先利用除法原理可知\(x^{2012}+1=(x^{2}+1)(x^{2}+x+1)Q(x)+px^{^{3}}+qx^{2}+rx+s\)
又可將除法原理改寫成\(x^{2012}+1=(x^{2}+1)(x-1)(x^{2}+x+1)\frac{Q(x)}{x-1}+ax^{^{4}}+bx^{3}+cx^{2}+dx+e\)
將\(x^{2}=-1\)及\(x^{3}=1\)帶入式子,求出(a,b,c,d,e)=(1,-1,1,-1,2),所以餘式為\(x^{^{4}}-x^{3}+x^{2}-x+2\)
但這是除以\((x^{2}+1)(x-1)(x^{2}+x+1)\)的餘式,但原式要求的為除以\((x^{2}+1)(x^{2}+x+1)\)之餘式,
所以再將\(x^{^{4}}-x^{3}+x^{2}-x+2\)除以\((x^{2}+1)(x^{2}+x+1)\),得到之餘式為\(-2x^{3}-x^{2}-2x+1\)就是答案了!!!
[ 本帖最後由 tacokao 於 2012-6-24 11:55 AM 編輯 ]