Question

竞赛题 1将多项式3x^3+2x^2+x-1按x-1得方幂展开

Answer 1

3x^3+2x^2+x-1

=3(x-1)^3+a(x-1)^2+b(x-1)+c

=3x^3-3x^2+3x-3+ax^2-2ax+a+bx-b+c

=3x^3+(a-3)x^2+(3-2a+b)x+c-b-3

a-3=2

a=5

3-2a+b=1

b=8

c-b-3=-1

c=10

3x^3+2x^2+x-1=3(x-1)^3+5(x-1)^2+8(x-1)+10