余式定理习题以x^2+2x+3除f(x)余x+12,以(x+1)^2除f(x)余5x+4,则以(x+1)( x^2+2x

f(x)=m(x)(x^2+2x+3)+(x+12)
f(x)=n(x)(x+1)^2+(5x+4)=n(x)(x^2+2x+1)+(5x+4)
=n(x)(x^2+2x+3)+x+12+4x-2n(x)-8
所以:m(x)=n(x)
4x-2n(x)-8=0
所以:m(x)=n(x)=2x-4
f(x)=(2x-4)(x^2+2x+3)+(x+12)
=(2x+2)(x^2+2x+3)+(x+12)-6(x^2+2x+3)
=2(x+1)(x^2+2x+3)+(-6x^2-11x-6)
所以,(x+1)( x^2+2x+3)除f(x)的余式为:-6x^2-11x-6