Question

11.已知多项式f(x) 除以x+2所得余数为1；除以x+3所得余数为－1，求f(

11.已知多项式f(x) 除以x+2所得余数为1；除以x+3所得余数为－1，求f(x)除以（x+2)(x+3)所得余式

Answer 1

解:11.由已知得f(x)=(x+2)g(x)+1=(x+3)h(x)-1(*)
即(x+2)g(x)+2=(x+3)h(x)
亦即(x+3)g(x)-g(x)+2=(x+3)h(x)
故g(x)=(x+3)[g(x)-h(x)]+2
令d(x)=g(x)-h(x),则
g(x)=(x+3)d(x)+2,代入(*)式,得
f(x)=(x+2)g(x)+1
=(x+2)[(x+3)d(x)+2]+1
=(x+2)(x+3)d(x)+2(x+2)+1
=(x+2)(x+3)d(x)+2x+5
故f(x)除以(x+2)(x+3)所得余式为2x+5.