因式分解...........i will thank a lot

交叉相乘：

１．x^2 + 9x + 8

　　ｘ　＼　／　８
　　ｘ　／　＼　１
－－－－－－－－－
　８ｘ　　＋　　ｘ　　　＝　　９ｘ

所以第一題係　(x+8)(x+1)

２．x^2 + 2x - 15

　　ｘ　＼　／　　５
　　ｘ　／　＼　－３
－－－－－－－－－－
　５ｘ　　－　　３ｘ　　＝　　２ｘ

所以第二題係　(x+5)(x-3)

３．3x^2 - 4x - 4

　　３ｘ　＼　／　　２
　　　ｘ　／　＼　－２
－－－－－－－－－－－
　　２ｘ　　－　　６ｘ　＝　　－４ｘ

所以第三題係　(3x+2)(x-2)

1)x^2+9x+8
=x^2+x+8x+8
=x(x+1)+8(x+1)
=(x+8)(x+1)

2)x^2+2x-15
=x^2+2x+1-16
=(x+1)^2-4^2
=(x+1-4)(x+1+4)
=(x+5)(x-3)

3)3(x^2)-4x-4
=3x^2-6x+2x-4
=3x(x-2)+2(x-2)
=(3x+2)(x-2)

If (x^2+9x+8) = (x+h)(x+k) for some integers h, k,
then ab=8, which implies {h,k} = {1,8} or {2,4} (Think: 1+8=9)
Try that (x+1)(x+8)=x^2+9x+8. Done.

Or you may also use the formula for finding the roots of a quadratic equation in the form
ax^2+bx+c=0, for any integers (in general, complex numbers) a,b & c.
The roots are [ -b ± sqrt(b^2-4ac)] / 2a.

For example in question (2), [ -b + sqrt(b^2-4ac)] / 2a = 3 and [ -b - sqrt(b^2-4ac)] / 2a = -5
so the factorization should be ( x-3 ) ( x-(-5) ) = (x-3)(x+5).

The third one is left to you as an exercise.

交叉相乘：

１．x^2 + 9x + 8

　　ｘ　＼　／　８
　　ｘ　／　＼　１
－－－－－－－－－
　８ｘ　　＋　　ｘ　　　＝　　９ｘ

所以第一題係　(x+8)(x+1)

２．x^2 + 2x - 15

　　ｘ　＼　／　　５
　　ｘ　／　＼　－３
－－－－－－－－－－
　５ｘ　　－　　３ｘ　　＝　　２ｘ

所以第二題係　(x+5)(x-3)

３．3x^2 - 4x - 4

　　３ｘ　＼　／　　２
　　　ｘ　／　＼　－２
－－－－－－－－－－－
　　２ｘ　　－　　６ｘ　＝　　－４ｘ

所以第三題係　(3x+2)(x-2)

１．x^2 + 9x + 8

=(x+8)(x+1)

２．x^2 + 2x - 15

=(x+5)(x-3)

３．3x^2 - 4x - 4

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