因式分解一條

(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....>

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