※ 引述《karta1103084 (Sky)》之銘言：
: 一多項式f(x)且deg f(x)≧3
: 且以(x-b)(x-c),(x-c)(x-a),(x-a)(x-b)除之
: 餘式分別為3x-1,x+1,2x+3
: 求 (1)3a+4b+5c=? (2)求以(x-a)(x-b)(x-c)除f(x)之餘數為?
f(x) = (x-b)(x-c)Q(x) + 3x-1 ...(1)
= (x-c)(x-a)Q'(x) + x+1 ...(2)
= (x-a)(x-b)Q''(x) + 2x+3 ...(3)
(2)(3) f(a) = a+1 = 2a+3 => a = -2
(1)(3) f(b) = 3b-1 = 2b+3 => b = 4
(1)(2) f(c) = 3c-1 = c+1 => c = 1
(1) 3a+4b+5c = -6 +16 +5 = 15
(2)
f(x) = (x+2)(x-4)(x-1)Q(x) + k(x+2)(x-1) + x+1
f(4) = 18k + 5 = 11 => k = 1/3
餘式為 1/3 (x+2)(x-1) + x+1
--
※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 220.129.71.148
※ 文章網址: https://www.ptt.cc/bbs/Math/M.1441429922.A.73A.html