F.2因式分解

(x)(x)-(y)(y)+4x+8y - 12點計到(x+y - 2)(x - y+6)
(x)(x)-(y)(y)+4x+8y – 12
= x^2 + 4x + 4 - y^2 + 8y - 4 - 12 (rearrange the question)
= [(x)^2 + 2(x)(2) + (2)^2] - [(y)^2 - 2(y)(4) + (4)^2]
= (x + 2)^2 - (y - 4)^2
= [(x + 2) + (y - 4)] [(x + 2) - (y - 4)]
= [x + 2 + y - 4] [x + 2 - y + 4]
= (x + y -2) (x - y + 6) (Ans.)

Hope can help.


2007-11-26 10:15:27 補充：
方法將數式組成兩組二次方: (x /- n)^2, (y /- n)^2然後用 (a^2 - b^2) = (a b)(a - b) 因式分解

(x)(x)-(y)(y)+4x+8y - 12
= x^2-y^2+4x+8y-12
= x^2+4x+4-y^2+8y-16 <-- +4 and -16 is equal to -12
= x^2+2(x)(2)+2^2 - y^2+2(y)(4)-4^2
= (x+2)^2 - (y-4)^2
= [(x+2)+(y-4)][(x+2)-(y-4)]
= (x+2+y-4)(x+2-y+4)
= (x+y-2)(x-y+6)

要知道D公式先做倒'

(x+y)^2 = x^2 + 2xy + y^2
(x- y)^2 = X^2 - 2xy +y^2
(a+b)(a-b) = a^2-b^2