已知x6+4x5+2x4-6x3-3x2+2x+l其中f(x)是x的多项式,则f(x)=________.
±(x3+2x2-x-1)
分析:由于x6+4x5+2x4-6x3-3x2+2x+l=[(x3+2x2)2-(2x4+6x3+4x2)+(x+1)2]=[(x3+2x2)2-2(x3+2x2)(x+1)+(x+1)2]=[(x3+2x2-x-1)2.从而得出f(x)的值.
解答:∵[f(x)]2=x6+4x5+2x4-6x3-3x2+2x+l
=[(x3+2x2)2-(2x4+6x3+4x2)+(x+1)2]
=[(x3+2x2)2-2(x3+2x2)(x+1)+(x+1)2]
=[(x3+2x2-x-1)2.
∴f(x)=±(x3+2x2-x-1).
故答案为:±(x3+2x2-x-1).
点评:本题考查了因式分解的应用,解题的关键是将x6+4x5+2x4-6x3-3x2+2x+l分解为[(x3+2x2-x-1)2.