WEBVTT mathematics/pre-calculus/selhorst-jones 00:00:00.000 --> 00:00:02.300 Hi--welcome back to Educator.com. 00:00:02.300 --> 00:00:06.200 Today, we are going to talk about the fundamental theorem of algebra. 00:00:06.200 --> 00:00:08.400 This lesson is not going to be like a standard lesson. 00:00:08.400 --> 00:00:12.900 So normally, we see some interesting ideas, and then we look at how numbers interact with those ideas. 00:00:12.900 --> 00:00:15.300 And then, we start working out examples. 00:00:15.300 --> 00:00:22.100 For this lesson, though, we are going to just see interesting ideas; there will be hardly any numbers, and there will be no examples whatsoever. 00:00:22.100 --> 00:00:27.000 Why? The ideas we are about to see are from advanced mathematics, and they are more 00:00:27.000 --> 00:00:32.100 about showing fundamental truths than just giving us a bunch of numbers and exercises to work with. 00:00:32.100 --> 00:00:41.300 Furthermore, since they don't really involve numbers, most teachers and textbooks can't and won't be able to test this stuff. 00:00:41.300 --> 00:00:43.400 And instead, they only briefly mention it. 00:00:43.400 --> 00:00:46.900 So, why are you still watching this lesson, if you are never going to be graded on it? 00:00:46.900 --> 00:00:50.000 It is because this stuff is cool--at least, I think so! 00:00:50.000 --> 00:00:55.300 You are about to see some amazing results from really high-level mathematics, and you are going to be able to understand it. 00:00:55.300 --> 00:01:00.500 You are actually going to be able to understand this stuff from what you have just seen in the previous lessons on polynomials. 00:01:00.500 --> 00:01:09.700 It is a really cool chance to have a culmination of understanding, and be able to get something pretty advanced from a level that is not super difficult to understand. 00:01:09.700 --> 00:01:13.200 If you have the time, humor me and watch this--you might find it interesting. 00:01:13.200 --> 00:01:16.800 I think it is pretty cool, and I hope you do, too; thanks. 00:01:16.800 --> 00:01:26.900 All right, previously we have seen that, while some polynomials seem to have no roots, they can actually have hidden roots, in a way, in the complex numbers. 00:01:26.900 --> 00:01:32.000 For example, if we look at x² + 1 = 0, there are no solutions to that. 00:01:32.000 --> 00:01:37.400 x² + 1 has no roots in the reals; if we look at its graph over here on the reals, 00:01:37.400 --> 00:01:42.100 we see that it never manages to cross the x-axis, so it has no roots. 00:01:42.100 --> 00:01:48.400 But if we switch into the complex numbers, we find roots--we find that i and -i work. 00:01:48.400 --> 00:01:56.700 For example, i² + 1 becomes -1 (i, the imaginary number, squared, becomes -1) + 1; so we get 0. 00:01:56.700 --> 00:02:00.900 Look, that is a solution; the same thing would happen if we took -i and squared that. 00:02:00.900 --> 00:02:06.300 So, this realization that there are roots, even though at first it doesn't seem like there are roots, 00:02:06.300 --> 00:02:11.900 might give us this sneaking suspicion that all polynomials have roots, if we allow for the complex numbers-- 00:02:11.900 --> 00:02:16.300 that there is some root out there that we might not be able to find using the real numbers, 00:02:16.300 --> 00:02:22.000 but that it is out there when we expand our search to the complex numbers--at least we would have this suspicion. 00:02:22.000 --> 00:02:28.000 It turns out that our suspicion was correct: this idea is expressed by the fundamental theorem of algebra. 00:02:28.000 --> 00:02:35.700 It is such an important, fundamental idea about how polynomials work that it gets this name, "fundamental theorem of algebra." 00:02:35.700 --> 00:02:43.200 And it says that every polynomial of degree n > 0 has at least one root in the complex numbers. 00:02:43.200 --> 00:02:48.500 Notice that it is "at least"; so it could have more; and it also says "complex numbers." 