餘式定理~~一個小問題

x^100+2x^55+5x+1除以x^2(x^2－1)所得的餘式為?
Sol
x^2(x^2－1)=0
x^4－x^2=0
x^4=x^2
x^100=>(x^4)*x^96=>x^98=>(x^4)^x^94=>x^96…..
=>x^6=>x^4*x^2=>x^2*x^2=>x^4=>x^2
x^55=>x^53.+x^51=>x^5=>x^3
x^100+2x^55+5x+1
=>x^2+2x^3+5x+1
=>2x^3+x^2+5x+1
餘式為2x^3+x^2+5x+1
or
設 f(x)=x^100+2x^55+5x+1=p(x))(x^4－x^2)+ax^3+bx^2+cx+d
f(0)=1=d
f(1)=1+2+5+1=9=a+b+c+d
f(－1)=1－2－5+1=－5=－a+b－c+d
f’(x)=100x^99+110x^54+5=p(x)(4x^3－2x)+p’(x)(x^4－x^2)+3ax^2+2bx+c
f’(0)=5=c
a+b+5+1=9
a+b=3
－a+b－5+1=－5
－a+b=－1
2b=2
b=1
a=2
餘式為2x^3+x^2+5x+1

令f(x)=x^100+2x^55+5x+1=q(x)[x^2(x^2-1)]+ax^3+bx^2+5x+1
f(1)=1+2+5+1=a+b+5+1=9
f(-1)=1-2-5+1=-a+b-5+1=-5
a+b=3
-a+b=1
b=2 ,a=1
ax^3+bx^2+5x+1=x^3+2x^2+5x+1