Question 17497
There is are several sections in my Basic Algebra: One Step at a Time that deals with facotring, and one of them specifically teaches this topic.  You can see my entire lesson on this free on my website by clicking on my tutor name "rapaljer" anywhere in algebra.com.  This goes to my website, then look for "Basic Algebra", click on that, then click on "Samples from Basic Algebra: One Step at a Time", then look for "Chapter 2" and in this chapter, it is section 2.06 ADvanced Factoring of Trinomials.  You might want to look at some of the earlier sections of factoring first, but this question specifically comes from what I call "Advanced Factoring of Trinomials."


For your particular question  {{{3x^2 + 7x + 2}}}, you need to find a product of two binomials where the first times first is 3x^2.  That would be 
{3x____)(x_____)


Next, the LAST times LAST has to come out to 2 in such a way that the middle term when you FOIL it out comes to 7x.  In this case, the last times last has to be 2*1 or 1*2.  In order to get it to add up to 7x, put the 2 as the LAST and the 1 in the middle:

(3x+1)(x+2) 


In this way, when you do the middle terms you have actually 6x + 1x, which is 7x.


{{{3n^2 +2n -1 }}}
(3n_____)(n______)


Last times last has to be -1, so try -1 for the inner sign and +1 for the LAST sign:
(3n-1)(n+1)


In this way, the middle terms will be 3n -1n = 2n.


R^2 at SCC