3. a) Yes. 6 | 2a(3b + 3) since 2a(3b + 3) = 6 · k for some integer k.
| 2a(3b + 3) | = | 6ab + 6a |
| = | 6(ab + a) |
b) No. If a = 1 and b = 1, then 2a(4b + 1) = 10 which is not a multiple of 4. Notice that if you had used the same approach in part b) as you did in part a), you would end up with 2a(4b + 1) = 8ab + 2a which does not have a common factor of 4.