问题
填空题
设a为常数,多项式x3+ax2+1除以x2-1所得的余式为x+3,则a=______.
答案
∵多项式x3+ax2+1除以x2-1所得的余式为x+3,
∴可设x3+ax2+1-(x+3)=(x2-1)(x+b),
整理可得:x3+ax2-x-2=x3+bx2-x-b,
∴
,a=b b=2
∴a=2.
故答案为:2.
设a为常数,多项式x3+ax2+1除以x2-1所得的余式为x+3,则a=______.
∵多项式x3+ax2+1除以x2-1所得的余式为x+3,
∴可设x3+ax2+1-(x+3)=(x2-1)(x+b),
整理可得:x3+ax2-x-2=x3+bx2-x-b,
∴
,a=b b=2
∴a=2.
故答案为:2.