 Hi, I'm Zor. Welcome to a new Zor education. Continuing talking about high-order polynomial equations, I would like to demonstrate a couple of cases when you really can solve the equation of higher or higher than two order. Just a couple of examples which I have here. What I do suggest to you is, first of all, try to solve these equations yourself. They are in the notes on Unisor.com website, notes for this lecture. So try to do it yourself, whether you do or you don't, successful or unsuccessful, only then listen to this lecture. It's basically about one particular technique, how you can solve certain equations. There is no general formula for solving these higher-order equations, as I was talking about in the lectures, which received this particular exercise. So basically, you have to think about particular case about some peculiarity, about some technique maybe, or invent your own whatever to come up with a solution. These are three examples of equations which have integer coefficients, and of this attempt can be made to find integer solutions. So let me just try one after another. The first one is xq minus 7x plus 6 equals to 0. First of all, even without going into this technique which I'm going to use in all these examples, I actually can see something right up front, just looking at this particular equation. Now, what do I see? Well, here, 7x can be represented as minus x minus 6x plus 6. Now, why did I decide to do this? Lucky guess, if you want. But actually, I see that this is something which is the way to simplify the problem. Now, why I see, I don't know. Maybe because of experience, maybe I solved probably hundreds of other equations, which look like this one in some way or another, and at some points, I basically came up with this particular technique. So first, I'm looking and I'm trying to find out, is there anything here which reminds me some way to ease the problem. This is actually the way to do it because obviously, x can be factored out. So I have x square minus 1. Now, 6 can be factored out. And this is x minus 1 times x plus 1. As you know, x square minus 1, difference of two squares is difference of the basis times sum of the basis. And what do I see now? I see that x minus 1 can be factored out from both places, leaving x x plus 1 minus 6 equals to 0. So obviously, I see there is a solution, x is equal to 1 of this original equation. Now, how about this? If this is equal to 1, this is also equals to 1. So solutions to this equation is also this. So x times x plus 1 equals to 6. This is where we are looking for solutions to this equation, where 6 is such a nice number that 2 times 3 would be equal to 2 to 6. So I already have another solution, x is equal to 2. And as you know, we have three different solutions for the equations of the third degree in the complex area granted. So my question is, are there any other solutions? Well, obviously others are when this one of these is equal to minus 3 and another is equal to plus 3. So minus 3 times minus 2 also equals to 6. So x is equal to minus 3. As you see, I have practically guessed all the solutions of this equation. And it's not because I knew everything up front. Actually, I just kind of thought about x is equal to 1. It's kind of an obvious solution. But I didn't know about 2 and minus 3. All right, so let's check it out. 1 is 1 minus 7 plus 6 is correct. Now 2, 2 to the power of 3 is 8 minus 14. It's minus 6 plus 6 is 0. Good, minus 3. Minus 3 cubed is minus 27. This is plus 21. So it's minus 6 plus 6 is equal to 0. So all three are good. There are no more than three solutions because this is an equation of the third degree. So basically I found everything. Now, this is not, I would say a general approach. Well, there is no general approach, but this is a very peculiar solution which is based on some peculiarity of these numbers. So half of this I guessed, half of this I basically also guessed I could have solved this quadratic equation as a quadratic equation. x squared plus x minus 6 is equal to 0. I can solve it using some formula. But again, I was using certain particular properties of these numbers and basically guessed it. So I wouldn't say this is a generalized approach, but what would be a generalized approach in this case? Well, probably the best way in this and the subsequent cases would be to look for integer solutions among divisors of the three member, the last one, the constant 6. So 6 has one, two, three, six, minus one, minus two, minus three, and minus six solutions. So if we will check all these eight, some of them will fit and we will find the three solutions and then we can stop. So we can basically go and sequentially check one. Does it work? Yes, it works. Two, does it work? Yes, it works. Three would not work here because this is 27 minus 21, 6 plus 6, 12. Six obviously will not be as well. Minus one will not fit, minus two will not, minus three will fit, and the minus six will not. So that's how we can find all our three solutions. On another hand, if we have found one, which is this, you know that you can always represent this particular equation as a product equals product like x minus one because one is already found a solution and some polynomial of the second degree. Second because this is the third, this is the first degree. So we have to have second degree here. So the sum of exponents should be equal to three. And obviously the first coefficient at x square should be equal to one because otherwise this will not be equal to one. I don't really need these parentheses, right? So we need equal to one. So this should be x square plus two, oops, plus some kind of a something like this. And now we can basically find b and c by multiplying this and comparing with this. Well, let's do it. Now x multiplied by this would be x square plus b, sorry, x cube plus bx square plus cx. Minus one would be minus x square minus bx minus c. And it's supposed to be equal to this, right? Now x cube and x cube is fine. Now x square, there is no x square here which means b should be equal to one. So bx square minus x square should be zero because there is no x square here. So b is equal to one. Now cx minus bx should be seven minus seven, right? So b is equal to one. So c must be equal to minus six, right? So c is equal to minus six. That plus c would be plus six which corresponds to this obviously because one is the root of this