00:02:48.500 --> 00:02:55.300 So, every polynomial that isn't just a constant--that has a degree of 1 or greater--has at least one root. 00:02:55.300 --> 00:03:00.500 We are guaranteed a root in the complex numbers, if we expand our search to the complex numbers. 00:03:00.500 --> 00:03:06.600 This seems like a pretty simple idea, probably; but it is actually very difficult to prove. 00:03:06.600 --> 00:03:11.200 The proof requires advanced mathematics, and so we won't be able to see it here in this course. 00:03:11.200 --> 00:03:18.400 But trust that it is true; and it can be proved if you get to a high enough level and have enough background. 00:03:18.400 --> 00:03:21.700 So, at this point, we might say, "Well, where are all the imaginary roots?" 00:03:21.700 --> 00:03:27.100 Considering the fundamental theorem of algebra, why haven't we seen more imaginary roots? 00:03:27.100 --> 00:03:35.000 If we are guaranteed what that says, why is it that x² - 4 = 0...why do we get only the roots of 2 and -2? 00:03:35.000 --> 00:03:42.100 Those don't use imaginary numbers; it seems that we are just fine without imaginary numbers; we can find roots just fine. 00:03:42.100 --> 00:03:45.600 So, what is up with the fundamental theorem of algebra? 00:03:45.600 --> 00:03:51.600 Well, remember: it didn't say imaginary numbers; it said complex numbers. 00:03:51.600 --> 00:03:58.700 It is the complex numbers; now, by definition, a complex number is of the form a + bi. 00:03:58.700 --> 00:04:05.700 So notice, if we felt like it, we could say b = 0; and if b = 0, then the complex numbers are all real, 00:04:05.700 --> 00:04:10.100 because we will be left with just the a portion; and a is just a real number. 00:04:10.100 --> 00:04:14.400 So, we see from this that the reals are a subset of the complex. 00:04:14.400 --> 00:04:17.100 The reals are contained in the complex numbers. 00:04:17.100 --> 00:04:23.900 The complex numbers have all of the real numbers inside of them, because we can just cut off that imaginary component by setting b to 0. 00:04:23.900 --> 00:04:32.100 So, the above solutions of x = 2 and -2...they are complex numbers, because they fit in the a portion. 00:04:32.100 --> 00:04:37.100 They fit in the real portion of a complex number, so it is still a complex number. 00:04:37.100 --> 00:04:42.900 We can show all real numbers as complex numbers; so that is what the fundamental theorem of algebra is telling us. 00:04:42.900 --> 00:04:47.300 It is telling us that the roots are going to be complex numbers. 00:04:47.300 --> 00:04:55.000 But since it is a complex number, it could just be a real number, just like all integers are contained in the real numbers. 00:04:55.000 --> 00:05:01.600 You can be a real number, but still be an integer; you can be a complex number, but still also be a real number. 00:05:01.600 --> 00:05:05.700 All right, with the fundamental theorem of algebra learned, we can now go on to the n roots theorem, 00:05:05.700 --> 00:05:11.100 which is also sometimes called the end zeroes theorem or the linear factorization theorem. 00:05:11.100 --> 00:05:16.700 A polynomial of degree n has exactly n roots in the complex numbers. 00:05:16.700 --> 00:05:22.200 And these roots are not necessarily distinct; so they are not necessarily going to be different from one another. 00:05:22.200 --> 00:05:24.500 And we will talk more about all of these things. 00:05:24.500 --> 00:05:33.700 Another way to say this is that, for any polynomial p(x) of degree n > 0, there exist n complex numbers, 00:05:33.700 --> 00:05:39.600 z₁, z₂, up until z<font size="-6">n</font> (n complex numbers, and each one is going to be some z). 00:05:39.600 --> 00:05:44.800 And they are also, once again, not necessarily distinct: z₁ is not necessarily going to be different than z₂. 00:05:44.800 --> 00:05:47.700 We are just guaranteed that there will be a total of n numbers. 00:05:47.700 --> 00:05:57.400 And a constant number a, such that the polynomial is equal to a(x - z₁)(x - z₂)(x - z<font size="-6">n</font>)-- 00:05:57.400 --> 00:06:02.500 that is a way of saying that we can break it into a bunch of linear factors. 