Recognizing quadratic factor methods

We have other videos on individual techniques for factoring quadratics, but what I would like to do in this video is get some practice figuring out which technique to use. So, I'm going to write a bunch of quadratics, and I encourage you to pause the video and try to see if you can factor that quadratic yourself before I work through it with you.

So the first quadratic is 6x² + 3x. So pause and see if you can factor this. This one might jump out at you that both of these terms here have a common factor. Both of them are divisible by three; six is divisible by three, and so is three, and both of them are divisible by x. So, you can factor out a 3x.

If you factor out a 3x, 6x² / 3x, you're going to have a 2x left over there, and then 3x / 3x, you're going to have a one. And that's about as much as we can actually factor it. You can verify that these two expressions are the same if you distribute the 3x. 3x * 2x is 6x², and 3x * 1 is 3x. And that's all we would do; we would be done. That's all you can do to really factor that.

As we'll see in this example, trying to factor out a common factor was all we had to do. But as we'll see in future examples, that's usually a good first step. We should always check whether the terms have a common factor, and if they do, it never hurts to factor that out.

So let's do another example. Let's say I have the quadratic 4x² - 4x - 48. Pause the video and try to factor that as much as you can.

All right, so the first thing you might have noticed is that there is a common factor among the terms. All of them are divisible by four. Four is clearly divisible by four, and 48 is also divisible by four, so let's factor a four out. This would be the same thing as 4 * (x² - x - 12). I just divided each of these by four and factored it out.

You can distribute the four and verify that these two expressions are the same. Now, are we done? Well, no, we can factor what we have inside the parentheses. We can factor this further. Now, how would we do that?

So over here, the key realization is: all right, I have a one as a coefficient on my second degree term. I've written it in standard form, where I have the second degree, and then there's a first-degree term, and then I have my constant term or my zero degree term. If I have a one coefficient right over here, then I say, "Okay, are there two numbers whose sum equals the coefficient on the first degree term on the x term?"

So are there two numbers that add up to -1? You didn't see a one here before, but it's implicit there; -1x is the same thing as -1x. So are there two numbers A + B that are equal to -1 and whose product is equal to -12? This is a technique that we do in other videos, and here the key is to realize that, hey, maybe we can use it here.

So A * B is equal to -12, and there's a couple of key realizations here. If I have two numbers and their product is going to be negative, that means one of them is going to be positive and one is going to be negative. If they had the same sign, then this would be positive.

So let's think about the factors of 12, especially think about them in terms of different sign combinations. You could think about 1 and 12, 12, and whether you're thinking about -1 and 12. -1 + 12 would be positive 11. If you went the other way around, if you went -12 and 1, that would be 1, but either way, that doesn't work.

2 and 6; -2 and 6 would be 4. -6 and 2 would be 4, so that doesn't work. 3 and 4; let's see, -3 and 4 would be positive 1, but 3 and -4 work out. You add these two together, you take the product, you clearly get -12, and then you add them together, you get -1.

So we can write inside the parentheses. Let me write: So this is going to be 4 times. We can factor that as two binomials. The first is going to be (x + 3) and then the next is going to be (x - 4), and we're done.

If any of this seems intimidating to you, I encourage you to watch the videos on the introduction to factoring polynomials. The key here is to recognize the method. So once again, at first, try to factor out any common factors. We did that in both examples. Then we saw here that, hey, if we have a leading one coefficient here on the second degree term and we have it written in standard form, well, let's think of two numbers that add up to this coefficient and whose product is equal to the constant term.

In this case, it was 3 and -4, and we were able to factor it this way. We proved that in other videos. Let's do another example; we can't get enough practice. And like always, pause the video and see if you can work through it yourself: 3x² + 30x + 75.

All right, I'm assuming you had a go at it. So you might immediately see that all of the terms are divisible by three, so let's factor three out. So it's going to be 3 * (x² + 10x + 25).

Oh, whoops, this should be an x here; my apologies. Pause the video again and see if you can do it now that I wrote the actual right thing there. All right, so as you imagine, it's nice to factor out a three first. So it's 3 * (x² + 10x + 25).

You might immediately say, "All right, well, let's use the technique we had here." We have a leading one coefficient; it's written in standard form. Can I think of two numbers that add up to 10? So A + B is equal to 10, and whose product A * B is equal to 25? This would work.

If you look at the factors of 25, you'd say, "All right, well, you know this thing here is positive; this is positive. So I'm dealing with two positive numbers." To get 25, it's either 1 and 25 or 5 and 5. 5 and 5 match; 5 + 5 is equal to 10 and 5 * 5 is equal to 25.

So just using the exact technique we just did, you would say, "Okay, this is 3 times," and the stuff in parentheses would be (x + 5) * (x + 5), or you could say 3 * (x + 5)².

Some of you might have immediately said, "Well, I don't have to do that exact technique; I could have immediately recognized this as a perfect square." Because I have a square constant right over here, and that's a good sign that, hey, maybe I should explore whether this is going to be a perfect square polynomial.

So if this is a perfect square, and if I were to take the square root of it, and this coefficient is twice that square root, well, that's a good sign that I'm probably dealing, or that I am dealing, with a perfect square. But either way, whether you recognize it as a perfect square or whether you use the technique that we used in the second problem, either one of those would get you to the appropriate answer.