math of f2 plz help quick

1) 3a^2-8bx+4ab-6ax
= 3a^2 + 4ab - 6ax - 8bx
= a(3a+4b) - 2x(3a+4b)
= (a-2x)(3a+4b)
2)2-8a+8a^2
= 2(1-4a+4a^2)
= 2[1^2 - 2(1)(2a) + (2a)^2]
= 2(1-2a)^2
3)25-(y-4)^2
= 5^2 - (y-4)^2
= (5+y-4)(5-y+4)
= (1+y)(9-y)

3999x4001
= (4000-1)(4000+1)
= 4000^2 - 1^2
= 16000000 - 1
= 15999999

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1) 3a2-8bx+4ab-6ax
= (3a2+4ab)-(8bx+6ax) <-----Grouping terms
= a(3a+4b)-2(4b+3a)
=(a-2)(3a+4b)
2)2-8a+8a2
=2(1-4a+4a2) <---- Taking common factors
=2[1-2(1)(2a)2+(2a)2] <-----Perfect Square
=2(1-2a)2

3)25-(y-4)2 <----Don't expand the perfect square
=52-(y-4)2
=[5+(y+4)][5-(y+4)]
=(5+y+4)(5-y-4)
=(y+9)(y-1)

3999x4001
=(4000-1)(4000+1) <---Different of 2 squares
=4000 2-12
=16000000-1
=15999999


2007-11-06 21:27:34 補充：
srry, 寫錯=(5 y 4)(5-y-4)=(y 9)(-y 1)=(y 9)(1-y)

1) 3a(a-2x)+4b(a-2x)
= (a-2x)(3a+4b)

2) 2(1-4a+4a^2)
= 2( (2a)^2 - 2*2a*1 + 1^2)
= 2(2a-1)^2

3) 25-y^2+8y-16
= -(y^2-8y-9)
= -(y+1)(y-9)

last question: 3999x4001
= (4000-1)(4000+1)
= 4000^2-1^2
= 15999999