 Hi, I'm Zor. Welcome to Unisor Education. Today we will talk about equations of higher order than 2. With quadratic equations, equations of the order 2, we did have basically purely theoretical material, how to derive general speaking, how to derive the formula for solutions of the quadratic equation. Now, with the equation of the third degree, the cubic equation, where the variable is in x cubed, the situation is more difficult. Actually, there are some formulas which, in theory, might be used, but they are quite complex, and these formulas express the solutions to a cubical equation in terms of coefficients. Now, when you go to even higher degrees, like the fourth degree, that's even more complex, and as far as the fifth and higher, there is actually a very interesting theorem in algebra that there is no general solution for equations of the fifth and higher degrees of its coefficients, which can be expressed in a regular algebraic expression. So, basically, if you have an equation of higher order, higher than 2, people don't usually solve these equations in general. However, since we are not really pursuing the idea of solving the equation, but rather equipping you with certain approaches and guessing and tricks, if you wish, just to develop the mathematical thinking and just to be able to face the problem which you might not really know the solution beforehand, I will use these higher order equations just to illustrate certain approaches which you can take, and hopefully they will be educational enough and just help you to develop certain logical thinking, etc. So, again, I will use the higher order equations just to develop the brain power, if you wish. All right. Now, first of all, what is the representation of the higher order equation? Well, in a concrete example, you can say something like 3x fourth minus 2x equals to zero. This is just an example of an equation of a fourth order which, well, it can be solved, and by the way, purely algebraically, but that's not the problem right now. Right now, this is just an illustration of the polynomial on the left and the equation of the fourth degree altogether. Now, if I wanted to express generally how any equation of the higher order looks like, I would use the following symbolics. Following on the case degree which is a function of x. What is this? Well, this is a sum of coefficient with x to the case degree plus another coefficient with x to k minus 1 degree plus, etc., etc. x to the power of k, x to the power of k minus 1, etc., down to a k minus 1 x plus a k. That's the polynomial of the case degree where x is in the power of k. And these are different coefficients. Generally speaking, coefficients are usually real numbers. But generally, they are complex. So the equation of k's order or k's degree is something like this. Some polynomial of x of degree k is equal to 0. So this is the general expression of the equation of the higher order, of any order actually, k, where r as a function of x with this index, if you wish, k is a representation of this sum of different elements. Each one of them is x to certain power and the maximum power is k, which corresponds to this particular index in my general expression of the polynomial. And r equals to 0, obviously. Sometimes this number, a kth, which has no x with it, usually it's called the free coefficients. And this one is usually the first coefficient with x to the power of k. So this is general polynomial expression of the equation of the higher order. And as I said, there is no such thing as a general solution of this for any k. At least for k greater or equal to 5, there are no general solutions. For k equals 3 and 4, as I said, there are, but they're too complex. So usually if people have a general solution with some coefficients, they go to computers basically to solve it. There are certain numerical algorithms which allow to write the programs for the computers and they will come up with certain quite precise solutions if these solutions exist, of course. Now, in our case, since we are pursuing a purely educational goal, I will emphasize two methods which might, in theory, be used to solve certain equations in certain cases. And again, the art of this is to find these methods, these specific methods to solve specific equations, not the general type of them. Okay, so what are specific methods which I would like to talk about today? Well, there are two. Method number one is the problem. Let's assume that our polynomial expression like this, that it can be expressed as a product of one polynomial expression times another polynomial expression. Now, this is a polynomial of the case order. This is of nth order and this is nth order. And what's important is that both m and m should be less than k. Well, obviously, they should be less than k because if you will think about the polynomial of the nth degree, so k, x is in the nth degree and there is some expression. And q is a polynomial of the nth degree, x had the maximum n. So if you will put this polynomial, complete expression in this