00:06:02.500 --> 00:06:11.600 Since x minus just some number is a linear factor, we can break any polynomial into n linear factors. 00:06:11.600 --> 00:06:15.600 Impressively, we will actually be able to prove this theorem from what we currently know. 00:06:15.600 --> 00:06:18.700 We are going to see a sketch of the proof after we discuss what the theorem means. 00:06:18.700 --> 00:06:24.500 And it is pretty cool that we can actually see how such an advanced piece of mathematics 00:06:24.500 --> 00:06:31.000 can be proven from fairly simple things, assuming we can assume the fundamental theorem of algebra. 00:06:31.000 --> 00:06:39.700 The first comment is multiplicity: we need to notice that a polynomial of degree n has n roots, but that they are not necessarily going to be distinct. 00:06:39.700 --> 00:06:46.500 For example, if we consider f(x) = x⁵ - 5x⁴ + 10x³ - 10x² + 5x - 1, 00:06:46.500 --> 00:06:51.300 and we factor it, we will eventually figure out that that is (x - 1)⁵. 00:06:51.300 --> 00:06:59.300 Now, that means that, when we look at it on the graph, it has just one root--just one x-intercept. 00:06:59.300 --> 00:07:07.800 But in a way, it is showing up 5 times; so f only has one distinct root at x = 1; that root repeats 5 times. 00:07:07.800 --> 00:07:17.600 We have it to the fifth; and so we can break f into five linear factors, (x - 1)(x - 1)(x - 1)(x - 1)(x - 1). 00:07:17.600 --> 00:07:26.000 They are just all the same linear factor; this idea is called multiplicity--that we can have the same root show up multiple times. 00:07:26.000 --> 00:07:28.600 We can have the same factor show up multiple times. 00:07:28.600 --> 00:07:33.300 In this case, we would say that f has a root at x = 1 with multiplicity 5, 00:07:33.300 --> 00:07:42.700 because the same linear factor, containing one root, shows up 5 times when we break it into its factors: multiplicity. 00:07:42.700 --> 00:07:46.900 Complex numbers are necessary: it shows up in the statement of the theorem, but it is really important 00:07:46.900 --> 00:07:53.500 to be aware that this n roots theorem thing is only true if we are allowing complex numbers. 00:07:53.500 --> 00:08:00.700 For example, f(x) = x³ + 1; we can break it into (x + 1)(x² - x + 1). 00:08:00.700 --> 00:08:11.400 So, (x + 1) has a solution in the reals, x = -1; but x² - x + 1...that thing is an irreducible quadratic. 00:08:11.400 --> 00:08:20.700 It doesn't bring any real solutions; so if we restrict ourselves to the reals, f has only one root at x = -1, if we are stuck just in the reals. 00:08:20.700 --> 00:08:28.800 But this theorem tells us that there must be an additional two roots, because we started at a degree of 3 in our polynomial. 00:08:28.800 --> 00:08:32.800 So, since our polynomial had a degree of 3, we are guaranteed a total of 3 roots. 00:08:32.800 --> 00:08:41.200 And if we found that there is one root already, then we know that we can subtract one off; so there must be two roots left that we haven't found yet. 00:08:41.200 --> 00:08:44.000 So, those two roots must be complex. 00:08:44.000 --> 00:08:51.200 In other words, the theorem guarantees n roots (allowing for multiplicity, as we just talked about), but only if we are using the complex numbers. 00:08:51.200 --> 00:08:57.100 If we restrict ourselves to the reals, we won't necessarily be able to find n roots. 00:08:57.100 --> 00:09:03.300 Complex coefficients are also allowed; it wasn't explicitly stated, but the n roots theorem is still true, 00:09:03.300 --> 00:09:07.000 even if the coefficients of the polynomial are complex numbers. 00:09:07.000 --> 00:09:18.300 For example, if we have this second-degree polynomial, this quadratic polynomial, (2 + i)x² + -15ix + (-7 + 49i), 00:09:18.300 --> 00:09:23.400 well, 2 + i, -15i, and -7 + 49i--those are all just complex numbers. 00:09:23.400 --> 00:09:30.300 And it turns out that the n roots theorem is true here; we can break it into two linear factors, 00:09:30.300 --> 00:09:33.800 because remember, we started with a degree of 2 on our polynomial. 