Explain this factoring problem

First of all, I am only in 10th grade honors, so please don't make fun of me! I'm a straight A student, but I wasn't here for the review of the homework, so I will greatly appreciate if anyone can explain this to me...
15. 3x(x+2)+5(x+2)
I know that you just combine the coefficients to each product to complete the factoring, like this:
(3x+5)(x+2)

This is the problem that's more advanced that I don't understand why the solution is this way.
The problem: 16a^2(b-4)^2 - 4a(b-4)

Why is the answer 4a(b-4) (4ab-16a-1) But why?
I know the common factor of 16a^2-4a is 4a, but I don't know how to go on from there. I feel so upset at myself :( I will give the 10 points to anyone who explains this to me. Thanks.

You're right...the problem 16a^2(b-4)^2 - 4a(b-4) isn't an easy one to factor, and the answer does seem kinda weird. I'll do it in slo-mo for you.

Let's list all the factors of each term:
1) 16a^2(b-4)^2: It has factors of 2, 2, 2, 2, a, a, (b-4), and (b-4).
2) 4a(b-4): It has factors of 2, 2 ,a, and (b-4)

So, we need to find the GREATEST common factor to fully factor this problem. Since EVERY factor on the bottom appears at least as often as it does on the top, the greatest common factor is actually 2*2*a*(b-4), or 4a(b-4) itself!

So, let's factor 4a(b-4) out of that top expression by rearranging the factors:
(2 * 2 * a * (b-4)) * (2 * 2 * a * (b-4))
4a(b-4) * 4a(b-4)

So, the first term has 4a(b-4) of the 4a(b-4) quantities, while the second term is just ONE of the 4a(b-4) quantity.

Thus, we can write that your original quantity is equal to:
4a(b-4) * [4a(b-4)] - 1 * [4a(b-4)]

See how both terms have 4a(b-4) in them? Now we factor!
(4a(b-4) - 1) * (4a(b-4))

Now we expand that thing on the left (which isn't exactly required):
(4ab - 16a - 1) * 4a(b-4)