✔ 最佳答案
1. f(x)=q(x)(x-1)^3(x-2)(x+2)+a(x-1)^3(x-2)+b(x-1)^3+3
f(2)=b(1)+3=6
b=3
f(x)=q(x)(x-1)^3(x-2)(x+2)+a(x-1)^3(x-2)+3(x-1)^3+3
f(-2)=a(-3)^3(-4)+3(-3)^3+3=30
108a-81+3=30
a=1
f(x)=q(x)(x-1)^3(x-2)(x+2)+(x-1)^3(x-2)+3(x-1)^3+3
2.f(1)=1+k+L+m=-4
m=-5-k-L
x^2+x+1=0
(x-1)(x^2+x+1)=0
x^3-1=0
x^3=1
w 為w^2+w+1=0之一根
f(w)=w^3+kw^2+Lw+m=5w-3
kw^2+(L-4)w+4=0
k=4,L=8,m=-17
3.設f(x)=q(x)(x^2+x+1)+ax+b
(x+1)f(x)=q(x)(x+1)(x^2+x+1)+(ax+b)(x+1)
=q(x)(x+1)(x^2+x+1)+ax^2+ax+bx+b
=q(x)(x+1)(x^2+x+1)+a(x^2+x+1)+bx+(b-a)
So
b=5,b-a=3
a=2
R2=2x+3
4.f(x)=x^50+2x^49+x^29+2x^28+1=q(x)(x^2+x-2)+a(x-1)+b
f(1)=1+2+1+2=7=b
f(x)=x^50+2x^49+x^29+2x^28+1=q(x)(x^2+x-2)+a(x-1)+7
f(-2)=2^50-2^50-2^29+2^29+1=-3a+7
a=2
2(x-1)+7=2x+5