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Sal factors p(x)=2x^5+x^4-2x-1 as (2x+1)(x^4-1) using grouping. Then he further factors (x^4-1) as (x^2+1)(x+1)(x-1) using the special product form of difference of squares. Created by Sal Khan.
Video transcript
Plot the real zeroes of the given polynomial on the graph below. And they give us p of x is equal to 2 x to the 5th plus x to the 4th minus 2x minus 1. And when they say plot it, they give us this little widget here. Where if we click at any point on this, we get our point. And we get as many points as we would like. And we can drag these points around. Or if we don't want these points anymore, we can just dump them in this little trash bin at the bottom right. So let's think about what the zeroes of this polynomial actually are. And to do that I'll take my scratch pad out. And this is a little daunting at first. This is a 5th degree polynomial here. Factoring 5th degree polynomials is really something of an art. You're really going to have to sit and look for patterns. If they're actually expecting you to find the zeroes here without the help of a computer, without the help of a calculator, then there must be some type of pattern that you can pick out here. So let me just rewrite p of x. So p of x is equal to 2 x to the 5th plus x to the 4th minus 2x minus 1. And one way that's typically seen when you're trying to factor this type of polynomial is to try to essentially undo the distributive property a few times. And if you want to relate it to techniques for factoring quadratics, it's essentially factoring by grouping. So for example, you see a 2x minus 1, or something that looks like a 2x minus 1 right over here. And over here you have a 2x to the 5th plus x to the 4th. So you have a 2x of a higher degree term plus a 1 x of a one degree lower. So there seems to be some type of a pattern. 2 times x of a higher degree-- this is a first degree term-- minus 1 times-- you could do this as x to the 0-- of a lower degree term. And so let's think about it a little bit. What happens if we essentially try to group these two terms, and we group these two terms right over here. And we try to factor out anything to essentially clean it up a little bit, to see if we can make sense of it. Well these two terms, their greatest common factor is x to the 4th. We could write this as x to the 4th times 2x plus 1. And this should get us excited because this looks pretty close to that, especially if we were to factor out a negative 1 here. So we could factor out a negative 1. And then this is going to be 2x plus 1. And that's exciting because now we can factor out a 2 x plus 1 from each of these terms. So you have a 2x plus 1. We're gonna factor both of these out, to get 2x plus 1-- which we just factored out. And if you factored it out of this term right over here, you're left with x to the 4th. And if you factor it out this term, you're left with just the minus 1. Minus 1. And now this is exciting because 2x plus 1, this is pretty easy to figure out when does this thing equal 0. And we'll do that in a little bit. And this is pretty easy to factor. This is a difference of squares. This right over here can be rewritten as x squared plus 1 times x squared minus 1. And of course we still have this 2x plus 1 out front. 2x plus 1. And once again we have another difference of squares. We have another difference of squares right over here. That's the same thing as x plus 1 times x minus 1. And let me just write all the other parts of this expression. x squared plus 1. And you have 2x plus 1. 2x plus 1. And I think I factored p of x about as much as could be reasonably expected. So p of x is equal to all of this business. And remember, the whole reason why I wanted to factor it is I wanted to figure out when does this thing equal 0? So if p of x can be expressed as the product of a bunch of these expressions, it's going to be 0 whenever at least one of these expressions is equal to 0. If any of these is equal to 0 then that's just going to make this whole expression equal to 0. So when does 2x plus 1 equal 0? So 2x plus 1 is equal to zero. Well you could probably do this in your head, or we could do it systematically as well. Subtract 1 from both sides, you get 2x equals negative 1. Divide both sides by 2, you get x is equal to negative 1/2. So when x equals negative 1/2-- or one way to think about it, p of negative 1/2 is 0. So p of negative 1/2 is 0. So this right over here is a point on the graph, and it is one of the real zeroes. Now we could try to solve this. x squared plus 1 equals 0, and I'll just write it down just to show you. If we try to isolate the x term on the left-- subtract 1 from both sides-- you get x squared is equal to negative 1. Now if we were to start thinking about imaginary numbers, we could think about what x could be. But they want us to find the real zeroes. The real zeroes. So there's no real number where that number squared is equal to negative 1. So we're not going to get any real zeroes by setting this thing equal to 0. There's no real number x where x squared plus 1 is going to be equal to 0. Now let's think about when x plus 1 could be equal to 0. We'll subtract 1 from both sides. You get x is equal to negative 1. So p of negative 1 is going to be 0. So that's another one of our zeroes right there. And then finally let's think about when x minus 1 is equal to 0. We'll add one to both sides. X is equal to 1. So we have another real 0 right over there. And so we could plot them. So it's negative 1, negative 1/2, and 1. So it's negative 1, negative 1/2 and 1. And we can check our answer, and we got it right. Now one thing that might be bugging you, is like hey Sal, you just happened to group this in exactly the right way. What if I try to group it in a different way? What if I tried to-- Well actually let's try to do that. That could be interesting. Just to show you that this isn't voodoo. And actually there's several ways to get there. There's several ways to get there. So what if instead of writing it like this, where you're writing it kind of in the highest degree term and the next highest degree and so on and so forth, you were to write it like this: p of x is equal to 2 x to the 5th minus 2x plus x to the 4th minus 1. Well actually, even this way you could do a fairly interesting grouping. If you grouped these two together, you see that they have the common factor 2 x. You factor 2x out, you get 2x times x to the 4th minus 1. And I think you see what's going on. And then this can be rewritten as plus 1 times x to the 4th minus 1. Minus 1. And now you can factor out an x to the 4th minus 1. And you're left with-- I'll do this in a neutral color-- x to the fourth minus 1 times 2 x plus 1, which is much easier to factor now. A difference of squares. Exactly what we did the last time around. So there are several ways that you could have reasonably grouped this, and reasonably undone the distributive property. But I do admit it is something of an art. You really just have to play around in see, well let's group the first two terms, let's see if there's a common factor here. Let's group the second two terms, let's see if there's a common factor here. Hey, once we factor out those common factors, it looks like both of these two terms have this common expression as a factor. And then you could start to factor that out.