Form 3 Maths 幫幫手 － Q 1

11. Factorize
(a) 3x^2 + 8x - 80
=(3x + 20)(x - 4)
(b) x^3 - 64
=(x - 4)(x^2 + 4x + 16)
(c) x^3 - 6x^2 - 16x + 96
=x^3 + 160 - 64 - 6x^2 - 16x (160 - 64 = 96)
=x^3 - 64 - (6x^2 + 16x - 160)
=(x - 4)(x^2 + 4x + 16) - 2(3x^2 + 8x - 80) (for the result of (b))
=(x - 4)(x^2 + 4x + 16) - 2(3x + 20)(x - 4) (for the result of (a))
=(x - 4)(x^2 + 4x + 16 - 6x - 40)
= (x - 4)(x^2 - 2x - 24)
= (x - 4)(x + 4)(x - 6)
or

x^3 - 6x^2 - 16x + 96
=x^2(x - 6) - 16(x - 6)
=(x - 6)(x^2 - 16)
=(x + 4)(x - 4)(x - 6) (for the identity (a + b)(a - b) = a^2 - b^2)

2008-02-09 15:18:54 補充：
上面全部錯左,條題目寫明"Hence"並無otherwise,即是話"由此"而利用(a)和(b)的結果,由於並沒有"otherwise"所以並不可以用其他方法,故樓主只得我是正確...

2008-02-09 15:20:09 補充：
係HKCEE唔跟題目做, 隨時無分囉呀....

11C) x^3-6x^2-16x+96
(x-4)(x^2-2x-24)
(x-4)(x+4)(x-6)
你put x=4的時候...條equation會等於0(即係表示佢係x-4係一個factor
之後你再factorize就會變成gum...你明唔明白?

11c) x^3-6x^2-16x+96
=x^2(x-6)-16(x-6)
=(x^2-16)(x-6)
=(x+4)(x-4)(x-6)

我又想不到點hence(a)同(b)..........

x^3-6x^2-16x+96
=x(x^2-16)-6(x^2-16)
=(x^2-16)(x-6)
=(x^2-4^2)(x-6)
=(x+4)(x-4)(x-6),,