標題:
[F. 2 Maths] Factorization of Simple Polynomials (10 points)
發問:
a) Expand (x-1)(x-4) and (x-2)(x-3). b) Factorize (x-1)(x-2)(x-3)(x-4)-48.
最佳解答:
a) Expand (x-1)(x-4) and (x-2)(x-3). (x-1)(x-4) = x2 - 4x - x + 4 = x2 - 5x + 4 (x-2)(x-3) = x2 - 3x - 2x + 6 = x2 - 5x + 6 ============================= b) Factorize (x-1)(x-2)(x-3)(x-4)-48. (x-1)(x-2)(x-3)(x-4)-48 = [(x-1)(x-4)][(x-2)(x-3)] - 48 = (x2 - 5x + 4)(x2 - 5x + 6) - 48 【From (a)】 = [(x2 - 5x + 5) - 1][(x2 - 5x + 5) + 1] - 48 = (x2 - 5x + 5)2 - 12 - 48 【From identity a2-b2 = (a-b)(a+b)】 = (x2 - 5x + 5)2 - 49 = (x2 - 5x + 5)2 - 72 = [(x2 - 5x + 5) - 7][(x2 - 5x + 5) + 7] 【From identity a2-b2 = (a-b)(a+b)】 = (x2 - 5x - 2)(x2 - 5x + 12) 2007-01-28 13:44:16 補充： 小小補充：(b) part 可以再分解為有 surd (√) 的因式。由於你題目為 simple polynomial，我覺得這答案已足夠。下面的分解只為供你參巧：= (x2 - 5x - 2)(x2 - 5x + 12)= [x - (5+√33)/2][x - (5-√33)/2](x2 - 5x + 12)
a) (x-1)(x-4) (x-2)(x-3) x -1 x -2 x -4 x -3 ------------------- ---------- -x-4x = - 5x = - 5x
[F. 2 Maths] Factorization of Simple Polynomials (10 points)
發問:
a) Expand (x-1)(x-4) and (x-2)(x-3). b) Factorize (x-1)(x-2)(x-3)(x-4)-48.
最佳解答:
a) Expand (x-1)(x-4) and (x-2)(x-3). (x-1)(x-4) = x2 - 4x - x + 4 = x2 - 5x + 4 (x-2)(x-3) = x2 - 3x - 2x + 6 = x2 - 5x + 6 ============================= b) Factorize (x-1)(x-2)(x-3)(x-4)-48. (x-1)(x-2)(x-3)(x-4)-48 = [(x-1)(x-4)][(x-2)(x-3)] - 48 = (x2 - 5x + 4)(x2 - 5x + 6) - 48 【From (a)】 = [(x2 - 5x + 5) - 1][(x2 - 5x + 5) + 1] - 48 = (x2 - 5x + 5)2 - 12 - 48 【From identity a2-b2 = (a-b)(a+b)】 = (x2 - 5x + 5)2 - 49 = (x2 - 5x + 5)2 - 72 = [(x2 - 5x + 5) - 7][(x2 - 5x + 5) + 7] 【From identity a2-b2 = (a-b)(a+b)】 = (x2 - 5x - 2)(x2 - 5x + 12) 2007-01-28 13:44:16 補充： 小小補充：(b) part 可以再分解為有 surd (√) 的因式。由於你題目為 simple polynomial，我覺得這答案已足夠。下面的分解只為供你參巧：= (x2 - 5x - 2)(x2 - 5x + 12)= [x - (5+√33)/2][x - (5-√33)/2](x2 - 5x + 12)
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其他解答:a) (x-1)(x-4) (x-2)(x-3) x -1 x -2 x -4 x -3 ------------------- ---------- -x-4x = - 5x = - 5x
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