一题：g(x)=x^2-2x+1,f(x)=x^3-3x^1-x-1求用g（x）去除f（x）所得的商及余式.二题：试确定a和b的值,使x^4+ax^2-bx+2能被x^2+3x+2整除.

Question

一题：g(x)=x^2-2x+1,f(x)=x^3-3x^1-x-1求用g（x）去除f（x）所得的商及余式.
二题：试确定a和b的值,使x^4+ax^2-bx+2能被x^2+3x+2整除.

Answer 1

1、
打错了,应该是f(x)=x^3-3x^2-x-1吧
f(x)/g(x)=(x^3-3x^2-x-1)/(x^2-2x+1)
=(x^3-2x^2+x-x^2-2x-1)/(x^2-2x+1)
=[x(x^2-2x+1)-(x^2-2x+1)-4x]/(x^2-2x+1)
=(x-1)-4x/(x^2-2x+1)——商是x-1,余式是-4x,还可以继续化简
=(x-1)-[(4x-4)+4]/(x^2-2x+1)
=(x-1)-(4x-4)/(x-1)^2-4]/(x-1)^2
=(x-1)-4/(x-1)+4/(x-1)^2——部分分式的形式,便于积分运算等
2、
x^4+ax^2-bx+2
=x^2(x^2+3x+2)-3x^3-2x^2+ax^2-bx+2
=x^2(x^2+3x+2)-3x(x^2+3x+2)+9x^2+6x-2x^2+ax^2-bx+2
=x^2(x^2+3x+2)-3x(x^2+3x+2)+[(9-2+a)x^2+(6-b)x+2]
要使(7+a)x^2+(6-b)x+2被x^2+3x+2整除,则
7+a=1 即a=-6
6-b=3 b=3
也可以使用待定系数法,设x^4+ax^2-bx+2=(x^2+3x+2)(x^2+mx+n),……