一位同学把被除式的二次项的系数写成a,结果的一次项系数写成b,即（x3+ax2+1）/(x+1)=x2-bx+1,求a,

Question

一位同学把被除式的二次项的系数写成a,结果的一次项系数写成b,即（x3+ax2+1）/(x+1)=x2-bx+1,求a,

Answer 1

(x3+ax2+1)/(x+1)
=(x3+1+ax2)/(x+1)
=[(x+1)(x2-x+1)+ax2-a+a]/(x+1)
=x2-x+1+[a(x2-1)+a]/(x+1)
=x2-x+1+[a(x+1)(x-1)+a]/(x+1)
=x2-x+1+a(x-1)+a/(x+1)
=x2+(a-1)x+1-a+a/(x+1)
因为(x3+ax2+1)/(x+1)=x2-bx+1
所以x2+(a-1)x+1-a+a/(x+1)=x2-bx+1
则有1-a+a/(x+1)=1,a-1=-b得
a=0,b=1