00:09:33.800 --> 00:09:39.200 So, we can break it into two linear factors, along with some constants at the front. 00:09:39.200 --> 00:09:48.700 2 + i times x minus 3 - i times x - 7i--the degree of the polynomial is 2, and it has 2 roots; 00:09:48.700 --> 00:09:53.600 or we can look at that, equivalent-ly, as being factored into two linear factors. 00:09:53.600 --> 00:10:00.500 So, the theorem holds, whether the coefficients of the polynomial are real or complex numbers--pretty cool. 00:10:00.500 --> 00:10:03.100 Existence theorem--this is a really important idea. 00:10:03.100 --> 00:10:08.400 The fundamental theorem of algebra and the end roots theorem are existence theorems. 00:10:08.400 --> 00:10:13.900 They guarantee the existence of roots; we know that there have to be n roots. 00:10:13.900 --> 00:10:20.100 However, they don't tell us how to find the roots; all they do is tell us that they are out there somewhere. 00:10:20.100 --> 00:10:25.700 They guarantee existence, but that is all; they don't tell us how to actually find what the roots are. 00:10:25.700 --> 00:10:32.000 For example, we know that x⁴ + 1 = 0; since we have a degree of 4, it must have 4 roots. 00:10:32.000 --> 00:10:35.100 But that doesn't actually put us any closer to figuring out what they are. 00:10:35.100 --> 00:10:40.400 All we know is that they are out there somewhere; but we don't know what they are going to be from these theorems. 00:10:40.400 --> 00:10:46.700 The theorem guarantees the existence, but it doesn't tell us how to actually get to them; that is an important point. 00:10:46.700 --> 00:10:55.500 All right, we can now actually see a proof sketch of how the n roots theorem can be proven. 00:10:55.500 --> 00:10:59.800 The fundamental theorem of algebra, like I said, requires some very advanced mathematics. 00:10:59.800 --> 00:11:05.600 We won't be able to see it here; but we can actually understand how the n roots theorem is being built out of the fundamental theorem of algebra. 00:11:05.600 --> 00:11:10.100 So, starting with the fundamental theorem of algebra, which we will not be able to prove, 00:11:10.100 --> 00:11:16.500 and the division algorithm, which we addressed earlier when we talked about polynomial division: the proof is within our graphs. 00:11:16.500 --> 00:11:22.000 Start by considering some polynomial, p(x), with degree n > 0. 00:11:22.000 --> 00:11:26.600 Now, the fundamental theorem of algebra, remember, said that there is at least one root. 00:11:26.600 --> 00:11:30.700 So, by the fundamental theorem of algebra, we are guaranteed that there is at least one root. 00:11:30.700 --> 00:11:37.500 So, let's give that one root that we are absolutely guaranteed a name; and we will call it z₁, because it is our first root. 00:11:37.500 --> 00:11:45.200 And we know that p(z₁) = 0; it is a root, because if we plug z₁ in, we get 0 out of the polynomial. 00:11:45.200 --> 00:11:51.600 So, there is going to be at least one root to our polynomial, p; and we will call it z₁. 00:11:51.600 --> 00:11:58.400 Now, from earlier work, we know that, if we have a root, that is the same thing as saying (if there is some root 00:11:58.400 --> 00:12:07.500 that causes our polynomial to give a 0) that there is a factor, (x - z₁) somewhere inside of our polynomial. 00:12:07.500 --> 00:12:14.900 So, by the division algorithm, we know that there exists some polynomial q₁ (we will just call it q, for quotient; 00:12:14.900 --> 00:12:19.200 and then we will need to give it a number, because there are going to be a bunch of these quotients coming up soon); 00:12:19.200 --> 00:12:23.900 so we will call it q₁(x), because it is the first time we have divided our initial polynomial. 00:12:23.900 --> 00:12:32.700 And it is going to be degree n - 1, because we are going to pull out x - z₁ so that we will get (x - z₁)(q₁(x)). 00:12:32.700 --> 00:12:37.100 We can pull out our factor, because we were guaranteed this factor, 00:12:37.100 --> 00:12:41.900 because we know there is a root of z₁, so there must be a factor of (x - z₁). 