 Hi, I'm Zor. Welcome to Unisort Education. I'd like to dedicate this lecture to higher-order equations. This is the second lecture. It's a continuation of the previous one and just a couple of words maybe before that. Why do we really need to solve higher-order equations? Well, usually we don't. It's just a challenge. It's an intellectual exercise. Actually, ability to solve higher-order equations goes back to history. Obviously linear equations and quadratic equations were quite simply to solve and the formulas were known for a very long time. So it was really challenging to come up with analogous formulas to solve equations of the third degree, fourth degree, etc. Well, there were certain successes and certain not that successful attempts to do this. There is a famous formula of Cardano to solve equation of the third degree. So the solutions are expressed as a very large formula where coefficients of this cubical equation, cubical, because it's third degree, participate. Now, I don't know if any successful attempt to come up with a formula for equation of the fourth degree was invented. But starting from the fifth degree there was a famous theorem proven by French mathematician Galois, if I'm not mistaken, that it's not really possible, the general equation of the polynomial equation of the fifth degree to express the solutions through this equation as a formula, however complex, where coefficients are involved. That's a very interesting though negative result. So in general, we obviously don't want to use any formulas. So the question is how can we solve equations of the higher order? Well, there are a few techniques. Sometimes they might work. Sometimes they don't work. And I'm going to talk about couple of techniques, couple of approaches to solve higher order equations. Again, it's purely intellectual exercise. So if you can find some interesting peculiarity about this or that particular equation to be able to solve it, well, that basically is the purpose of the whole thing of solving these equations. So if you see some problem in the textbook or anywhere else where the higher-order polynomial equation participates, most likely they are asking you to find what's specific, not general, about this particular equation, which you can use to find the solutions. And that's exactly what I'm going to talk about. What kind of specific cases exist where you can solve the equation of the higher order. And to recognize these specifics, looking at any particular equation, is actually the challenge, is actually the problem is all about. All right, so this is a generalized polynomial equation of the nth degree. So it's polynomial because this is the polynomial, nth degree because the coefficient at x to the nth power, a0, is not equal to 0. So it's a true nth degree polynomial, and it's equal to 0. It means that a0, a1, etc., an are known coefficients, x is an unknown variable, and we have to solve it. And obviously, we are not talking about this general case because there is no general speaking, especially with n greater than 4, there is no formula which would help you. So you have to find a approach, some specific, what is particular about this particular equation, which can help you to solve the equation, to solve it. All right, so the first thing which comes to mind is the following. Let's assume that this particular polynomial, which I can just for grammatical a polynomial, because the coefficients are a0, a1, etc., can be represented as a product of two other polynomials. So a0, x to the n plus, etc., plus an, maybe, maybe, I'm not talking about every polynomial in general, I'm talking about any particular case where this is possible to the, let's say, lth degree multiplied by another polynomial of nth degree. It's supposed to be l, and this is supposed to be m in this case. From b0 to bl, from x to the power of l to the l to the x to the 0's power, same thing here. So if this is true, now why am I talking about this? Well, think about it this way. For instance, we have this representation. A polynomial is a product of b polynomial and c polynomial. Well, number first, the very important thing which you can observe is that these two polynomials are of lower degree than n. Why? Here's why. If you will open the parenthesis, you will multiply this by this plus this plus this plus this plus this, multiply by this, this, this, etc., etc. What is the element in the obtained sum of different little products of nth degree What is the element with the higher power? Well, that's obviously the higher power from here and higher power, higher power, higher statue, not higher, higher power, from here and from here. So the whole thing is equal to b0, c0, x to the power of l plus m, and only this coefficient, b0 times z0, z0, is with x to the power of l plus m. Ever anything else, any other product of elements of this times elements of this would have a lower degree. This times this would be l plus m minus 1. This plus this would be also l plus m minus 1. So everything else will be lower power. And the only thing which is the highest power is this one, which necessitates that l plus m is equal to n. This is supposed to be equal to this, and b0 times z0 should be equal to a0. Well, this alone actually tells us that both l and m are less than n. We are not talking about zero degrees, we are not talking about constant. It's a true representation of polynomial as a product of two real polynomials of the power greater than zero, so it's not constants. Okay, so number one, the observation which we made is that the power of any one of those is smaller than m. Well, smaller power means easier to solve, right? Well, at least that's what people think about. The lower the degree of the polynomial, the easier it is to solve to find the solutions of polynomial equation. All right, so now let's assume that after we have represented our polynomial, original polynomial in this way, let's assume that after that we can solve an equation which involves only one of those polynomials. Let's assume we can solve this equation, and there is a solution to this equation. So