標題:
因式分解一條
發問:
因式分解: (x-1)(x-2)(x-3)(x-4)-48 計左好耐,唔係太明 圖片參考：http://l.yimg.com/f/i/tw/ugc/rte/smiley_30.gif 更新: #2 真係犀利 有d自卑,我自己搞唔掂 你點做架? 更新 2: 你係唔係做得多因式分解,先對呢種題目易上手?
最佳解答:
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(x-1)(x-2)(x-3)(x-4)-48 =[(x-1)(x-4)][(x-3)(x-2)]-48 =(x-5x+4)(x-5x+6)-48 Let u=x-5x (x-5x+4)(x-5x+6)-48 =(u+4)(u+6)-48 =u+10u+24-48 =u+10u-24 =(u+12)(u-2) =(x-5x+12)(x-5x-2) Hence,(x-1)(x-2)(x-3)(x-4)-48=(x-5x+12)(x-5x-2) 2009-04-03 14:53:15 補充： Let u=x^2-5x (x^2-5x+4)(x^2-5x+6)-48 =(u+4)(u+6)-48 =u^2+10u+24-48 =u^2+10u-24 =(u+12)(u-2) =(x^2-5x+12)(x^2-5x-2) Hence,(x-1)(x-2)(x-3)(x-4)-48=(x^2-5x+12)(x^2-5x-2) 2009-04-03 19:04:41 補充： 其實呢D都係要展開先得...但係唔會展開哂,因為對分解唔會有幫助,反而係留意到[(x-1)(x-4)]同[(x-3)(x-2)]2舊野都會整到x^2-5x呢個式出黎,於是設u=x^2-5x令整條式變左做2次方程再做分解就方便得多了!^^ 2009-04-04 13:47:54 補充： 做呢D其實都係要TRY囉....>
其他解答:
x^4-10x^3+35x^2-50x-24 2009-04-03 14:39:51 補充： first step= (x^2-2x+2)*(x-3)*(x-4)-48 second step=(x^3-3x^2-2x^2+6x-6+2x-6)*(x-4)-48 final answer =x^4-10x^3+35x^2-50x-24
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