and multiply them, the highest order of this one, which is m, and the highest order of this one, which is n, and when you multiply them together, you will have m plus n as a degree. Why? Very simple. Let's just do it in one simple example. Let's substitute. This is a0 times x to the case plus something. This, let's say, is b0 x to the nth maximum plus something times, and this will be c0 x to the nth plus something. Now, if you will open parentheses and multiply, what's the maximum power the x will be in when you have the maximum of this, which is m, and the maximum of this, which is n? So it will be b0 times c0 times xm times xn, which is b0 c0 x to the nth plus n. And if they are equal, the maximum power of this should be equal to the maximum power of this, which means a0 x to the case equals to b0 c0, let's be accurate, x to the nth plus n. That's what we will have. This is the maximum power of x, and this is the maximum power of x. So if these expressions are the same, these members should be the same, which means a0 should be equal to b0 times c0, and k should be equal to n plus n. Since it's identity, it's supposed to be equal for any x, if this is a true identity, true representation of r as a product of p times q. So obviously, if m plus n is equal to k, and we are not considering trivial cases when m is equal to 0 or n is equal to 0, because that actually makes the whole expression just a constant. So obviously, m plus n is equal to k, from which follows that both m and n are less than k. By the way, they are all positive and integer numbers. I didn't mention it, but it kind of seemed that this is understood. So m plus n is equal to k, so m is smaller and n is smaller. So if you can represent a polynomial of, let's say, third degree with a product of the polynomial of the second degree and the first degree, then you have a chance, basically. Then you can have the multiplication of the second and the first degree. So m is 2 and n is 1. You will have k is equal to 3, and that will be the polynomial of the third degree. Now, why do we have to represent r as a product of two polynomials of smaller, of lower degree? For a very, very simple reason. Let's say you know how to solve this particular equation and this particular equation separately. Well, look at it this way. If you found an x when this is equal to zero, then obviously this is equal to zero because it's a multiplication. So any solution to this equation, which is easier than original because the m is slower than k, any solution to this equation will be a solution to the original equation r k x equals to zero. This is what you have to solve. If this is what you have to solve and r is represented as a product, then any solution of the p is equal to zero and any solution of the q is equal to zero would be a solution to r is equal to zero. So one problem of solving an equation of let's say third degree, you might reduce to solving two problems. One is the first degree and another is the second degree. Just as an example. Which is easier? Obviously the lower degree, the lower order of the equation is, the easier it is to solve and if you can reduce it down to first and second degree polynomial, well, you know the formulas and you definitely can solve a quadratic or linear equation without any problems. So what's important is to reduce solving the equation of the higher order to solving the two or more equations of the lower order. That's the approach which you can take. That's one thing and let me illustrate it as an example. So if you are given an equation, let's just think about it this way. Whether it's a math exam or anything like that you have given an equation x to the third plus 5x squared plus x minus 15 is equal to zero. Obviously you cannot solve this equation using any kind of a formula because there is no formula, at least you don't know the formula for the cubicle equation. Which means you have to find something which will simplify your job and again my first recommendation is to find some representation of this polynomial of the third degree as a product of two polynomials of the lower degree, like first and second. Well, absolutely the most difficult problem is how? Yes, you know that maybe it is possible but you don't know how. Well, here it is. Here is my recommendations. Let me just say that somehow, magically, I have guessed the following thing. Let's represent 5x squared as 3x squared plus 2x squared. So x cubed stays. How I guessed it is not really the question right now but let's consider that I have guessed it and I will represent 6x as 6x minus 5x. So this is the same as this and this is the same as this minus 15 equals to 0. So this is the identical transformation. Now, why do they do it? Here is why. Now let's consider pairs. One pair, second pair and third pair. You can factor out x squared here and you will have x squared times x plus 3. You can factor out 2x from the second pair to x squared plus 6x factor out 2x, what will be left from 2x squared you will have x and from 6x if you factor out 2x you