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標題:
Math question factorization
發問:
Factorize (x-1)(x-2)(x-3)(x-4)- 48
最佳解答:
其他解答:
Math question factorization
發問:
Factorize (x-1)(x-2)(x-3)(x-4)- 48
最佳解答:
此文章來自奇摩知識+如有不便請留言告知
(x-1)(x-2)(x-3)(x-4)- 48= (x - 1)(x - 4) (x - 2)(x - 3) - 48= (x^2 - 5x + 4) (x^2 - 5x + 6) - 48令 x^2 - 5x + 5 = y ,則(x^2 - 5x + 4) (x^2 - 5x + 6) - 48= (x^2 - 5x + 5 - 1) (x^2 - 5x + 5 + 1) - 48= (y - 1)(y + 1) - 48= (y^2 - 1) - 48= y^2 - 49= (y - 7)(y + 7)代回 x^2 - 5x + 5 入 y := ( x^2 - 5x + 5 - 7) ( x^2 - 5x + 5 + 7)= (x^2 - 5x - 2) (x^2 - 5x + 12)其他解答:
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