Question

求f(x)的余式，需要解题过程

若f(x)除以(x-2)²余式为31x-51，除以(x+3)²余式为-4x-6，则
(2) f(x)除以(x-2)(x+3)²所得余式为：

Answer 1

f(x)=(x-2)^2•q(x)+31x-51 (1)
f(x)=(x+3)^2•q(x)-4x-6 (2)
由(1):f(2)=11
由(2):f(-3)=6
设f(x)=(x-2)(x+3)^2•h(x)+[a(x+3)^2-4x-6]
令x=2
f(2)=25a-14=11
a=1
所以余式为x^2+2x+3

Answer 2

f(2)=31*2-51=11 f(-3)=-4*-3-6=6
f(x)除以(x-2)(x+3)²所得余式为ax^2+bx+c
f(2)=4a+2b+c=11 f(-3)=9a-3b+c=6->a-b=-1 a,b为自然数，a/=0
-->a=-1,b=0,c=15时本方程成立。余数为-x^2+15

追问

a/=0是什么意思呢？

追答

a不等于0

Answer 3

f(x) = q(x).(x-2)^2 + (31x-51) (1)
f(x) = p(x).(x+3)^2 + (-4x-6) (2)
f(x) = r(x)(x-2)(x+3)^2 + (ax^2+bx+c) (3)
from (3) : f(2)= 4a+2b+c
from (1) : f(2) = 31(2) -51 =11
4a+2b+c =11 (4)
from (3) : f(-3)=9a-3b+c
from (2) : f(-3) = -4(-3) -6 = 6
9a-3b+c = 6 (5)
from (3)
f(x) = r(x)(x-2)(x+3)^2 + ( ax^2+bx+c)
f'(x) =r'(x)(x-2)(x+3)^2 +r(x)(x+3)^2 + 2r(x)(x-2)(x+3) + (2ax+b)
f'(-3) = 2a(-3) +b = -6a+b
from (2)
f(x) = p(x).(x+3)^2 + (-4x-6)
f'(x) = p'(x).(x+3)^2 + 2p(x)(x+3) -4
f'(-3) =-4
-6a+b = -4 (6)
(4)-(5)

-5a+5b = 5
a-b = -1 (7)
(6)+(7)
-5a=-3
a=3/5
from (6)

-6a+b = -4
-18/5+b=-4
b=-2/5
from (4)
4a+2b+c =11
12/5-4/5+c=11
c=47/5
f(x)除以(x-2)(x+3)²所得余式为：ax^2+bx+c =(1/5)(3x^2-2x+47)

追问

你好，请问f'(x)是导数的意思吗？这道题要用到导数的知识吗？
另外标准答案是x²+2x+3？是答案错了吗？