Re: [中學] 餘式定理請教
※ 引述《shanewang43 (WSG)》之銘言：
: f(x)除以(x-2)^2為13x-17
: 除以(x-1)^2為3x-1
: 求除以[(x-1)^2][(x-2)^2]的餘式
: 想不出來 跪求想法
令 f(x) = [(x-1)^2][(x-2)^2]p(x) + a(x-1)(x-2)^2 + b(x-2)^2 + 13x - 17
又 f(1) = 2 = b - 4 => b = 6
=> f(x) = [(x-1)^2][(x-2)^2]p(x) + a(x-1)(x-2)^2 + 6(x-2)^2 + 13x - 17
= [(x-1)^2][(x-2)^2]p(x) + a(x-1)[(x-1)(x-3) + 1] +
6[(x-1)(x-1) - 2x + 3] + 13x - 17
= [(x-1)^2][(x-2)^2]p(x) + a(x-3)(x-1)^2 + 6(x-1)^2 +
(1 + a)x + 1 - a
= G(x)(x-1)^2 + (1 + a)x + 1 - a
=> 1 + a = 3, 1 - a = -1 得 a = 2,
所求為 2(x-1)(x-2)^2 + 6(x-2)^2 + 13x - 17
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10/01 10:51, 1F
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