there is some number, whatever the number is, let's say x is equal to p, which if substituted instead of x will turn the whole polynomial into zero. What does it mean? Well, considering this representation, if you will substitute p into this equation, and knowing that this polynomial is equal to product of this, so p is substituted into each of these, and this is equal to zero because p is the solution of this equation. So what does it mean? It means this is also equal to zero. Since this component is zero, multiplied by whatever it is, would be zero. So this is zero, which means what? It means that the p is also a solution of this original equation. It's the root of the polynomial, as they say. The root means it's the value which converts, transforms polynomial into zero. So if you can solve a lower power, lower power polynomial equation, then its root is also the root of the original equation. So that actually paves the way to at least approach to solve certain, not all of them, but certain higher order polynomial equation. And what's the recipe? Well, first of all, try to represent this polynomial as the product of two polynomial of a lower degree. And then try to solve one of these. Maybe it's easier. So if this is the polynomial of the third degree, for instance, it might be represented as a product of two polynomial, one of the first degree and another is a second degree. This is linear, this is quadratic. We both, in both cases, linear equation we can solve and quadratic equation we can solve. So instead of solving equation of the third degree, if we manage to represent it as a product of two polynomials of the first and the second degree, then we can actually solve the whole problem. We will see all the solutions of the original equation. So this is the first approach, represent our polynomial as the product of two polynomials of the lower degree and try to solve polynomial equations of lower degree. Okay, fine. Done. Next. Next is a very interesting observation. Let's assume that somehow we have guessed one particular solution to this equation. We found the root of this equation. So this is the root. Doesn't matter how. Maybe somebody told us. Maybe we have just looked at the equation and realized, okay, this is the solution. Does it help us? Well, the purpose is to find all the solutions, right? Whenever we are talking about quadratic equation, we usually are looking for two solutions. Sometimes it can be one, sometimes it can be zero, but we are not stopping at finding one particular solution and saying, okay, that's it. The problem is solved. No, we're trying to investigate completely what other solutions might exist. So here is exactly the same problem. And what I'm going to say is that knowing one particular root at least helps us. I'm not saying it solves completely the problem of finding all other roots, but it helps us to find other roots. And here is why. So what I'm going to say is the following. For instance, x equals p is a root of this particular polynomial. What does it mean? Well, it means that a, zero, p to the nth, plus, et cetera, plus a, zero, is really equals to zero. So p is a number and a, zero, plus a, zero, and a, n, and a, one, and a, two, et cetera, and a, n are all some real numbers and p is the coefficient. Well, maybe real, maybe complex. I'm not really restricting myself to the real numbers, complex as well. All right, so let's assume that somehow we know that this is the fact. Now, what I'm staging right now is the following. That the a polynomial original one can be represented as a product of x minus p times some kind of b polynomial of n minus one's degree, from b zero to b n minus one, and x to the power of n minus one down to x zero, which is a constant. So you can always divide the original polynomial by x minus p, where p is the root. And then we have to prove it, actually. There are certain, a little bit more difficult theorems, which I'm living without proof, but that's a rare case. I'm trying not to do it too often. But this is something which can be proven, and relatively simply. All right, how can I prove it? Very simply, let me consider this as existing equation between whatever is known and whatever is unknown. What is known here? Well, coefficients from a zero to a n are known. p is, again, a known root of the original equation, which means this is also a known equality. It's not equation, it's equality, because p is some number. Now, x can be anything. So this equation I presume, if it exists, it must be valid for any x, anything, any tool. Now, b zero, b one, etc, b n minus one are not known coefficients, and I'm going to find, explicitly, find these coefficients. How can I do it? Okay, here is how. Let's just multiply this by this. And as a result of this multiplication, I will get on the left side my original a zero x to the n, etc. And on the right, I will also have some polynomial with x to a certain power. So, x, I will multiply by each of those, and then minus p, multiply by each of those. Now, what will be the result of multiplication of x times these? It will be, obviously, b zero x to the n's, right, x to the first degree and x to the n minus first. So, when I multiply one by another, my exponents are added together, so it will be n. This is one, this is n minus one. Similarly, next one will be n minus one, etc., and n minus one. So, let's say, n minus two, before that was x, it would x square plus b n minus one. That would be, if I multiply x by all of these. Now, minus, minus, minus, p multiply by all of them. So, p multiplied by this would be p b zero x n minus one plus p b one x minus two plus, etc. Now, the one before that was b n minus two x. So, it would be here, p b n minus two x minus p b n minus one. So, that's the result of multiplication. Now, let's do some multiplication. Now, I know that this polynomial is supposed to be equal to this polynomial with any, for any x. x is a variable. So, if polynomials are equal, their coefficients are supposed to be equal. Which means that the coefficient at x to the n's degree, so let me just wipe out this. We don't need