polynomial as we were already saying. So right now I can say that my original equation can be replaced with x minus one times x square plus x minus six equals to zero. So obviously first solution is this which we have already found. And the second solution is this which we can actually use the formula if we don't want to guess. So this is what? One plus 24 divided by two, right? So it's five plus one is four divided by two so it's x is equal to two. And if it's minus five, it's minus six, minus three. The same three solutions which I have kind of guessed before. So this is a little bit more scientific approach. Once you found one particular root of this polynomial x is equal to one just by guessing. Then use this to find the other two. All right, fine. So let's remember both approaches. One a little bit less scientific which I presented the first where I was guessing lots of different things. Another is a little bit more scientific. Now which one is better by the way? Hard to say. I think that the element of guessing is very, very important. It actually is the result of your intuition, your mathematical intuition which is basically the result of your experience. So there's nothing wrong with guessing. A little bit more procedural approach which is the second one. Well, first of all, it's still based on the guessing of the first particular root. And again, we kind of assumed that maybe there was an integer root so we were just trying to check all the divisors. And we were lucky enough that on the first one we already got hit. But then everything else after that was relatively procedural like presentation as a product of polynomial of the first degree and the second degree, et cetera, et cetera. So I would say that one is better than another. Try to be prepared for any kind of approach whatever you need to solve the equation. It's your ingenuity. The whole exercise of solving these equations is to develop your creativity or ingenuity, your analytics, your logic, whatever else. So that's fine if you guess it. That means that you have this mass intuition. That's okay. Now, next one. X cube minus five X square plus nine X minus nine equals to zero. Well, again, I am kind of assuming that there are integer roots. So integer roots must be plus or minus one, plus or minus three, plus or minus nine. These are all divisors of this particular equation. Well, let's do the same thing as I did before. Plus or minus one. Let's check them. That's easier, right? If it's one, it means one minus five. It's minus four. This is zero, so that's not good. Minus one, that's minus one and minus five and minus and minus. Too many negatives. Not good, not zero. Let's try three. 27 minus three square is nine, 45. So 27 minus 45, that's what? Minus 18, right? Minus 18. Now, minus 18 plus 27, that's nine. And minus nine is zero. Okay, so three is good, which means we can always represent this as x minus three times x square plus vx plus c. Again, I'm using the coefficient one here because the product should have the coefficient one at x cubed. All right, fine, that's easy. Now let's multiply this and find b and c, right? That's easy. Actually, c I can find right now. Minus three times c, that's the only free member without any x and it's supposed to be equal to minus nine, right? So minus three times c is equal to minus nine, so c is equal to three. We got that, all right? Now let's do the multiplication. x times this would be x cubed plus bx square plus cx, but x is equal to, so it's plus three x, right? I'm using c is equal to three already. Now, minus three x square, minus three bx and minus nine. Again, I'm using c as three. And this is supposed to be equal to x cubed minus five x square minus nine x minus nine. So what can we do from here? Well, bx square minus three x square is b minus three and it's supposed to be equal to minus five. So b minus three is supposed to be equal to minus five. So b is equal to minus two. Now, how about this one? Three and minus two, that's six, that's nine x. This is minus nine, nine x, but no, something is wrong. What's wrong here? Minus nine, c is equal to three, right, something is wrong. This is nine, three times b, that's six, that's nine. So why am I not... Okay, let me check it, let me check again. Is three the root? Okay, it's 27 minus 45, it's minus 18. Minus 18 plus 27, that's nine. Yes, three is the root. Okay, fine. Now, minus three times c should be equal to minus nine. That's correct. So x cube plus bx square plus cx, which is three x, right, minus three x square, minus three bx and minus nine. So x cube is fine, b minus three x is equal to minus five, x square again. All right, so b is equal to minus two, that's strange. How come it's not working? I'm supposed to have minus nine x. Oh, I'm sorry. This is plus and this is plus. I have made this little mistake by writing the wrong thing here and then transferred it here. It's plus, of course, and minus two is correct. All right, well, as you see, I'm making some mistakes as well. Well, although this is just a very simple one. All right, so basically we have found our equation, b equals minus two and c equals to three, which means we can rewrite the right part as x minus three times x square minus two x plus three. All right, fine. Now, is our life easier right now? Oh, absolutely. Because one solution we have already found, which is x is equal to three, and another solution is to solve this one, right? And the roots of that thing is, well, it's obviously one plus minus square root of one minus whatever, 12 or something. Well, obviously this doesn't have any roots at all. Well, obviously it's seen from it's x minus one square plus two, right? x minus one square is x square minus two x plus one. So this is positive plus two, I mean, non-negative plus two. So there are no solutions to this equation. It cannot be equal to zero. So in this particular case, in the realm of real numbers, we have only one solution. In the realm of complex numbers, we have three solutions, one from here and two complex one with real and imaginary part from here. And well, let's just do it. It's what, it's minus, it's two plus minus comma is four minus comma is four minus four 12. So it's one plus minus eight, four times two, square, square root of two, right? So that's two other solutions which are supposed to be solutions of the third degree polynomial equation. I hope I