Hi Jim.

Here's what I would do with your questions:

12ac+28bc-3ad-7bd

First, I look for "common factors", which means something that goes into
every term. There's nothing of that here.

Then I look for good groupings. By this I mean perhaps this breaks down
into two factors each with two terms. As I look at the co-efficients, I
see that 7 and 3 are prime numbers, so I look to the others (12 and 28)
to see if they are multiples of the other two. Yes. Both are 4 times
those primes. So I start here:

4(3ac + 7bc) - 3ad - 7bd

Now I make the signs the same by factoring out a minus one from the last
two terms:
4(3ac + 7bc) - 1(3ad + 7bd)

The first parentheses contains two terms that both have c, so I can
factor that out:
4c(3a + 7b) - 1(3ad + 7bd)

And the d comes out in the back:
4c(3a + 7b) - 1d(3a + 7b)

Now I am relieved because the two parentheses sets contain the same
thing, so I can factor it out:
(3a + 7b)(4c - 1d)

The 1 isn't needed here, so the final form is
(3a + 7b)(4c - d)

For the second question, I see a common factor.
7a² - 14ab + 7b²
is
7(a² - 2ab + b²)

There are many approaches to factoring a quadratic trinomial like this.
If the first term's co-efficient is 1 (it is), then you look for two
numbers that, when multiplied together, give the co-efficent of the last
term and when added give the co-efficient of the middle term.

So we are ignoring the 7 for now and looking at the co-efficients 1, -2,
and 1 in a² - 2ab + b². So I want to think of two numbers that
multiply to +1 (the last term) and add to -2 (the middle term). The
numbers I think of are -1 and -1. So the trinomial breaks down into (a
- 1b) and another (a - 1b). This are the same, so we are really just
squaring it. The final factored form is
7(a - b)².

Hope this helps. Look for more worked examples by searching for
"factoring" in our Quick Search.

Cheers,
Stephen La Rocque.