will have 3. Minus 5x minus 15 the third pair I will factor out minus 5 and I will have x plus 3. You see, so far I just did a couple of identical transformation and I have seen that you see x plus 3, x plus 3, x plus 3. Now I can factor out x plus 3 and what will be left x squared plus 2x minus 5 equals to 0. So what did I did actually? I have represented my original polynomial in the third degree with the multiplication of two polynomials of the first and second degree. Obviously you can solve each of these equations x plus 3 is equal to 0 separately that's x equals to minus 3 and this x squared plus 2x minus 5 just a plain quadratic equation you can solve that too. Now solving these equations basically again as I told you in the general case gives you solution to original equation x cubed plus 5x squared, etc. equals to 0. So what was the important part of it? The most important is not really solving these equations because this is trivial but how to guess basically is these coefficients like 5 x 3 plus 2 1 x 6 minus 5, etc. That's the most important guessing game which you have to accomplish because after that everything is just trivial. It's like going along the paved road. So first you have to climb the mountain and then you see everything, how to do it. Now is there a general recommendation which I can give in this particular case? Well, yes there is. But again it's just a general recommendation it doesn't mean it will always work or something like this. Think about it this way. If you are given an equation of this type and somebody really asked you solve it. Obviously you understand that most likely solutions are relatively simple because otherwise you would not be asked to do it. Nobody would give you some crazy coefficients with big rational numbers, etc. So usually it's integer numbers and if they are integer to look for a solution would probably be very easy among factors of the three coefficient. The one which has no x, y. But here is why. Look at this. If you have some linear equation and quadratic equation. Linear expression and quadratic expression. Three member will only be obtained if you have no x here which is 3 and no x here which is minus 5. And that would actually multiply them. That's the only two numbers which will give you the free member of the original equation, 15. Everything else will contain x in some degree. So if you open all these parentheses all different products, x times x square or 3 times 2x, etc. All of them will contain x except one. Three member from here and three member from there. So only one gives you the multiplication of these three members will give you the free member of the free coefficient in the original equation. So you might just suspect that if there is some kind of a linear expression which can be factored out it should be x plus some factor of the free member. Now how many factors does this guy have? Well, 5 and 3 and minus 5 and minus 3. So these four different variations should actually be tried. In this particular case the factor minus 3 is the one which is used here. So x plus 3 would be something which you can look for. So you can try another one. You can try for instance minus 5 which means let's try to do x plus 5 and try to parenthesize this out to factor it out. Let's just try and see if it works. So let's consider I would like to factor out x plus 5. Well, then from these two I have to factor out x square I will get x plus 5, right? Now this is not x plus 5. This is x plus 5 but I have to have minus 15 so I have to have minus 20. This is definitely something extra. So x plus 5 doesn't really work in this particular case. Well, let's try x minus 5 for instance. Well, if I want to have x minus 5 and I have x cube plus 5x square what should I have? I should have x square but now I have minus 5x square I need 5 plus x square so I need to add 10x square. Now I have converted plus x minus 15 equals to 0. Now, I need x minus 5 again. Now, if I will factor out x and get x minus 5 what do I have? I have 10x square so this is fine but now I have plus x but here I have minus 50x so I have to add 51x to satisfy this and minus 15. Again, x minus 5 x minus 5 but this is something which is completely extra. So x minus 5 also doesn't work. So by trying a few divisors of this free coefficient I can just eliminate a few and finally if I will get x plus 3 let's just think about again how do they do it? Okay, I know that this thing has 15 has 3 and minus 3 among divisors. So let's try x plus 3. Okay, if this is x plus 3 now this is supposed to be factored out 2x square. Now I have x cubed plus 3x square but I need 5x square so I have to add 2x square. Okay, so far so good. Now I have plus x minus 15 equals to 0. So now this is already x plus 3x square now I have 2x square so I have to factor out 2x and I need the same x plus 3, right? So what do I have here? I have 6x, sorry yes, 2x square plus 6x. Now, if I have 6x but I need only 1x so I have to subtract 5x and lo and behold if I will factor out minus 5 I will have x plus 3 my guess is