factorization

6x^2-23xy-13y^2
=(2x+y)(3x-13y)

let a+2=x and b-3=y
6(a+2)^2-23(a+2)(b-3)-13(b-3)^2
= [2(a+b)+(b-3)] [3(a+2)-13(b-3)]
= (2a+2b+b-3) (3a+6-13b+9)
= (2a+3b-3) (3a-13b+15)


2010-10-29 01:23:36 補充：
SORRY... 下part搞錯左d野....

Sub a+2=x and b-3=y
6(a+2)^2-23(a+2)(b-3)-13(b-3)^2
= [2(a+2)+(b-3)] [3(a+2)-13(b-3)]
= (2a+4+b-3) (3a+6-13b+39)
= (2a+b+1) (3a-13b+45)

6x2 - 23xy - 13y2 = (3x - 13y)(2x + y)

Sub x = a + 2 and y = b - 3, we have:

6x2 - 23xy - 13y2 = (3x - 13y)(2x + y)

6(a + 2)2 - 23(a + 2)(b - 3) - 13(b - 3)2 = [3(a + 2) - 13(b - 3)][2(a + 2) + (b - 3)]

= (3a + 6 - 13b + 39)(2a + 4 + b - 3)

= (3a - 13b + 45)(2a + b + 1)