Identifying quadratic patterns | Polynomial factorization | Algebra 2 | Khan Academy

We're told that we want to factor the following expression, and they ask us which pattern can we use to factor the expression. U and V are either constant integers or single variable expressions. So we'll do this one together, and then we'll have a few more examples where I'll encourage you to pause the video.

So when they're talking about patterns, they're really saying, "Hey, can we say that some of these can generally form a pattern that matches what we have here?" And then we can use that pattern to factor it into one of these forms. What do I mean by that? Well, let's just imagine something like (u + v) squared. We've squared binomials in the past. This is going to be equal to (u^2 + 2 \cdot (uv) + v^2).

Now, when you look at this polynomial right over here, it actually has this form. If you look at it carefully, how could it have this form? Well, if we view (u^2) as (9x^8), then that means that (u) is, and let me write it as a capital (U), (U) is equal to (3x^4), because notice if you square this, you're going to get (9x^8). So this right over here is (u^2), and if we said that (v^2) is equal to (y^2), so if this is capital (V^2), then that means that (v) is equal to (y).

Then this would have to be (2 \cdot (uv)). Well, see, if I multiply (u) times (v), I get (3x^4y), and then (2) times that is indeed (6x^4y). So this right over here is (2uv). So notice this polynomial, this higher-degree polynomial can be expressed in this pattern, which means it can be factored this way.

So when they say, "Which pattern can we use to factor this expression?" Well, I would use the pattern for (u + v) squared, so I would go with that choice right over there. Let's do a few more examples.

So here once again, we're told the same thing. We're given a different expression, and they're asking us what pattern can we use to factor the expression. So I have these three terms here. It looks like maybe I could use— I can see a perfect square here. Let's see if that works. If this is (u^2), if this is (u^2), then that means that (u) is going to be equal to (2x^3) and if this is (v^2), then that means that (v) is equal to (5).

Now, is this equal to (2 \cdot (uv))? Let's see. (2 \cdot (uv)) would be equal to— well, you're not going to have any (y) in it, so this is not going to be (2uv). So this actually is not fitting the perfect square pattern, so we could rule this out.

Both of these are perfect squares of some form; one just has a, I guess you say, adding (v), the other one is subtracting (v). This right over here, if I were to multiply this out, this is going to be equal to— this is a difference of squares, and we've seen this before. This is (u^2 - v^2), so you wouldn't have a three-term polynomial like that. So we could rule that one out.

So I would pick that we can't use any of the patterns. Let's do yet another example, and I encourage you to pause the video and see if you can work this one out on your own. So the same idea, they want to factor the following expression, and this one essentially has two terms. We have a term here, and we have a term here. They both look like they are the square of something, and we have a difference of squares.

So this is making me feel pretty good about this pattern, but let's see if that works out. Remember, (u + v) times (u - v) is equal to (u^2 - v^2). So if this is equal to (u^2), then that means that capital (U) is equal to (6x^6). That works, and if this is equal to (v^2), well that means that (v) is equal to (y + 3).

So this is fitting this pattern right over here, and they're just asking us to say what pattern can we use to factor the expression. They're not asking us to actually factor it, so we'll just pick this choice. But once you identify the pattern, it's actually pretty straightforward to factor it because if you say this is just going to factor into (u + v) times (u - v), well (u + v) is going to be (6x^6 + (y + 3)) times (u - v) which is (6x^6 - (y + 3)).

Or we could write it as (6x^6 - y - 3) or we could distribute the negative sign, but either way, this might make it a little bit clearer what we just did. We used the pattern to factor this higher-degree polynomial, which is essentially just a difference of squares.

Let's do one last example. So here once again, we want to factor an expression: which pattern can we use? Pause the video.

All right, so we have two terms here, so it looks like it might be a difference of squares. If we set (u) equal to (7), then this would be (u^2). But then what can we square to get (10x^3)? Remember, we want to have integer exponents here.

The square root of (10x^3), if I were to take the square root of (10x^3), it'd be something a little bit involved like the square root of (10) times (x) times the square root of (x^3). And I'm not going to get an integer exponent here, so it doesn't look like I can express this as (v^2). So I would go with that we can't use any of the patterns, and we're done.