题目

一个三次多项式,除以(x^2-x+2)余数为(5x+4),除以(x^2+x-1)余数为(12x-1),求这个多项式.

提问时间：2020-12-30

答案

ax^3+bx^2+cx+d-(5x+4)=ax^3+bx^2+(c-5)x+(d-4) 整除(x^2-x+2)
(x^2-x+2)(ax+e)=ax^3-ax^2+2ax+ex^2-ex+2e=ax^3+(e-a)x^2+(2a-e)x+2e
ax^3+bx^2+cx+d-(12x-1)=ax^3+bx^2+(c-12)x+(d+1) 整除(x^2+x-1)
(x^2+x-1)(ax+f)=ax^3+ax^2-ax+fx^2+fx-f=ax^3+(a+f)x^2+(f-a)x-f
e-a=b
2a-e=c-5
2e=d-4
a+f=b
f-a=c-12
f=-d-1
a=1 b=-2 c=8 d=2 y=x^3-2x^2+8x+2
我觉得还有点复杂