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數條MATHS >V< 快入黎幫解吧XD
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(5) By the remainder theorem: -a/8 - 1/4 - b/2 + 1 = 0 a + 4b = 6 ... (1) a/27 - 1/9 + b/3 + 1 = 0 a + 9b = -24 ... (2) Solving, we have a = 30 and b = -6 (6) Just view from the graph: x = -0.25, 0.65 or 1.6 (7) Equating the graphs together: x2/2 - 3 = x + 1 x2 - 2x - 8 = 0 So the points of intersection give the ans which are x = -2 or 4. (8) 2x2 - x + 6 = 0 x2 = (x - 6)/2 Ans = C (1) Remainder = 1093 (By remainder thm) Quotient = 5x3 + 17x2 + 68x + 275 (2) Sub x = -2/3 into the dividend: 3(-2/3)4 +2(-2/3)3 - 3(-2/3)2 + 2(-2/3) - 7 = -29/3 (3) Sub k = -1: -k + 6 - 33 = -35 -k = -8 k = 8 (4) Sub x = 2/5 into the polynomial: 5(2/5)3 - 7(2/5)2 - 8(2/5) + 4 = 0 So 5x - 2 is a factor, by long division, the quotient is: x2 - x - 2 So after factorization, it will be (5x - 2)(x - 2)(x + 1)
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數條MATHS >V< 快入黎幫解吧XD
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[IMG]http://i293.photobucket.com/albums/mm67/zaza520/0006-27.jpg[/IMG] [IMG]http://i293.photobucket.com/albums/mm67/zaza520/0005-33.jpg[/IMG] 唔該>v<最佳解答:
(5) By the remainder theorem: -a/8 - 1/4 - b/2 + 1 = 0 a + 4b = 6 ... (1) a/27 - 1/9 + b/3 + 1 = 0 a + 9b = -24 ... (2) Solving, we have a = 30 and b = -6 (6) Just view from the graph: x = -0.25, 0.65 or 1.6 (7) Equating the graphs together: x2/2 - 3 = x + 1 x2 - 2x - 8 = 0 So the points of intersection give the ans which are x = -2 or 4. (8) 2x2 - x + 6 = 0 x2 = (x - 6)/2 Ans = C (1) Remainder = 1093 (By remainder thm) Quotient = 5x3 + 17x2 + 68x + 275 (2) Sub x = -2/3 into the dividend: 3(-2/3)4 +2(-2/3)3 - 3(-2/3)2 + 2(-2/3) - 7 = -29/3 (3) Sub k = -1: -k + 6 - 33 = -35 -k = -8 k = 8 (4) Sub x = 2/5 into the polynomial: 5(2/5)3 - 7(2/5)2 - 8(2/5) + 4 = 0 So 5x - 2 is a factor, by long division, the quotient is: x2 - x - 2 So after factorization, it will be (5x - 2)(x - 2)(x + 1)
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