 Hi, I'm Zor. Welcome to a new Zor education. Today I would like to talk about one particular application of the fundamental theorem of algebra, how to solve certain equations of the higher order. Now, we do know about the second order equations, but it's kind of difficult to deal with third order, fourth order, etc. So that's about what I'm going to talk. Okay, so now this lecture is called division of the polynomial. Basically polynomials can be divided like similar to a long division, something like this. Now, this lecture is part of the course called mass proteins. That's incorrect physics protein, because there is also a physics protein, but I have decided to return to this particular application of the fundamental theorem of algebra. So there is a mass proteins course, and this lecture belongs to that course. And it's all presented on Unizor.com, both mass proteins and physics proteins, and something else. Now, the site is completely free, there are no ads, so just pure knowledge, and I do recommend you to watch all lectures of both courses, mass and physics, from the website, because every lecture has a very detailed, textual part, like notes. Like parallel to the video, you have these notes, like a textbook basically. Now, in addition, the website basically contains the whole course, not just one particular lecture which you can find on YouTube or somewhere else. So that's important, and also there are exams. Also, you can take them as many times as you want, but it kind of verifies whether you really know the subject. So, mass proteins, algebra, division of polynomials, that's what today's lecture is about. Now, let me just remind you that, again, it's very easy, we know the formula to solve the quadratic equation. If this is your quadratic equation, there is a formula which we have derived, and many other people did, and if you want two roots of these equations are minus b plus minus square root of b square minus 4ac. Okay, so we know the formula and we can use it without any problems. If b square minus 4ac is negative, then we are talking about complex solutions. But according to the fundamental theorem of algebra, there are always two solutions. Sometimes they can coincide, two solutions can coincide. That's when the graph of this function, which is parabola, just touches, for instance, 0 in one particular place. So these are two real roots, and this is basically two, but coinciding with each other. That's how it's basically supposed to be explained. Alright, so we know this. Now, what if my equation is slightly more complex? This is the third order equation. Well, there is a formula. It's that big, and it's called cardano formula. Now, but it's really so big that you cannot really practically use it. I mean, maybe somebody does, but I've never seen. But the formula is really very impressive. And there is basically the special way how to derive this, how cardano derived and some other derivations of this formula. There is also a formula for the fourth order of an equation like this. It's even bigger than cardano formula. And that's it, basically. Starting from the number five, the fifth order, if this is x to the fifth degree, the highest member, then there is no formula, and as far as I know, there is a proof that there cannot be a general formula, which helps in the equation of the fifth and higher order equations. So how do we do in these cases? Well, the general brute force approach is numerical. I mean, we have computers and we can always have certain algorithms which really help us to derive with the roots to a very, very high precision. But that's not what we are talking about. I would like to talk about certain techniques which might actually help you to solve the equation of the higher order. So that's about these techniques. And the technique, actually, which I'm talking about today is division of the polynomials. Now, let's just recall that according to the fundamental theorem of algebra, if you have some kind of a polynomial, let's say, of nth degree of x, then it can always be represented as, well, some kind of a coefficient probably, and then multiplication of these particular members. Well, let me explicitly put 1x minus xn. So the polynomial has exactly n complex, generally speaking, complex roots, and it can be represented as this particular product where x1, x2, and xn are roots of this polynomial. So the polynomial would look then as the general polynomial of the nth degree, would be a0, let's say, x to the n minus l, a1, a1. Or I will start with nc. Yeah, I will start with n. So 0 plus a1x to the n minus 1 plus et cetera plus a1n to the first plus a nc. This is 1 times. So this, in this case, would be a0. So this polynomial of the nth order can be represented in this way. This is basically the corollary of the fundamental theorem of algebra. The fundamental theorem is that there is always a root, complex root of any polynomial. And as a corollary from this follows