didn't make mistake here. Well, forgive me if I did. In any case, this is just a square, this is just a quadratic equation, so it's okay. Principle is more important than the details of this. And the last but not least, we have the equation of the fourth degree, x to the fourth plus four x square, no, x cube, sorry, minus two x square minus 12x plus nine equals to zero, all right. Now, this is, again, the polynomial equation of the fourth degree. And we are looking for integer solutions first, which means, again, we are looking among plus or minus one, plus or minus three, and plus or minus nine, which are divisors of the ninth. Well, let's check it out. I mean, we don't have any better approach. Okay, start from one, it's easier, right? So everything is basically x is equal to one. So one plus four is five, minus two is three, minus 12, minus nine and plus nine. Well, we're so lucky, one is the root. Okay, that's good, but that's basically not the complete pass to solution, because one will allow you to reduce from the fourth to the third degree, all right? Okay, let's do that. So we have x minus, all right, we need something else, how about this? So we have x minus one times, now the coefficient at x cubed should be one, otherwise I will not get one when I multiply to get x to the fourth, plus, well, let's use bx square plus cx plus b, all right? So that is supposed to be equal to this. Since one, x one, is a solution, it's supposed to be represented in this way. All right, let's find out dc and d. Well, d is easier, right? Because multiplied by minus one, it should give nine, which means d is equal to minus nine, all right, fine? Let me make my job easier, and I will put minus nine here, that's easier, right? Fine, now we have only two variables. Let's think about x cubed, x cubed can be either minus one, or when you multiply b times x, so it's plus b. So these are only members when they open the parentheses which give you x to the power of three, right? Minus x cubed and plus bx cubed, everything else will be lower degree, and I should get four. So b is equal to five, right? b is equal to five, and I will substitute it here because it's easier for me, great. Now we have only one left, c. All right, let's think about x square, how can x square be obtained? First of all, it's x times cx, which will be cx square, and secondly, minus one times plus five x, which is minus five. So we have c minus five, coefficient at x square, it's supposed to be equal to minus two. So c is equal to three, c minus five, so it's three. Okay, the only thing which is left is the coefficient at x, minus 12. We have to check if we really have it. Well, indeed, x to the first degree can be obtained from x times minus nine, which is minus nine, and also minus nine, minus one times three x, which is minus three, so it's minus 12. So this is a correct representation, fine. So we found one particular root, x is equal to one, and we have reduced our equation to the third degree. So now we have to solve this. So let me get rid of this. It already played its role, and let's solve this equation. Found one root and reduced the power of the equation. Okay, what should we do with this one? Well, let's try again. What kind of roots this particular equation has? Again, we're looking for integers, right? So minus nine, again. It's nine, so it's plus minus one, plus minus three, and plus minus nine. Well, let's try again. Let's start with one, again. One plus five, six plus three, six, nine, minus nine, zero. We're lucky again, x is equal to one is a solution. So we have another solution, x is equal to one. It's a double root, if you wish. Yes, a particular solution can be represented twice in the polynomial, which means that this again can be divided into x minus one times x square plus bx plus c. Do exactly the same thing. First, we start from this free member. Minus one times c should be minus nine, so c should be equal to nine, right? Okay. Now, so x cube is fine, and minus nine is fine. We have these two. Well, let's start from five x square. x square can be either x times b, x. So we will have bx square, or minus one. And that should be equal to five. So b is equal to six. b is equal to six. And all I have to do to check if I'm right, I have to verify that coefficient at x is what I'm really looking for. Well, coefficient at x is plus three. Here, x can be multiplied by free member, which is nine x, or minus one can be multiplied by six, which is minus six. So nine x minus six, x is three x, so that corresponds. Everything is fine, we don't make any mistakes. So now, we have another root, which is x is equal to one, and another equation to solve. Well, again, I'm guessing, but look, this is obviously x plus three square, right? The x square plus double x times three, which is six x plus square of this. But, I mean, if you want to really solve it, use the formula or whatever else. Now, what does this mean? It means that, again, we have two roots. One root is equal to x equals to minus three, and another root is also equal to x minus three. So we have four roots, but these are actually two different values, each of them double. We have a one double root and minus three double root, and which is fine. Altogether, we have four solutions to the equation of the fourth degree, which is correct, which is the way how it's supposed to be. These are all happened to be integer solutions. That's what it is, that's what it is. Okay, basically, I think you understand the schema which I was using here. It's one of the approaches which you can use if you would like to solve some equation of the high degree. Try to find some root, and then using this representation as a product of x minus this root times polynomial of the lower degree, you basically make your life easier. Okay, there are some others, and I might actually present some other further lectures, but this is it for today. Try to go through these solutions yourself again, and don't get scared if somebody offers you to solve the polynomial equation of a higher order, because most likely solutions are relatively simple. You just have to approach it properly, and if it's a problem given to you, most likely it has a relatively easy solution, not a generalized solution. There is no generalized solution in this particular case. Nobody would give you this problem. All right, that's it for today. Thank you very much.