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Sum of squares of polynomial roots

Video transcript
In the last video, we figured out how to figured out the sum of the roots of a polynomial. What we're going to try to attempt to do in this video is think about the sum of the squares of the roots of a polynomial. So let's think about that a little bit. So let's say that I have a second degree polynomial that looks like this. So it's x squared plus a1x, plus a2 and this is equal to 0. We saw-- and actually, I'll just redo some of what we did in the last video, because it's going to be useful over here. This thing is going to have two roots, r1 and r2. Two roots, and that tells us that this thing can be rewritten as x minus r1, times x minus r2 is equal to 0. And if you expand this out, you get x squared minus r1 plus r2, x, plus r1r2. So that's what you get right over there. Now, let's think about it. We figured out from the last video that r1 plus r2, which is right over here, that needs to be equal to the negative of this coefficient. This a1 over here is equal to the negative of r1 plus r2. So r1 plus r2 is equal to the negative of a1. So let's see if we can use that information, right over there, to figure out what-- let me do this in a new color-- to figure out what r1 squared plus r2 squared is. So let's see how we could get that. Well, one place that we could start off with, we could start off with r1 plus r2 squared. So r1 plus r2 squared, which is, of course, the same thing as negative a1 squared, which would be the same thing as a1 squared. a1 squared-- I could put a negative here, but a negative 1 times negative 1 is obviously a positive one-- this is going to be equal to r1 squared plus 2r1r2, plus r2 squared. So this has the sum that we care about in it. Let's call this sum 1-- we're just raising the roots to the first power. Let's call this sum 2. So this over here, we've written, essentially, that sum 1 squared-- that's sum s sub 1 squared is equal to-- this plus this is s sub 2-- is equal to s sub 2 plus 2r1r2. Now, can we figure out what r1r2 is based on looking at the actual original polynomial? Well, you look over here, if you just did an expansion-- of course, I forgot to write here that this is equal to 0-- r1r2 is essentially represented by a2 over here. So to take the first degree sum, or the roots raised to the first power sum, we looked at this coefficient. Now all of a sudden, we're involving the coefficient to the right of that. We're going two degrees below the degree of the polynomial. So this, right over here, is a2. So r1r2 is a2. That's a2. This right here-- this whole expression-- is a1 squared. So if we want to solve for this sum-- if we want to solve for s sub 2, we would get s sub 2 is equal to a1 squared-- that's this over here-- minus 2 times a sub 2. So we just figured out a fast way to figure out the sum of the squares of the root of just a second degree quadratic right now. So for example, if I were to give you-- let me give you something strange. So if I were to give you 7x squared minus pi x, plus e is equal to 0. And someone said, I want you to figure out the sum of the squares of the roots of this polynomial, first you want to make the coefficient in front of the x squared 1. So divide everything by 7. So you get x squared minus pi over 7 x, plus e over 7 is equal to 0. And then you could just use the result that we just found. The sum of the square of the roots-- is this s2 over here. Whatever the roots are, you square each of them, you take their sum, and it's going to be equal to a1 squared. So it's going to be pi squared over 49. I'm just squaring this right over here, or I could just square that. The negative doesn't matter what I'm squaring. minus 2 times e over seven. And that's about as simplified as you could get this expression. So it's pi squared over 49, minus 2 e over 7, which by itself a pretty neat result. Because it would be very hard to find the roots of this. You'd get some crazy numbers, and then you'd have to square them and sum them. But we just figured out a very fast way of doing it. Now, let's see if we can extend this to the third degree. And it actually turns out you can-- not only to the third degree you can extend this to the nth degree. And I might do an induction proof it gets a little bit messier than when you just take the first degree sum, right over here, where you take the sum of the roots to the first power. But let's just see if it works for the third power. So let's say, I have the polynomial x to third plus a1x squared, plus a2x, plus a3 is equal to 0. We have three roots. r1, r2, r3, which tells us that we can rewrite this polynomial as x minus-- or this equation as x minus r1, times x minus r2, times x minus r3 is equal to 0. Now we already figured out what this part right here is, it's this thing over here. We already multiplied it out. It's this thing. It is this over here. So essentially, to get this whole expression, we just have to multiply this times this thing over here. We can just use the distributive property-- multiple this thing times each of these terms. So first, let's multiply this whole thing by x. So you get x to the third, minus r1 plus r2 times x squared, plus r1r2x. So I just multiplied this thing over here times x. Now I can just multiply this thing up here times negative r3. So negative r3 times x squared. Negative r3 times this term