this and that was in the previous lecture, which I have proven. Okay, now, what does it mean? Well, it means that, let me talk about, let's say, the third degree, that would be better. So if you have a polynomial of the third degree, x, and that's a x to the third plus b to the second plus c to the first plus d, I can always represent it as a times x minus a times x minus b times x minus c, where a, b, and c, generally speaking, complex, roots of the equation, this polynomial equals to 0. Okay, this is the nth order. This is the third order. Just easier to explain basically. Actually, this division of the polynomial talks about. Let's assume that I know one particular root, let's say c. How? Well, there are different techniques, maybe, to guess, basically, the root. So if you have guessed, okay, you look at the equation and you kind of feel, okay, let me just check x equals 1 minus 1, 2 minus 2, 3 minus 3, and one of them probably fits. Well, great. If you have one particular root, then the question is how to find the other roots. Well, if you can make this division a x cubed plus b x square plus c x plus b, if you can divide it by x minus c, if you know c, if you can just algebraically divide to come up with some other polynomial, let's say, how can I do it, mx square plus mx plus q. Now, if you will be able to do this, now, this is already a quadratic equation. Quadratic equation we can solve. My point is that if you know one particular root of the equation, you can always reduce the order of the equation from the third to the second or from the nth to the n minus ones, all right? Which is, it's easier equation, all right? The lower the order, the easier it is. Maybe we will be able to guess then another root and another division will help us to reduce the order even further. But in this particular case, I would like to stop basically on the third degree which I have converted into the second degree, second order. And this, we have a formula, so we can basically solve it. And I'm going to talk about this particular technique, how to divide one polynomial onto another on a concrete example and you will understand the whole technique. Okay, so if I follow this. Okay, so what is my example? My example is x cube plus 4x square minus 11x minus 30. Okay, this is my equation. So the first thing is, well, maybe I can guess the root. Now, if you consider the representation of this polynomial as x minus a times x minus b times x minus c. This is the third order, so there are three roots. Then what is the free member? Well, obviously it's minus a times minus b times minus c, right? When you will multiply x minus a, x minus b, x minus c. The only free without any x member, if I will multiply it together, would be, well, it's something which depends on x or even I should really put maybe even x times something plus or rather minus in this case. Minus a times b times c, right? a, b and c is the only free member of this multiplication, of the product of this multiplication. So we have to have this. So a, b, c should be equal to 30. Okay, well, let me try, maybe there are some integer which fits this particular equation. An integer should be found among divisors of the free member. Well, 30 has like two, three, that's six and five, all prime numbers. Well, let me try these. Okay, let me try, let's say five. Five to the cube 125, five to the square 25 by four plus 100 minus 55 minus 30. No, definitely 100. Definitely it's not zero. Okay, let me try minus five maybe. Minus five to the third degree would be minus 125. This will be still 25, so it's plus 100. Minus five, that would be plus 55 and minus 30. Minus 155 plus 155. Okay, that's zero. Great, so minus five is the root. If minus five is the root, then I should put here minus minus five, which is plus five. All right. Okay, so I know that my polynomial can be represented as x minus a times x minus b times x plus five. I don't know a, b, but if I will be able to divide this polynomial by this polynomial and have the result of this multiplication, then I will be able to obtain the quadratic equation and that will solve it. So how to divide this by this? All right, let's just do it. x cube plus 4x square minus 11x minus 30. I will divide by x plus five. Okay, now, if I have to have some kind of a polynomial which multiplied by this gives this, then my first member should be x square, right, to get the x cube. Well, let's see what happens. If I will multiply x square by x plus five, I will have x cube plus 5x square. Okay, I've got that, but I need this. So if I will subtract from this, this, my result would be this. So this is unsatisfied members yet. Okay, now, I will add something to my result of my division. If I need to get x square and I have x plus five, so I have to multiply by minus x, right, so minus x. So what will be as a result? Minus x times x plus five plus minus x square minus 5x. Oh, I'm sorry, I didn't put x here. Okay, I've got this. So I've