设f(x)=A(x+2)(x+3)+B
B=C(x+2)+1
B=D(x+3)-1
C(x+2)+1=D(x+3)-1
Cx+(2C+1)=Dx+(3D-1)
所以:
C=D
2C+1=3D-1
解得:C=D=2
B=2(x+2)+1=2x+5
多项式f(x)除以(x+2)(x+3)所得余式是2x+5
设f(x)=A(x+2)(x+3)+B
B=C(x+2)+1
B=D(x+3)-1
C(x+2)+1=D(x+3)-1
Cx+(2C+1)=Dx+(3D-1)
所以:
C=D
2C+1=3D-1
解得:C=D=2
B=2(x+2)+1=2x+5
多项式f(x)除以(x+2)(x+3)所得余式是2x+5