over here, it will give us-- well it's going to be a negative times a negative, so it's going to be a positive-- so it's positive r3 three times this. We can distribute them. We're going to get r1-- so let me make sure I'm doing. Negative r3 times this over here is going to be negative-- let me just write it like this. I don't want to skip steps. So it's going to be-- the negative and the negatives cancel out. So negative r3 times r1 plus r2 x. And then finally, you have the negative r3 times r1r2. So minus r1r2r3. And all of that's going to be equal to 0. But let's simplify it. So this thing is going to be equal to x to the third minus r1 plus r2 plus r3-- we did that in the last video. That shows us that the sum of our roots is this second coefficient here, the coefficient on the second degree term, x squared. And then, we go over here. We have r1r2, x. Then you have r1, r3 x. Plus r1r3-- I'm just distributing this. And then you have r3 times r2. So plus r2r3. All of that times x, and then minus r1r2 and r3. Now if we want to take the sum of the squares of the roots. So if we want to take r1 squared plus r2 squared plus r3 squared, well, we could try to do the same thing. We could take r1 plus r2 plus r3, which we know how to calculate-- we can square that. And we're essentially going to get-- When you square this you're going to get-- let me see if I can write it neatly. This is going to be equal to r1 times-- So I'm just going to-- let me just rewrite it. This is equal to-- just to clarify, this is equal to r1 plus r2 plus r3, times r1 plus r2 plus r3. So it's going to be-- I could start over here. r1 times r1 is r1 squared, plus r1 times r2-- so r1r2-- plus r1 times r3. And then, plus-- now we're in the r2. r2 times r1 is plus r1r2-- same thing, I'm just switching the order so it looks the same. r2 times r2 is r2 squared. And then, r2 times r3 is plus r2r3. And then let's go to the r3, r3 times r1 is just another r3r1-- I just switched the order. r3 times r2 is another r2r3. And then, r3 times r3 is r3 squared. So what did we just get here when we squared just the straight up sum? You got r1 squared plus r2 squared plus r3 squared, plus-- you essentially got all of the different combinations, but you got them twice. You have two r1r2, so you have 2 times r1r2. Then you have two r1r3, so plus r1r3-- I have the 2 out front. Plus r2r3. And this over here is equal to-- remember what we did-- this is equal to r1 plus r2 plus r3 squared. So we got essentially the same result. Remember, this thing right over here-- let's make things clear. This thing over here r1-- all the combinations of the products of the roots. If you look over here, that's exactly what this thing is over here, which must be what our a2 coefficient is. So this is our a2 coefficient. This r1 plus r2 plus r3, that is-- we figured out multiple times-- that's this thing right over here, which is going to be a1. So this is going to be a negative, or it's equal to negative a1. But you take negative a1 squared, and that's the same thing as a1 squared. So this is a1 squared. And this is the sum that we care about. So we get r1 squared plus r2 squared plus r3 squared is equal to this business over here. It's equal to a1 squared minus 2 times a2. So we got the exact same result that we got for the second degree case. And it actually turns out this will be true of any degree. I haven't proven to you yet, although I can make an induction argument now-- we've proven some base cases. But just to make it clear how to apply it, I give you crazy polynomial. I give you-- Let's do a third degree case, let's say, 10x to the third minus 5 x squared, plus 7x-- I'm making this up on the fly-- plus 2 is equal to 0. And if I were to ask you the sum? If I were to ask you r1 squared plus r2 squared plus r3 squared-- If I were to ask you the square of the sum of the roots, you just take-- well, let me be careful. You have to make sure that you have a one in front of the highest degree term. So this thing over here, has to be rewritten. Divide both sides by 10 x to the third minus 5 over 10, so minus 1/2 x squared, plus 7 over 10 x, plus 2 over 10, which is one fifth is equal to 0. So I just divided both sides by 10, and yup, I did that part right. And now, we can apply this. I almost made a careless mistake. You have to have a one coefficient out here, at least, the way that we've derived this. So this sum is going to be equal to a1 squared. So it's this-- 1/2 squared. So a1 is actually negative 1/2, but negative 1/2 half squared is just one fourth. So it is going to be one fourth minus 2, times a2. So minus 2 times 7 over 10. So this is equal to one fourth minus 14 over 10. We could find a common denominator here. This is the same thing as 5 over 20 minus 14 over 10. And this is equal-- or sorry, minus 28 over 20, I should do, which is equal to negative 23 over 20. And you might say, hey, Sal, hang on. I'm taking the square of a bunch of roots and adding them up, and I'm getting a negative number. I thought squares-- I thought y squared numbers I always get a positive number that would be true if you're dealing with real numbers, but remember the words of a polynomial can be complex and complex numbers squared can be negative numbers. So this obviously is involving some type of complex roots. But we know that the sum when you take the sum of the squares of those complex roots you get negative 23 over 22.