多項式一條

g(x)=(x-1)Q1(x)+5-----------1
g(x)=(2x+1)Q2(x)+2---------2
把1代入1,得g(1)=5-----------3
把-1/2代入2,得g(-1/2)=2------4
設g(x)除以2x^2-x-1的餘式為(ax+b)[少於除式一次方]
g(x)=(2x^2-x-1)Q3(x)+(ax+b)
g(x)=(2x+1)(x-1)Q3(x)+(ax+b)-------------5
把3代入5
g(1)=a+b=5--------------6
把4代入5
g(-1/2)=-a/2+b=2---------7
6-7:
3a/2=3
a=2 , b=5-2=3
所以g(x)除以2x^2-x-1的餘式=(2x+3)

g(1) = 5
g(-1/2) = 2
g(x) = Q(x)(2x^2-x-1) + Ax + B
g(x) = Q(x)(2x+1)(x-1) + Ax + B
g(1) = A + B = 5 --------- (1)
g(-1/2) = -A/2 + B = 2 --------- (2)
(1) - (2), 3A/2 = 3, A = 2
2 + B = 5, B = 3
餘式 = 2x + 3


2007-12-26 13:01:49 補充：
oh, 慢左, 不過我做得幾清楚, 投我一票啦