高中數學中餘式定理與因式定理的應用

Question

f(x)為三次多項式
f(x)+1整除(x+1)^2
3-f(x)整除(x+2)且商式是一個一次式的完全平方[為(ax+b)^2的樣子]
求f(x)

請救救我
感激不盡

Answer 1

應該是(x+2)|3-f(x)
==>3-f(x)=(x+2)(ax+b)2 => f(x) = 3- (x+2)(ax+b)2 ---(A)
(x+1)2|f(x)+1
==>f(x)+1=(x+1)2(cx+d) => f(x) = (x+1)2(cx+d)-1 ---(B)
由(A)得f(-2)=3代入(B)==>c+d=3..........(1)
由(B)得f(-1)=-1代入(A)==>(a-b)2=4
a-b=2 or a-b=-2............(2)
將(A)兩邊微分==>f'(x)=-(ax+b)2-2a(x+2)(ax+b)..........(C)
將(B)兩邊微分==>f'(x)=2(x+1)(cx+d)+c(x+1)2.............(D)
由(D)得f'(-1)=0代入(C)==>-(-a+b)2-2a(-a+b)=0
(-a+b)(a-b-2a)=0==>a-b=0 or a+b=0..........(3)

由(2)(3)得
若a-b=2 ,a-b=0(不合)
若a-b=2,a+b=0==>a=1,b-1.........(4)
若a-b=-2,a-b=0(不合)
若a-b=-2,a+b=0==>a=-1,b=1......(5)
又f(x) = 3- (x+2)(ax+b)2 ---(A)
不管是a=1,b=-1 或a=-1,b=1代入都得
f(x)=3-(x+2)(x-1)2
=-x3+3x+1

Answer 2

1. 題目錯了吧!
應該是 f(x)+1被(x+1)2整除, 3-f(x)被(x+2)整除,商為(ax+b)2
2.
(1) 3-f(x)=(x+2)(ax+b)2 => f(x) = 3- (x+2)(ax+b)2 ---(A)
(2) f(x)+1=(x+1)2(cx+d) => f(x) = (x+1)2(cx+d)-1 ---(B)
(3) 比較(A)(B)之首項x3係數 => c=-a
(4) 比較(A)(B)之常數項 => d= 4-2b2
故 f(x)=(x+1)2(-ax+4-2b2) -1 ---(C)
(5) 令x=-1代入(A)(C)=> 3-(-a+b)2 = -1 => (a-b)2 =4 => a-b = 2 or -2
(6) 令x=-2代入(A)(C)=> 3= 2a+4 - 2b2 -1 => a = b2
(7) a-b= -2, a=b2 => b2-b+2=0,無實根(不合)
(8) a-b=2 , a=b2 => b2-b-2=0 => b=2 or -1, a=4 or 1
故f(x)=3-(x+2)(4x+2)2, 或 3-(x+2)(x-1)2