已知4x2-3x+1=a(x-1)2+b(x-1)+c對任意數(shù)x成立,則4a+2b+c=________
-2
分析:將a(x-1)2+b(x-1)+c展開后合并同類項與4x2-3x+1各項的系數(shù)相同,進(jìn)而求得a、b、c的值,代入4a+2b+c求出即可.
解答:∵a(x-1)2+b(x-1)+c
=a(x2-2x+1)+bx-b+c
=ax2-2ax+a+bx-b+c
=ax2-(2a+b)x+a-b+c
=4x2-3x+1
∴a=4、-(2a+b)=-3、a-b+c=1,
解得:a=4、b=-5、c=-8,
∴4a+2b+c
=4×4+2×(-5)+(-8)
=16-10-8
=-2
故答案為:-2.
點評:本題考查了有理數(shù)的混合運算,解題的關(guān)鍵是將多項式展開后合并同類項,兩個二次三項式相等,就是他們的各項的系數(shù)相等.