(a+b+c+d) (a+b) ? simplfy

(a+b+c+d) (a+b)
i tried to do this but think its wrong.
a(a+b) + b(a+b) + c(a+b) + c(a+b) + d(a+b)
a^2 +ab^2 +ab + bd + ac + bc + ad +bd

the answer is a^2 +b^2 +2ab + ac +ad .. how do you get this answer


16a^2 (b-4)^2 - 4a(b-4) ?
ok i know that the common factor is 4 so how do you do this...

a^2+ab+ba+b^2+ca+cb+da+db
=a^2+2ab+b^2+ac+bc+ad+db

First you multiply the a in the first parenthesis by the a in the second parenthesis. Then you multiply the a in the first parenthesis by the b in the second parenthesis. Then you multiply the b in the first parenthesis by the a in the second parenthesis. Then you multiply the b in the first parenthesis by the b in the second parenthesis and on and on....until you've multiplied each variable in the first parenthesis by each variable in the second. Finally, you combine like terms. For example, ab +ba means 2ab because 1 apple plus 1 apple is 2 apples.

Edit: Okay, the second problem:

16a^2(b-4)(b-4) - 4a(b-4)
16a^2(b^2-4b-4b+16) - 4ab + 16
16a^2(b^2-8b+16) - 4ab + 16
16a^2b^2 - 128a^2b + 256a^2 - 4ab + 16

If you want to factor it, the greatest common factor is 4a(b-4).

4a(b-4) [4a(b-4)-1]

a(a+b+c+d) + b(a+b+c+d)

a^2 +ab + ac + ad + ab + b^2 + bc + bd

a^2 + b^2 + 2ab + ac + ad + bc + bd


16a^2(b-4)^2 - 4a(b-4)
The common factor is: 4a(b-4)
4a(b-4)(4a(b-4) - 1)

A(a+b)+b(a+b)+c(a+b)+d(a+b). I'm pretty shire