Everyone, we've already seen one method of factoring polynomials, which is by factoring out a greatest common factor. For example, if I had x3 minus 2x2, once I wrote out the factor tree, I noticed that x2 was common in both terms. So, I could take that and pull it to the outside, and that's my factorization. I'm going to show you in this video that sometimes, for some polynomials, that's not going to work. You're not going to find one common factor that works for all of the terms. In this situation, I can't pull out an x2. I can't pull out a number like 2 or 4 because whatever I try to do, it won't work for all terms. I'm going to show you in this video that we're going to need a new method called grouping to solve these types of problems. And they're very similar to greatest common factor, but there are a couple of differences. Let me show you how it works here.
The first thing I want to do is talk about when you use these two different methods and, basically, it just comes down to whether you can identify a common factor that works for all the terms, like x2 over here, or whether you can't. Usually, what's going to happen in these types of grouping problems is there's going to be four terms. So if you see a four-term expression that you can't identify one greatest common factor for, it's usually a good indicator that it's going to be a grouping problem. Also, what you'll notice is that a lot of these problems, you'll see numbers that are multiples of each other, like 24 and 8.
Let me just show you how it works. Basically, the whole idea here is I'm going to take this four-term expression, and I'm going to try breaking this thing up into two different groups. That's why we call it grouping. And the hope is that I can factor out a greatest common factor from each one of the groups. Let me show you a step-by-step process. The first thing you want to do is make sure your polynomial is written in standard form just in case it isn't already. So in this situation, I have x3, 2x2, 4x, and 8. So this is actually already in standard form, and I don't have to do anything to it. The next thing you want to do is the grouping. Basically, you're going to group terms into pairs, and almost always, you're just going to do the first two and the last two terms. It's pretty much 99.9% of the time going to work.
So, basically, what I like to do is I like to drop parentheses around the first two and the last two terms. And now what I'm going to try to do is I'm going to try to factor out a one-term greatest common factor from each one of the groups. Basically, what I'm going to try to do is now that I've split these things off into groups, I'm going to turn them into problems where I just try to pull out a greatest common factor out of each one of them independently. So let's try to do this. In fact, I already know what the factorization for this expression is. If I try to do this, it's going to be x2 times (x minus 2) times x2. So I notice that the x2 is common. I draw the little parentheses, and I can pull the x2 out to the outside. This just becomes x2, and then I have (x minus 2).
What happens with the second group over here? Well, for the second group, what I notice is that the 4 is a multiple of eights. So when I write out the factor tree, this is just 4 times (x minus 2). So now what's common in this group? The common item in this group is the 4. So I can take the 4 and move it to the outside of the expression, and what I end up with is I end up with a 4 here, and then I end up with (x minus 2). You're going to factor out a one-term greatest common factor from each group. If you'll notice what's happened here, we've ended up getting the same exact thing in both of the expressions. We've gotten the (x minus 2) term.
So if you think about it, now what happens is if you wrap this whole thing in one parenthesis, it's actually like the (x minus 2) is now actually the common thing for both of them. So that leads us to the last step, which is now you're going to factor out this two-term greatest common factor out of both of the groups. So it's the same thing I did with the x2 and stuff like that. I take this (x minus 2), pull it to the outside over here, and then just write everything that's inside of the parentheses that remains. So this actually becomes (x minus 2), and then you have (x2 plus 4). And this is the complete factorization of this polynomial. If you go ahead and foil this out, what you'll see is that you'll end up getting back to your original expression. Alright?
So again, just to summarize, if you ever see a polynomial and it's four terms, and you notice that you can't find a common factor that works for all of them, but you notice that some of the numbers are multiples of each other, try splitting up into two groups, and what you're going to see a lot of the time is that you'll very coincidentally end up with the same factor that you can pull out of both groups. This is a very specific type of problem, but it's good to know. Anyway, that's it for this one, folks. Thanks for watching.