f(x)
=(1+a)x^4+x^3-(3a+2)x^2-4a
=(x^4+x^3-2x^2)+(ax^4-3ax^2-4a)
=(x^2+x-2)x^2+a(x^4-3x^2-4)
=(x+2)(x-1)x^2+a(x^2-4)(x^2+1)
=(x+2)(x-1)x^2+a(x+2)(x-2)(x^2+1)
=(x+2)[(x-1)x^2+a(x-2)(x^2+1)]
对任意的实数a,存在x0,恒有f(x0)≠0,即对任意的实数a,存在x0,恒有f(x0)与a值无关,其f(x0)≠0
显然有x+2=0和(x-2)(x^2+1)=0,恒有f(x)与a值无关,但又f(x)≠0,故只有(x-2)(x^2+1)=0符合
当(x-2)(x^2+1)=0时,f(x)与a无关,此时x=2为什么只有x+2=0和(x-2)(x^2+1)=0？为什么不能有(x-1)=0？请注意，f(x)=(x+2)*[(x-1)x^2+a(x-2)(x^2+1)]x-1=0时，f(x)=f(1)=3*[a*(-1)*2]=-6a