correct. So by just trying to guess which one of the divisors of the free coefficient fits this type of a schema you will be fine. So if there is a solution you will find it. But again we are talking about simple cases when solutions are looked for among integer numbers obviously and hopefully you will not be given more complex equations because it's much more difficult to solve and hopefully people don't really try to solve these equations in some kind of a formulas even if there are the possibilities obviously. So again if you are looking for a solution of equation which contains integer coefficients and you are looking for a solution which is also integer it's helpful to look at the factors of free coefficients. In this case it's 5 and 3 minus 5 and minus 3 so you have to plus x plus 3, x minus 3, x plus 5, x minus 5 only 4 cases you have to just go one by one and you might find some if it exists. It might not exist in which case, well, sorry there is no such a good solution. Alright so that's the first example which I wanted to present. The first way of solving equations of the higher order so you have to somehow guess in certain particular cases you can how to represent it as a product of two expressions of a lower order. Now the second approach which I might suggest you is the following. What if you can represent this particular equation of the case degree with a substitution. Let's say you can substitute something like some kind of a solution of some kind of expression sorry x squared plus 2x plus 3 for instance. You can represent it as some new variable y and if you can express this particular equation in terms of y but it will be of a different degree degree t which is smaller than this one then what you can do is you can try to solve this equation in terms of y and then having the value of y you can solve this equation in terms of x. Now I am not clear enough in this particular general case let me just express it in one particular example and it will be quite well understood what I mean. Let's say you have an equation x4 minus 2x square minus 3 equals to 0. Well this is an equation of the fourth order x is in the power of 4 but you notice that if you will substitute x2 equals to y in terms of y the equation looks like y2 minus 2y minus 3 is equal to 0. Well in this case it's very easily visible that x2 can be substituted and you will get a quadratic equation for y. Now how to solve this quadratic equation for y? Firstly I don't remember the formula so I will represent it in terms of a full square which is y2 minus 2y plus 1 minus 4 equals to 0 so I represent it minus 3 as 1 minus 4 and this is a full square this is y minus 1 square equals to 4 minus 1 square so y minus 1 absolute value is equal to 4 so it's y minus 1 is equal to 4 or y minus 1 is equal to minus 4 2 solutions to this right oh sorry it's square root square root of 4 so it's 2 2 and minus 2 2 and minus 2 ok which means y is equal to 3 and y is equal to 1 goes here minus 1 so we have 2 solutions for y and we know that x2 is equal to y so now we can find what the x is in terms of real numbers square root of 3 and minus square root of 3 will be solutions to x and in terms of real numbers this one doesn't produce any real solutions because x2 cannot be negative in terms of complex numbers if you are solving the problem in the demand of complex numbers you can additionally x2 is equal to minus 1 so it's plus i and minus i as solutions to original equation of the force order I have represented as this equation of the second order substituting x2 with y well obviously there are more different substitutions maybe it's not just straight forward x2 equals to y maybe it's something like x2 minus 1 is equal to y but again you have to guess this type of thing and that's not easy I understand that express certain x to certain degree some kind of a polynomial like x to the second degree plus something in terms of the substitution of the variables then you can reduce it you can reduce the solution of this guy this higher order equation 2 solving this particular equation which is of the lower order and then once you have obtained the y you can solve the next one in a little bit simpler way because it's in this case it's just a quadratic equation well basically these two methods one is representation as a product and another is a substitution of certain expression for x in such a way that it gives you a better handle over the equation this particular thing gives you a really good two methods to solve equations of the higher order and they are very specific these methods there is nothing general about them and they are working only in certain specific cases alright well that's basically all I wanted to talk about today I would like to take a look at the Unisor.com website there are many other interesting things there and as far as the higher order equations I'll try to provide certain examples and problems which some of them I will solve on the web and you are definitely invited to take a look at them thank you very much and good luck