Sum of polynomial roots (proof)

What I want to do in this video, is figure out if there's any fast way to figure out the sum of the roots of any polynomial. And actually, there actually is. And so that's why I'm doing this video. So let's start with a second degree polynomial. So let's say it's x squared plus a1x, plus a2 is equal to 0. So this is just a standard set quadratic equation, right here, second degree equation. And you might be saying, hey, but you put the coefficient on the x term being equal to 1. And in general, you can always convert any polynomial. And I'll do it with a second degree, but if you have ax squared plus bx plus c is equal to 0, you can just divide both sides of this equation by a. And you're going to have something that's in this form, where the coefficient on the x squared term is going to be equal to 1. And you can do that with any polynomial. You can do that with any polynomial that's set equal to 0. So with that out of the way, let's think about what the sum of the roots of this are going to be. So this is a second degree polynomial. It's a quadratic equation. So it'll have two roots. They could be real or complex. So let's call the roots r1 and r2. And that tells us, since these are roots, that x minus r1 times x minus r2 is going to be equal to 0. And if we multiply this out, we get x times x is x squared. x times negative r2 is negative r2x. And then we have negative r1 times x, so negative r1x. And then we have negative r1 times negative r2, which is plus r1r2 is equal to 0. Then we can simplify this middle term a little bit. It becomes x squared minus r1 plus r2 x plus r1r2 is equal to 0. So let's think about what the sum of the roots are. So based on what we're looking at right over here, what is r1 plus r2? Well, we see it right over here when we multiplied out these two expressions. r1 plus r2 is the negative of this second coefficient right over here. Or it's the negative of the coefficient on the first degree term. So if we look at the first degree term over here, a1 must be the same thing as negative r1 plus r2. Or another way to think about it is, r1 plus r2 must be equal to negative a1. So that wasn't too bad for the second degree case. Let's try the third degree case. Let's see, if we have x to the third plus a1x squared plus a2x, plus a3 is equal to 0. Let's think about what the sum of its roots might be. Well, this guy is now going to have one more root. I'm not saying it's the exact same equation, but let's say we're keeping everything general. So now we could say we have roots r1, r2, and r3. Or we could say that x minus r1 times x minus r2 times x minus r3 is equal to 0. Now we could multiply all of this out, but we already figured out what x minus r1 times x minus r2 is. It's this business over here. So we just have to multiply this times x minus r3. And actually, I'm not even going to do the full expansion, because we have a hunch here that it always deals, it seems, with the coefficient on the term that's one degree lower than the degree of the polynomial. This was a second degree, we looked at the coefficient on the first degree term. Maybe to find the sum, we only have to look at this coefficient over here. So I'm only going to figure out this product up to this point, and then we can just ignore the rest of it and see if get something that is useful. So let's do the expansion. Let's multiply this right over here times this over here. And what do we get? To get the x to the third term, the only way to get that is to multiply this x times this x squared. So we're going to get x to the third. And that's the only way to get the x to the third term. That's the only way to get the third degree term. Now how do we get this term over here? How do we get-- let me do this a new color-- how do we get the x squared term over here? Well, we could multiply this x times this over here, because the x times the x is going to give us x squared. So it'll be negative r1 plus r2 x squared times that term. And then if you multiply this x times that, you'll get r1r2 x, and all of that. But I'll just write, so on and so forth. And then, what other ways can we get an x squared term here? There's no other way to get x to the third term, but what other ways could we get an x squared term? Well, when we multiply this times this, you have the negative r3. Let me do this in another color. You have the negative r3 three times this. Your obviously going to take this term multiplied by everything. And then add it to this term multiplied by everything. But we only care about the things-- I did this just to show you that it's there-- but I only care, because we have a hunch that what matters is the coefficient on the x squared term. So we just want to see, how do you build the x squared term in our, I guess, our expanded polynomial? The other way to get an x squared is to multiply the negative r3 times this x squared over here. So this is going to be negative r3 times x squared. And then when you multiply this times this, you're going to get something else. Then when you multiply this times this, you're just going to get the negative product of all the roots. So that's just something else. And so when we add everything together you get x to the third. And then what's the sum of these two things? It's negative r1 plus r2 plus r3 x squared. And then you're just going to have a bunch of other stuff that we didn't take the time to add. But it looks like our hunch paid off, because what is now the sum of the roots? r1 plus r2 plus r3. What is that equal to? Well, that's sitting right over here. That's equal to the negative of this coefficient right over here. That's once again equal to negative a1. It's equal to the negative of the coefficient on the degree term that's one less than the degree of our polynomial. So we already see a pattern. Let's say if we can prove it generally. So we've already proven, essentially, two base cases. We proven it for a degree of two, a degree of three. Let's assume that it's true for degree of n, and then we can prove it for degree of n plus 1. And essentially, this will be a proof by induction for polynomials of any degree. So let's just say we have an nth degree polynomial. So let's just assume this. So we're going to assume this step, and this gives a little proof by mathematical induction practice. So let's just