it anymore. Coefficient at x to the power of n is what? In this case, it's a zero. And in this case, it's just b zero. Now, coefficient at x to the power of n minus one. This is a one. And this is b one minus p b zero. B one minus p b zero. Which is equal to, I know what b zero is already. So, it's b one minus p a zero. From which I can derive that b one is equal to a one plus p a zero. A two, this is the power, this is the coefficient at x to the power n minus two. n minus two is next to this one. It would be b two, which is here, plus p b one. Now, I know what b one is. So, it's b two plus p a one plus p square a zero. From which b two is equal to, I mean minus, I'm sorry, this is supposed to be minus. This is plus, this is minus. So, b two minus p b one. Well, I put minus here, but I didn't really put minus here. That's my mistake. And that's why I'm kind of confused. This is minus p. So, everything should be with a minus sign. So, everything here should be with a minus sign. Which means whenever I resolve it for b two, it would be a two plus p a one plus p square a zero. Well, I hope you see the way how it goes. So, without going into the details and without a rigorous proof by induction, I hope that you will just believe me that any coefficient b i is equal to a i minus two plus p a one plus p square. Well, let me just do it differently. Plus p a minus one plus p square m minus two plus etcetera plus p to the i a zero. This is i. Just sequentially doing exactly the same steps which I did before. So, this is a generalized expression for b i's. Now, well, great. Basically, I found all my b i's. There is one little detail. Whenever I multiply, let me write it again, x minus p times b zero x to the n minus one plus etcetera plus b n minus one. I compared all the coefficients of this product to the corresponding coefficients of the original a polynomial. But what I didn't do, I did not specify the very last one. I specified with x to the n minus one, x to the n, x to the n minus one, x to the n minus two, etcetera, x to the first degree. And I did not really use this free constant, which is minus b times b n minus one, which is supposed to be a n's. This looks like, this is yet another equation for b n minus first, because the last one in this row would be b n minus one equals to a n minus one plus p a n minus two plus p square a n minus one. Three plus etcetera plus p to the n minus one a zero. So, it looks like I have two different expressions for b n minus one. And this one shouldn't contradict this. Well, otherwise, I'm basically coming up with a contradictory set of equations for b, for b n minus first, at least. Now, let me show that this is not a contradiction. If b n minus first is equal to this, then what's important is, don't forget that p is a root of the original equation. Let's multiply minus p by this expression for b n minus one. What do we have? Well, we have this equation. a n equals to minus p a n minus one minus p square a n minus two minus etcetera minus p to the n's a zero. What is this? Well, move everything to the left side and you will see the original equation, which is p to the n's times a zero. This is this one. It's on the left. Plus a one p to the n minus one etcetera plus this one p a n minus one p n plus a n equals to zero. And we know that this is true because p is the root of the original equation. This is our original assumption. If p is the root, then the whole polynomial can be represented in this format. So, the fact that p is the root actually is the only significant piece which is required here, and only this makes our system of equations where we define, where we derive it actually with b first, b second, etcetera, etcetera. Only this makes it really a true solution because if p is not the root of this original polynomial, if this is not an equation of the original high order polynomial equation, then this is not true. And if this is not true, then this would be a contradictory two equalities, which means there is no solution. So, solution exists with concrete b zero etcetera b n minus one expressed as this particular set of equations. It exists only in case when p is a true root. That's why we are saying that if p is a root, and we don't know, maybe it doesn't exist, but if it exists, this is very important. If it exists, then the original polynomial can be represented in this way. Now, what's good about this? Well, as I was saying, if you represent the polynomial as a product of two other polynomials, these are of a lower degree. And obviously, this polynomial has a root x equals p, which we already know, and this polynomial is of n minus first degree, which is easier to solve than the polynomial of the nth degree. So, this is basically the conclusion of these logical steps. Again, if you know somehow that there is a solution of the original polynomial equation x equals p, then the original polynomial can be represented in this way. And these are the way how you derive the coefficients b zero, b one, etcetera, up to b n minus one. You get the second equation to be solved, which is a lower power, lower degree, and that makes your life easier, which means if you know one root to equation, it allows you to reduce the problem of solving this original equation to find other roots into an easier problem to solve the polynomial equation of the lower degree. Alright, that's what guessing the root gives you. Well, how can you guess? Well, again, we are not talking about any practical problems. We are talking about problems which you are given as an intellectual challenge. And whenever we are talking about intellectual challenges, in many cases, people who invent the problems try to operate with integer numbers. So, let's consider you have an equation which has integer coefficients and maybe, just maybe, there is an integer root to this equation, integer solution to polynomial equation. Now, it's worth to try, so if you have some equation, it's worth to try some integer solution. But where is this integer solution? Here is a very important thing. I just wiped it out, but the last equation which I had, if