satisfied x square. If I will subtract, it will be minus 6x minus 30. Unsatisfied. But I have my x plus 5 multiplied by what? By minus 6. Then I will have minus 6x minus 30. Zero. So I have to multiply my x plus 5 by x square and minus x and minus 6 and add the results together to get this, which means that the result of my multiplication is x square minus x minus 6. So this is this. Now, this is a square. This is a quadratic equation, x to the square. All right, so it's easier to solve it. So let me just have this solution, which would be, according to the formula, would be 1 plus minus square root of 1 square minus x plus 24 divided by 2, which is 1 plus minus 5 divided by 2, which is 3 and minus 2. So besides minus 5, which I have guessed in the very beginning, I have 3 and 2 are the roots and the whole thing, cubic equation is solved. Again, how? By guessing one of the roots and then dividing using this polynomial thing. Okay, so as you see, it's just a technique. I mean, this lecture is not really very theoretical, just a technique. How to do reduction of the order of the equation in case you know one particular root. Okay, so let me just, as an example, do one more thing and that would be the end of it. Because technique is really very simple. If you have the root, then you try to find out all the components, you have to multiply this root to get the original equation. One by one, member by member. Okay, in this case, my equation is of the fifth order. Don't get scared. It's really simple. Okay, now, again, my free member is 6. What if I have some integer which is this particular solution to this particular equation? I mean, if you are in school, then most likely the problem that you are given is something supposed to be like simple, right? Relatively simple. So, relatively simple in this case to look for the root of this equation among divisors of 6. And there are only 2, 2 and 3. Well, actually, plus or minus, 2 and plus or minus 3. Well, I checked it and it seems to be that x is equal to minus 3 is a solution. Well, you can check. Which means it should be divisible by x plus 3, right? Minus minus 3. So, let's divide this by x plus 3. We have to satisfy x5 first. So, what's my first member is x4. If I'm allowed to multiply x plus 3 by x to the fourth, I will have x to the fifth plus 3x to the fourth. And if I will subtract what's left, this is 0. This is 0. Left is this minus x to the third minus 3x square minus 2x minus 6. Okay. I have to divide it by x plus 3. And that would be minus x square. So, minus x cube minus 3x square. And if I will subtract nothing here, minus 2x minus 6. Now, x plus 3 is my divisor. Obviously, it's minus 2. The minus 2x minus 6, 0. So, these are components. So, the result would be x plus 3 times x to the fourth minus x square and minus 2. So, that's the result of my division. Now, I have reduced my original equation of the fifth order to the fourth. So, now I have to solve this, x to the fourth minus x square minus 2 is equal to 0. Well, in the previous case, the result of the division was quadratic equation. In this case, it's the fourth order. But, thankfully, it's actually by quadratic equation. Because if I will substitute x square equals to y, I have a quadratic equation. Which has solutions 1 plus minus square root of 1 plus 8 divided by 2. Which is 1 plus minus 3 divided by 2. Which is minus 1 and 2. Right? Two solutions. So, this is either minus 1 or 2. And this is easy. If x square is equal to minus 1, then solutions are plus minus i. Right? Where i is the imaginary unit, remember this? i square is equal to minus 1. And if x square is equal to 2, then x is equal to plus minus square root of 2. So, we have five solutions. Two solutions, two solutions and one solution solved. Okay, again, all these kind of techniques are good in good cases. Like in this particular case, I have guessed the first root, integer root. And the result was equation of the fourth order. Which kind of the same as equation of the second order. Because it's only square in the fourth degree. There is no x to the first order, first degree and no x to the third degree. Well, we are lucky. Okay, but these are the problems which you probably might be given by your teacher or exam or something like this. So, that's why these particular techniques are important. Again, my purpose was just to demonstrate how to divide one polynomial by another. And obviously, you understand that you can divide not only using this technique, you can not only divide the polynomial of some higher degree by linear expression. In this case, x-cluster. You can divide it into polynomial of the second degree as well. The technique is exactly the same. Okay, that's it for today. I suggest you to read the notes for this lecture. You go to the math for teens chapter part of this course. And it's algebra. And it's part of the topic called fundamental theorem of algebra. That's it. Thank you very much and good luck.