中二factorization~~~~(20點)

1. (a-5)^2 - (b+3)^2
= (a-5+b+3)[(a-5)-(b-3)]
= (a+b-2)(a-b-2)
2. (a-2b)^2 - (2a+b)^2
= (a-2b+2a+b)[(a-2b)-(2a+b)]
= (3a-b)(-a-b)
= -(3a-b)(a+b)
= (b-3a)(a+b)
3. (c+d)^2 - 9(2d-c)^2
= (c+d)^2 - [3(2d-c)]^2
= [(c+d)+3(2d-c)][(c+d)-3(2d-c)]
= (c+d+6d-3c)(c+d-6d+3c)
= (7d-2c)(4c-5d)
4. n^2+3n+3p-p^2
= n^2-p^2+3n+3p
= (n+p)(n-p) + 3(n+p)
= (n+p)(n-p+3)
5. 8q^2+2r+4q-2r^2
= 8q^2-2r^2+2r+4q
=2(4q^2-r^2)+2(r+2q)
=2(2q+r)(2q-r) + 2(r+2q)
=2(r+2q)(2q-r+1)

2009-10-06 20:43:43 補充：
1.
= (a-5+b+3)[(a-5)-(b+3)] < not [(a-5)-(b-3)] plz correct it

= (a+b-2)(a-b-8)

2009-10-06 20:52:04 補充：
The guy 002, please respect my ans.
You can make it as a reference, but not copying it
OK?

2009-10-07 18:35:28 補充：
The guy 002 plz dont copy my ans.
The evidence comes from no.1 You copied the wrong step
(a-5+b+3)[(a-5)-(b-3)]

1. (a-5)^2 - (b+3)^2

= (a-5+b+3)[(a-5)-(b-3)]

= (a+b-2)(a-b-2)/(a+b-2)^2



2. (a-2b)^2 - (2a+b)^2

= (a-2b+2a+b)[(a-2b)-(2a+b)]

= (3a-b)(-a-b)

= (b-3a)(a+b)



3. (c+d)^2 - 9(2d-c)^2

= (c+d)^2 - [3(2d-c)]^2

= [(c+d)+3(2d-c)][(c+d)-3(2d-c)]

= (c+d+6d-3c)(c+d-6d+3c)

= (7d-2c)(4c-5d)




4. n^2+3n+3p-p^2

= n^2-p^2+3n+3p

= (n+p)(n-p) + 3(n+p)

= (n+p)(n-p+3)(p)




5. 8q^2+2r+4q-2r^2

= 8q^2-2r^2+2r+4q

=2(4q^2-r^2)+2(r+2q)

=2(2q+r)(2q-r) + 2(r+2q)

=2(r+2q)(2q-r+1)