Recognizing quadratic factor methods part 2

In the last video, we looked at three different examples. It really is a bit of a review of some of our factoring techniques and also to appreciate when we might want to apply them.

We saw in the first example that it was just a process of recognizing a common factor. Once we factored that out, we were done. In the second example, there was a common factor of four, but then after that, we used our most basic factoring technique, or one of our more basic factoring techniques. We said, "Okay, what two numbers add up to this middle to the first degree coefficient, and then their product is the constant?" We were able to factor the expression.

In the third example, we once again started off by factoring out a common value, which in this case was 3. We could have done it the same way we did the second one, or we could have immediately recognized that this is a perfect square polynomial. But either way, we were able to factor the expression.

Let's keep going to see if we can tackle some other types of polynomials that might require some other techniques. So, let's say we have the expression 7x squared minus 63. Like always, pause this video and see if you can factor that.

All right, well, I've intentionally designed all of these so that you can check whether there's a common factor across the terms, and here they're all divisible by seven. So if you factor out a 7, you're going to get 7 times x squared minus 9. Now, you might immediately recognize this as a difference of squares; you have x squared minus this right over here is 3 squared minus 3 squared.

If the term "difference of squares" or how to factor them is completely foreign to you, I encourage you to watch the videos on factoring difference of squares or do a search on Khan Academy for difference of squares. But you will see when you have a difference of squares like this, it can be factored as 7. This is just a 7 out front, and then this part right over here can be written as x plus 3 times x minus 3.

It is x squared minus 3 squared. Now, one thing to appreciate: this really isn't a different technique than the one that we saw in the previous video. If we just focused on x squared minus 9, you could view this as x squared plus 0x minus 9. In that case, you'd say, "Okay, what two numbers get me a product of negative nine and add up to zero?" Well, if I need to get a product of negative nine, that means that they must be different signs—a positive and a negative. Otherwise, they're the same sign; you'd get a positive here.

So they're different signs, and nine only has three factors: 1, 3, and 9. If you make one of these threes negative, that does add up to zero, so you say, "Okay, well, my two numbers are going to be negative three and three." Therefore, it's going to be x minus three times x plus three. Once again, I'm just focusing on what was inside the parentheses right over here; you would put that 7 out front.

If we were doing this exact same expression, but if you recognize it as a difference of squares, it might happen for you a little bit faster. Let's do one more example. So, let's say that I have 2x squared plus 7x plus 3. In general, when my coefficient on the second-degree term here is not a one, I try to see if there is a common factor here, but 7 isn't divisible by 2, and neither is 3.

So I can't use the techniques that I used in the last few videos or even over here, where I say, "Oh, there's a common factor," and get a leading coefficient of 1. If you see a situation like that, it's a clue that factoring by grouping might apply here. Factoring by grouping, on some level, everything that we've just done now, you can view as special cases of factoring by grouping.

But factoring by grouping, you say, "Okay, can I think of two numbers that add up to this coefficient?" So a plus b is equal to 7, and a times b—instead of just saying it needs to equal to 3, it actually needs to be equal to 3 times this—3 times the leading coefficient, the coefficient on the x squared term. So it needs to be equal to 3 times 2.

If you think about it, we've always been doing that, but in the other examples we gave, the leading coefficient was a one. Whenever you took the constant term and multiplied it by one, you're saying, "Oh, well, a times b needs to be equal to that constant term." But if we want to talk about it more generally, it should be that a times b should be the constant term times the leading coefficient.

In the introduction to factoring by grouping, we explain why that works. You should never just accept this as some magic formula; it makes sense for a very good mathematical reason. But once you accept that, then it's useful to be able to apply this technique.

So can we think of two numbers that add up to 7 and whose product is equal to 6? They're going to have to be the same sign since this is a positive value, and they’re going to be positive because they're adding up to a positive value. They’re both going to be positive.

Well, let's see: 1 and 6 seem to work. 1 plus 6 is 7; 1 times 6 is 6. So in factoring by grouping, we rewrite our expression where we break this up between the a and the b. I can rewrite this as 2x squared plus 6x plus 1x plus 3.

As you can see, the 7x has been broken up into the 6x and 1x. That whole exercise I just did is to see how we can break up this first-degree term right over here. But then what's useful about this is now we can essentially apply the reverse of the distributive property twice.

For these first two terms, you see a common factor: 2x squared and 6x. They're both divisible by 2x, so let's factor out a 2x out of those first two terms. If you do that, 2x squared divided by 2x, you're just going to have an x left over, and 6x divided by 2x, you're just going to have a 3. And then you have plus.

Over here, this is a special situation where x plus 3—there is no common factor between x and 3, so we'll just rewrite that as x plus 3. But when I put a parenthesis on it, which is equivalent to writing it without a parenthesis, you might see something else: well, I can undistribute or I can factor out an x plus 3.

So what happens if I do that? I'm going to get an x plus 3, and then I'm going to have left over in this term. If I factored out an x plus 3, I'm just going to have a 2x left over: 2x, and then this term, if I factor out an x plus 3, well, I'm just going to have a 1 left over—plus 1. Let me do that same color. Having trouble switching colors today, 2x plus 1.

And we are done! So as I said, these are all various techniques. On some level, factoring by grouping is sometimes viewed as the hardest one, but I'll say hard in parentheses because everything we did is just a variation, really a special case of factoring by grouping.

As you can see, it's all about finding two numbers that add up to that middle coefficient, that on the first degree term when it's written in standard form, and their product is equal to the product of the constant and the leading coefficient. If you do that and break it up, it works out quite nicely, where you keep factoring out terms.

This one, in some levels, was a little bit more subtle because you have to recognize that this x plus 3 has a 1 coefficient out there implicitly. 1 times x plus 3 is the same thing as x plus 3. Then see that you can factor out an x plus 3 out of both of these terms, and then once you do that, you're going to be left with a 2x plus 1.

But all of these, if you really feel comfortable with this arsenal of techniques, you're going to do pretty well. And frankly, if none of these work, well, you might already be familiar with the quadratic formula or you might soon learn it. That's when the quadratic formula might be effective if none of these techniques work.