Question

已知x^10-px+q能被（x+1）^2整除,求p和q的值.

Answer 1

x^10-px+q

=(x+1-1)^10-px+q

=(x+1)^10-10(x+1)^9+……-10(x+1)+1-px+q

=(x+1)^10-10(x+1)^9+……-10x-10-+1-px-q

=(x+1)^10-10(x+1)^9+……-[(10+p)x+(9-q)]

只有最后一项不能被(x+1)^2整除,

所以最后一项必须为0,即

10+p=0

9-q=0

解得

p=-10,q=9.

满意请采纳.