[10點] F.4 Maths 急~
2011-02-09 9:03 pm
1. 設p和q為常數。當f (x) = 4x^2 + 9x + p除以x – 1時，商式是qx + 13，餘式是7。(a) 試以x和q表示f (x)。(b) 求p和q的值。(c) 由此，解方程f (x) – 3 = 0。 2. 已知多項式f (x) 除以x + 3時，商式是x^2 + 5x – 4，餘式是 – 6。(a) 求多項式f (x) 。(b) 若g (x) = 2 f (x) – ( x^2 + x – 6)，求g (x)。(c) 求g (x)除以x + 3的商式和餘式。
回答 (1)
2011-02-09 9:24 pm
✔ 最佳答案
1(a) f(x) = (x - 1)(qx + 13) + 7(b) 代x = 1, 4 + 9 + p = 7 => p = -6
f(x) = 4x^2 + 9x - 6 = (x - 1)(4x + 13) + 7 => q = 4
(c) f(x) - 3 = 0
4x^2 + 9x - 9 = 0
x = (-9 + 15)/8 或 x = (-9 - 15)/8
x = 3/4 x = -3
2(a) f(x) = (x + 3)(x^2 + 5x - 4) - 6 = x^3 + 8x^2 + 11x - 18
(b) g(x) = 2f(x) - ( x^2 + x - 6) = 2x^3 +15x^2 + 21x -30
(c) g(-3) = -12。 g(x) = (x + 3)(2x^2 +9x - 6) - 12
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