00:12:41.900 --> 00:12:46.400 So, we can pull that factor out; and we will be left with that factor, times some quotient. 00:12:46.400 --> 00:12:53.000 So, we will have (x - z₁), the factor that we know must be inside, because of the fundamental theorem of algebra; 00:12:53.000 --> 00:12:57.300 and then, we pull it out through the division algorithm, and we must be left with some q₁(x). 00:12:57.300 --> 00:13:01.100 There can't be a remainder, because we know that that factor must be cleanly in there; 00:13:01.100 --> 00:13:05.100 it must be able to divide out evenly; otherwise it couldn't be a factor. 00:13:05.100 --> 00:13:13.000 So, we have that our initial polynomial is equal to (x - z₁) times some other quotient polynomial, q₁(x). 00:13:13.000 --> 00:13:18.400 And since we pulled out a degree of 1 from a polynomial that initially started at n, we started at n; 00:13:18.400 --> 00:13:24.700 and then we pull out 1; and we know that we are going to have a degree of n - 1 in our quotient polynomial. 00:13:24.700 --> 00:13:26.700 So, we can keep going with this procedure. 00:13:26.700 --> 00:13:31.600 We did it once, and we pulled out a root, and then we broke our polynomial up. 00:13:31.600 --> 00:13:37.000 We knew that there was a root, and then we were able to factor out that linear factor, because of that root. 00:13:37.000 --> 00:13:46.400 So, if we have n - 1 > 0 (remember, n - 1 is the degree of our first quotient), we can use the fundamental theorem again. 00:13:46.400 --> 00:13:50.900 The fundamental theorem says that as long as your degree is greater than 0, there is a root in there. 00:13:50.900 --> 00:13:55.300 So, if that is the case, we know that there must be some root, z₂, where, 00:13:55.300 --> 00:14:01.300 when we plug it into our quotient (remember, since n - 1 > 0, we can use the fundamental theorem of algebra 00:14:01.300 --> 00:14:08.100 to guarantee that there is a root z₂, so we plug it into our quotient), q₁(z₂) is equal to 0. 00:14:08.100 --> 00:14:16.200 Cool; but we also know that z₂ has to be a root of p(x), because the way that we got q₁(x) was by dividing it out. 00:14:16.200 --> 00:14:22.400 So, we have (x - z₁)(q(x)) is the same thing as p(x). 00:14:22.400 --> 00:14:28.700 So, if we plug in z₂ into p(x), then we are going to have z₂ plugged in for our x. 00:14:28.700 --> 00:14:35.800 And so, we will get z₂ - z₁; and that is some number; but we are also going to have q₁ plugging into z₂ here. 00:14:35.800 --> 00:14:40.800 So, we knew that q₁ of z₂ equals 0; so that means that the whole thing has to come to 0, 00:14:40.800 --> 00:14:44.600 because we take out a 0 here, and 0 will knock out whatever else we have. 00:14:44.600 --> 00:14:53.100 So, the whole thing comes to 0, which means that p(z₂) has to equal 0--pretty cool. 00:14:53.100 --> 00:14:55.300 Now, we can just use the division algorithm again. 00:14:55.300 --> 00:14:58.700 We know that q₁ of z₂ equals 0; 00:14:58.700 --> 00:15:05.800 so it must be the case that (x - z₂) is a factor of q₁(x), since it is able to be pulled out. 00:15:05.800 --> 00:15:10.000 So, we can pull out a factor, (x - z₂), from q₁, which gives us, 00:15:10.000 --> 00:15:14.900 once again, another quotient--our second quotient, so we will call it q₂(x). 00:15:14.900 --> 00:15:18.700 And q₂ will be a polynomial of degree n - 2, because remember: 00:15:18.700 --> 00:15:24.100 q₁ started at n - 1; and then we are going to subtract 1, because we are pulling out a degree of 1. 00:15:24.100 --> 00:15:30.900 So, we subtract 1; so we are going to get n - 2 as the degree of q₂(x). 00:15:30.900 --> 00:15:39.000 Now, this means that we can also express p(x) as (x - z₁)(x - z₂)(q₂(x)). 00:15:39.000 --> 00:15:43.300 Since q₁(x) = (x - z₂)(q₂(x)), and we know that p(x) 00:15:43.300 --> 00:15:51.600 equals (x - z₁)(q₁(x)), we take this right here, and we swap it in for q₁(x) right here. 00:15:51.600 --> 00:16:00.000 And we are going to get this thing right here, (x - z₁)(x - z₂) times our quotient 2, our second quotient of x. 00:16:00.000 --> 00:16:07.000 And we just keep going with this method, until eventually, with the quotient polynomial, we get down to some q<font size="-6">n</font> 00:16:07.000 --> 00:16:11.800 that is going to just be a constant; we will be stuck at degree 0. 