Question

Xuan

為什麼圈起來那裡要那樣假設？

JU)=
· a ( x = x + 1 ) + bx-b-w
(x-1) f(x) = (x²-x+1) (X-1)Q00+ (ax+b)(x-1) a b = 3,
(X+1)(X-1)
5. 設 f(x)為一多項式,以x+1除之餘5;以x-1除之餘-3,則f(x)除以x²-1之餘式為_
f(x)=(x+1)Q(x)+5
二
ax²-ax+bx-b
3-0=5 8=a
(X-1)Q(X)-3
(x+1)(X-1)G2(1)+ax+b
f(-1)= 5 = -a+b atb--3
f(1)=-3=a+b+1-2+b=5
-4X²+4X+8
=
2b=2
『b=1_
a+1=-3
a=-4
+
6. f(x)為一多項式,分別以x²+x+1、x−1除之依次幹式8x+12、8,則以(x-1)(x²+x+1)除f(x)之餘式為
f(x)=(x+x+1)Q(0x+12)
= (x-1) Q2
=(x-1)(X²- DQx)+Q(X²+x+1)+8x+12