assume that we have a polynomial x to the n-- it's nth degrees-- so x to the n, plus a1x to the n minus 1, and it just goes on and on and on all the way down to the 0th degree term. It has roots, obviously, r1, r2, all the way to r sub n. And we're going to assume that r1 plus r2 plus all the way to r sub n is equal to negative a1. Or another way to think about it, if we were to multiply x minus r1 times x minus r2, and just keep multiplying all the way to x minus r sub n is equal to 0, this should give us x to the n minus r1, plus r2, plus all the way to rn x to the n minus 1, and then plus a bunch of other stuff that we're not going to calculate. So this tells us this. That this coefficient right over here is going to be equal to this right over here. So that is what we are going to assume. So now let's think about-- and we're not assuming this part-- given that, let's think about the situation where we have a polynomial x to the n plus 1, plus a1x to the n, plus-- and it just keeps going. You get all the degree terms there. So now this is going to have the roots r1, r2, all the way to rn. And then it'll have rn plus 1. It'll have these n roots, and then it'll have an rn plus 1. So essentially, that tells us that we're going to have, essentially, this product times x minus rn plus 1 is equal to 0. Let me write it out. I don't want to skip steps. So this tells us, if these are all the roots, that x minus r1 times x minus r2, blah, blah, blah, all the way to x minus rn times x minus rn plus 1 is going to be equal to 0. It has one more root than the previous example, than the one where we made the assumption. So if we were to expand this thing out, how can we do that? Well, this thing, if we were to multiply it out, it's just this thing over here times this new binomial, because this thing over here is-- let me set it up, let me use color coded-- this thing, let me do a different color. This thing over here we already established. We assumed is this thing over here, which is this thing over here. So if we want this entire expansion, we just have to multiply this times this. So how can we get an x to the n plus 1 term? How can we get to the x to the n plus 1 term? Well, there's only one way to get the x to the n plus 1 term. And that's by multiplying this x times this x to the n. So that will give us x to the n plus 1. That's the only way to do it. That's the only way this highest degree term. Now how can we get-- let me do it in purple-- how can we get at the x to the n term? Well, we can multiply this x times this second term over here. So if you multiply this x times this business over here, you're going to get negative r1 plus r2, plus all the way to rn. x times x to the n minus 1, is just going to be x to the n. And then obviously, you're going to multiply this x times all the other terms, so you're just going to get a bunch of other stuff. So that's how we can get these two terms using the x. And then can we get either of these degree terms using the r sub n plus 1? Well, we can multiply r sub n plus 1, or negative r sub n plus 1 times x to the n. And then we'll also get a similar degree term. So we'll have negative r sub n plus 1 times x to the n. And then obviously, we're going to multiply this times all of the other characters in this polynomial. So you're going to have all this plus so on and so forth. But this will suit our purposes, because when you add these two things, what do we get? We get x to the n plus 1, and then we get minus-- well, this and this are the coefficients on the x to the n term. So this is going to be minus r1 plus r2 plus all the way to plus rn. And then this is minus rn plus 1. We have the minus out front. So this is plus r sub n plus 1 x to the n. And then we're going to have a bunch of other lower degree terms that don't matter. But we've just proven our case. What is r1 plus r2 plus all the way to plus r sub n plus 1? What is this equal to? Well, in our expansion we have it right over here. That's equal to that over there. That is equal to the negative of this coefficient. It's equal to the negative of a1. Now the reason why this works out no matter what-- you say, hey, if I have complex roots, how does it always end up being a real number? And that's because the imaginary parts cancel out when you take their sum. So let's just apply it. We've proven it. We've proven, actually, two base cases. We've proven it for a degree 2, for degree 3. And then we showed that if it's true for any degree n, then it's definitely true for any degree n plus 1. So we know if it's true for degree 3, then it's definitely true for degree 4, which means it's definitely true for degree 5. So A mathematical induction is kind of this domino proof that if it's true for 3, which makes it true for 4, then assuming it's true for 4 makes it true for 5. And it just keeps going to any n. So in general, if someone gives you a polynomial-- let me think of some crazy polynomial. If I were to give you x to the seventh, minus pi x to the sixth plus ex to the fifth, minus square root of 2 x to the fourth, minus-- I won't write out all the terms. So minus 3 is equal to zero. If someone were to give you this polynomial, and they were say, what do the seven roots of this polynomial add up to? You say, oh, well, the sum of these roots is going to be the negative of this coefficient right over here. So r1 plus r2 plus all the way to r7, all seven roots are going to add up to the negative of this coefficient. They're going to be equal to pi. And one other thing, you might be saying, hey, wait. Every polynomial you did had a 1 coefficient in front of the highest degree term. You know, not all polynomials have that. And the answer there is-- let's say, if I were to give you a polynomial, I don't know, 7x to the fifth, minus 6x to the fourth, plus, I don't know, pi x to the third, plus a bunch of stuff other stuff is equal to 0-- and you wanted to find the sum of this polynomials roots. You'd first divide everything by 7. So it becomes x to the fifth, minus 6 over 7x to the fourth, plus pi over 7x to the third, plus so on and so forth, is equal to 0/7 which is just 0. And now you look at this polynomial, well the sum is going to be the negative of this coefficient here. So it's going to be the sum of all five roots. Or it's going to be 6/7. Anyway, hopefully you enjoyed that.