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。
F.4 Maths...急!!!
2011-12-11 8:52 pm
回答 (2)
2011-12-11 9:14 pm
✔ 最佳答案
1 3x^2 + ax - 6 = (3x - b)(x - 2) + 2Sub. 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
2011-12-13 10:43 pm
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
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
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