Re: [解題] 高一數學 多項式
: 4.題目：多項式f (x)除以(x – 1)^2 , (x – 2)^2之餘式分別為2x + 1與x + 3，
: 試求
: 1)f (x)除以(x – 1)(x - 2)^2之餘式。
: 2)f (x)除以(x – 1)^2(x - 2)^2之餘式
: 答案:1)- x^2 + 5x - 1
: 2)-x^3 + 4x^2 – 3x + 3
: 不同章節或主題之題目請份篇發表
1. f(x) = (x-1)(x-2)^2*Q(x) + a(x-2)^2 + x+3
f(1) = a + 4 = 3 => a = -1
得 r1(x) = -(x-2)^2 + x+3 = - x^2 + 5x - 1
2. f(x) = (x-1)^2(x-2)*Q(x) + b(x-1)^2 + 2x+1
f(2) = b + 5 = 5 => b = 0
得r2(x) = 2x + 1
f(x) = (x-1)^2(x-2)^2*Q(x) + c(x-1)(x-2)^2 + (- x^2 + 5x - 1)
= (x-1)^2(x-2)^2*Q(x) + d(x-1)^2(x-2) + (2x + 1)
c(x-1)(x^2 -4x +4)-x^2 +5x -1 = d(x-2)(x^2 -2x +1)+2x +1
=> cx^3 +(-5c-1)x^2 +(8c+5)x +(-4c-1) = dx^3 +(-4d)x^2 +(5d+2)x +(-2d+1)
比較係數可得 c = d = -1 所以餘式為 -x^3 +4x^2 –3x + 3
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