x2
因为存在x0,恒有f(x0)≠0.根据上式可知当
x4-3x2-4=0,x4+x3-2x2≠0时满足条件
x=2,x=-2; x=0,x=-2,x=1
所以当x0=2时f(x0)≠0

(1)f(x)=(1+a)x4+x3-(3a+2)x2-4a
=(x4+x3-2x2)+a*(x4-3x2-4)
=x2*(x2+x-2)+a*(x2-4)*(x2+1)
=x2*(x+2)*(x-1)+a*(x+2)*(x-2)*(x2+1)
=(x+2)*[x2*(x-1)+a*(x-2)*(x2+1)