Hi Jim.
Here's what I would do with your questions:
12ac+28bc3ad7bd
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 coefficients, 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.
7a^{2}  14ab + 7b^{2}
is
7(a^{2}  2ab + b^{2})
There are many approaches to factoring a quadratic trinomial like this.
If the first term's coefficient is 1 (it is), then you look for two
numbers that, when multiplied together, give the coefficent of the last
term and when added give the coefficient of the middle term.
So we are ignoring the 7 for now and looking at the coefficients 1, 2,
and 1 in a^{2}  2ab + b^{2}. 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.