00:16:11.800 --> 00:16:17.200 At this point, we will have a total of n roots, because we have pulled out one root each time we take a step down. 00:16:17.200 --> 00:16:23.700 And so, we will have taken n steps down; we will have managed to pull out z₁, z₂, all the way up until z<font size="-6">n</font>. 00:16:23.700 --> 00:16:28.500 Effectively, what we are doing is whittling down the degree of the polynomial we started with. 00:16:28.500 --> 00:16:34.100 So, we are whittling down the degree: for every step we lower the degree by, we manage to get another root. 00:16:34.100 --> 00:16:40.800 If we start at degree n, that gives us n steps to go down; so we have n steps to pull from--we manage to get n roots out of it. 00:16:40.800 --> 00:16:49.000 For example, let's consider if p(x) is degree 5; we start at degree 5 with p(x), our initial polynomial. 00:16:49.000 --> 00:16:56.300 By the fundamental theorem of algebra, we are guaranteed that it must have one root; z₁, we will call it. 00:16:56.300 --> 00:17:02.500 Then, we use the division algorithm to break it into q₁(x). 00:17:02.500 --> 00:17:11.000 Now, q₁(x) is going to have degree 4; and so, by the fundamental theorem of algebra, we are guaranteed another root, z₂. 00:17:11.000 --> 00:17:18.300 And since z₂ is the root of q₁, and q₁ is contained inside of our initial polynomial p, 00:17:18.300 --> 00:17:22.400 z₂ must also be a root for p(x); great. 00:17:22.400 --> 00:17:25.900 Now, we use the division algorithm again; we can break this down, 00:17:25.900 --> 00:17:31.000 and we are able to get to saying that q₂(x) is able to come out of q₁(x). 00:17:31.000 --> 00:17:37.900 And it is going to have a degree of 3; by the fundamental theorem of algebra, we are guaranteed that it must have a root, z₃. 00:17:37.900 --> 00:17:42.100 And since q₂ is contained inside of q₁, which is contained inside of p, 00:17:42.100 --> 00:17:47.600 we know that z₃ is, once again, a root for our initial polynomial that we started with. 00:17:47.600 --> 00:17:51.500 We then break down q₂(x); we get to q₃(x). 00:17:51.500 --> 00:17:56.200 q₃(x) will now have a degree of 2, because we are stepping down once each time. 00:17:56.200 --> 00:18:03.600 Since it has a degree greater than 0, we are guaranteed a root there, which is once again going to be a root of our initial polynomial. 00:18:03.600 --> 00:18:14.000 We break it down once again; we get to q₄(x); it has a degree of 1, so it is guaranteed to have a root of z₅. 00:18:14.000 --> 00:18:18.200 We plug that in; and so, that is going to once again be yet another root. 00:18:18.200 --> 00:18:26.000 And finally, we use the division algorithm one last time, and we get down to q₅(x), which now has a degree of 0. 00:18:26.000 --> 00:18:29.500 And since we have a degree of 0, we are not able to get a root out of it. 00:18:29.500 --> 00:18:35.400 So, we have used up all of our roots; we have managed to pull out 5 roots, because we started at degree 5. 00:18:35.400 --> 00:18:40.700 And each step, we are able to pull out one root as we subtract by 1. 00:18:40.700 --> 00:18:43.200 And so, we are finally left with some constant polynomial. 00:18:43.200 --> 00:18:50.400 And that is where our a, the a that multiplies the whole thing--that is going to be our fifth quotient polynomial at the very end. 00:18:50.400 --> 00:18:59.900 It is going to be a(x - z₁)(x - z₂)(x - z₃)(x - z₄)(x - z₅)--pretty cool. 00:18:59.900 --> 00:19:04.200 This is really, really deep stuff from advanced mathematics that we can actually be pretty good at understanding, 00:19:04.200 --> 00:19:07.800 just from what we have managed to learn about polynomials so far--pretty impressive. 00:19:07.800 --> 00:19:09.000 All right, we will see you at Educator.com later--goodbye!