you remember that was it. So, if the result solution x equals p of the original equation, as you remember, then we can represent as a product of x minus p times some polynomial of n minus first degree, right? And now, since these are two polynomials, there is a constant a nth with, it's a coefficient at x to the power of zero, if you wish. Now, what is here the coefficient which is actually a coefficient at x to the power of zero, the constant? Well, it's only if you have this constant minus p and this constant bn minus first, because everything else or other members contain x to some power greater than zero, first, second, etc. So, only when you multiply these two members, you can get this, I'm sorry, this is nth. So, that's actually how I derive this. It's only minus p times bn minus one is equal. So, if there is an integer solution p to a polynomial equation with integer coefficients a, then this p must be one of those which are one of the divisors of a nth. So, if you have this equation which contains only integer coefficients, look at the very last one, the free member of this equation, the constant, and look at the divisors of this particular number. Maybe one of those divisors with a plus or a minus sign can be the root. And, well, the number of divisors is limited, obviously. For instance, if this is nth, then what do we have? We have 3, we have 2, we have 4, we have 6, and we have minus 2, 3, 4, 6. So, there is a certain limited number of integers which are divisors of this constant triad. If it works, great, because if you found one of them, then you can always represent your original equation in this format, and this will be an equation. Now, to solve this particular equation would be easier because it's a lower order equation. So, if integer coefficients, and you're looking for integer solutions, look for them among divisors of the free member of the constant which participates in the whole polynomial. Now, and the last one which I wanted to talk about is the following. We were just talking about if x equals p is a solution, then it can be represented in this way. Now, if this has a solution, if equations for p be polynomial, if it has a solution, then it can be represented as a product of x minus q, let's say x equals q is a solution. So, it can be represented and coefficient, let's say c, 0. Now, x to the n minus 2, this is b. So, we have lowered one more time the polynomial. So, with every solution, we can represent original polynomial as a product of x minus b times x minus q times the c polynomial which is of n minus 2nd degree. And this in turn can be done many times. How many times? Well, apparently no more than n times. On each time, we reduce the power, the degree of the polynomial by 1. So, every time, if we find the root, then we are reducing this remaining polynomial after these differences, linear differences by 1. Which means that the polynomial of nth degree cannot have more than n roots because every new root found will reduce the power of this polynomial by 1. And we can only reduce it n times because our original polynomial is of the nth degree. Which means that polynomial of nth degree cannot have more than n different, n solutions. Different is not really the right word because sometimes you can have the same solution repeated twice. So, n solutions. This is a very interesting observation, by the way. So, quadratic equation cannot have more than two solutions, by the way. And linear equation cannot have more than one. And the equation of the third degree cannot have more than three different roots. Can it have less? Well, that's a very interesting question. We know that there are no more than n solutions for polynomial of the nth degree. And I'm talking about complex numbers. In case of complex numbers, by the way, the quadratic equation always has some solutions, right? All right. So, that actually is a subject of the next lecture. Next theoretical material. Not the problems which I will spend some time on in the next lectures or two. So, there is something which is called fundamental theorem of algebra which actually deals with existing of the solutions. I will definitely go into the details in that lecture, but right now we have already learned that we cannot have more than n solutions. The interesting is that the number of solutions always is equal to n. Although some roots can repeat multiple times and should be counted multiple times. And this is actually the fundamental theorem of algebra just formulated slightly different, but I'm definitely going to the details of this in the corresponding lecture. So, just going a little bit ahead of myself, we have proven that there is no more than n solutions to a polynomial equation of the nth degree. But the fundamental theorem of algebra helps us to basically formulate that there are always n exactly n solutions for the polynomial of the nth degree. It doesn't mean that we can find these solutions using some kind of big formula using some kind of multiplication division roots from the coefficients. Now, generally speaking, we don't have it with polynomials of the fifth degree and more. But the existence of these n solutions is the fact. This is a theorem which is proven, which contains basically partially this fundamental theorem of algebra, which means that there is at least one solution. That's what it says actually. And this, there is no more than n solutions. So, the combination of these gives you really this interesting statement that the polynomial of the nth degree has exactly n solutions among the complex numbers. Well, that ends this particular lecture and I encourage you to go through this material again and don't forget the problem solving. We are making all this presentation for one and only one purpose. So, you will learn how to solve different problems and you will train your mind, your creativity, your analytical abilities, your logic and your intelligence. So, that's the purpose. And please register to Unisor.com with your supervisor or parent or teacher and go through exams. Always welcome. And that's it for today. Thank you very much.