標題:

F.4 Maths...急!!!

aa.jpg

 

此文章來自奇摩知識+如有不便請留言告知

發問:

除了x以外的未知數皆為常數 1. 當3x^2+ax-6除以x-2時,商式是3x-b,餘式是2。求a和b的值。 2. 當px^3+QX^2-14X-37除以4X^2-5時,商式是3X+8,餘式是X+3。求p和q的值。 3. 當x^3-px^2-qx+r除以x^2-2x+3時,商式是x+4,餘式是2x-1。求p,q和r的值。

最佳解答:

1 3x^2 + ax - 6 = (3x - b)(x - 2) + 2 Sub. x = 2. 12 + 2a - 6 = 2 => a = -2 Sub. x = 0. -6 = 2b + 2 => b = -4 2 px^3 + qx^2 - 14x - 37 = (4x^2 - 5)(3x + 8) + x + 3 Sub. x = 1 and x = -1 p + q - 51 = -7 -p + q - 23 = -3 So, 2q - 74 = -10 q = 32, p = 12 3 x^3 - px^2 - qx + r = (x^2 - 2x + 3)(x + 4) + 2x - 1 Sub. x = 0. r = 12 - 1 = 11 Sub. x = 1 and x = -1 - p - q + 12 = 11 -p + q + 10 = 15 So, -2p + 22 = 26 => p = -2, q = 3

其他解答:

1 3x^2 + ax - 6 = (3x - b)(x - 2) + 2 x = 2. 12 + 2a - 6 = 2 a = -2 x = 0. -6 = 2b + 2 b = -4 2 px^3 + qx^2 - 14x - 37 = (4x^2 - 5)(3x + 8) + x + 3 x = 1 and x = -1 p + q - 51 = -7 -p + q - 23 = -3 So, 2q - 74 = -10 q = 32, p = 12 3 x^3 - px^2 - qx + r = (x^2 - 2x + 3)(x + 4) + 2x - 1 x = 0. r = 12 - 1 = 11 x = 1 and x = -1 - p - q + 12 = 11 -p + q + 10 = 15 So -2p + 22 = 26 p = -2, q = 3
arrow
arrow
    創作者介紹
    創作者 rll33xb99t 的頭像
    rll33xb99t

    rll33xb99t的部落格

    rll33xb99t 發表在 痞客邦 留言(0) 人氣()