因式分解點做?

首先見到前面2個係正既後面2個係負既
寫番靚仔佢先
變成 x^3 - x + x^2 -1
之後見到前面一pair同後面一pair係差個x就一樣
前面就fact個x出黎
變成 x(x^2 - 1) + (x^2 - 1)
最後再 fact 相同既野出黎
變成 (x^2 - 1)(x + 1)
最後由於 x^2 -1 = (x + 1)(x - 1)
所以最後答案就係 (x + 1)^2 (x - 1)

我有1元3次計數機program,要既搵我,希望幫到你, :)

2011-08-12 14:53:12 補充：
計都可以教埋你
首先抽左factor先
試正負1, 試出1代入條式度係=0
即係( x - 1) 係factor,
咁就用長除 除個 (x - 1)
之後就= (x - 1)( 8x^2 - 2x - 1)
之後見到後面果舊仲有得fact
就fact埋佢
最後變成 = (x - 1)(2x - 1)(4x +1)

再有唔明可以再問,希望幫到你 :)

x^3+x^2-x-1
=x^3+x^2-(x+1)
=x^2(x+1)-(x+1)
=(x^2-1)(x+1)
=(x^2-1^2)(x+1)
=(x+1)(x-1)(x+1)
=(x-1)(x+1)^2
8x^3-10x^2+x+1
=(x-1)(8x^2-2x-1)
=(x-1)(2x-1)(4x+1)
hope it can help you!

x^3 + x^2 - x - 1
=x^2(x+1)-(x+1)
=(x+1)(x^2-1)
=(x+1)(x-1)(x+1)
=(x-1) (x+1)^2

x^3+x^2-x-1
=x^2(x+1)-(x+1)
=(x^2-1)(x+1)
=(x+1)(x-1)(x+1)
=(x+1)^2 (x-1)

2011-08-12 17:57:37 補充：
8x^3-10x^2+x+1
=(2x-1)(4x^2-3x-1)
=(2x-1)(4x+1)(x-1)