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Current time:0:00Total duration:9:56

2003 AIME II problem 9

Video transcript

consider the polynomials P of X is equal to X to the 6 minus X to the fifth minus X to the third minus x squared minus X and Q of X is equal to X to the fourth minus X to the third minus x squared minus 1 given that Z 1 Z 2 and Z Z 3 and Z 4 are the roots of Q of x equals 0 find P of Z 1 plus P of Z 2 plus P of Z 3 plus P of Z 4 so a good place to start and Union this you might have already realized this and if these guys are the roots of Q of x equals 0 that means that Q of Z 1 is equal to Q of Z 2 which is equal to Q of Z 3 which is equal to Q of Z 4 which is equal to 0 there the roots there are solutions to this equation here now given that and we want to evaluate P of Z 1 plus P of Z 2 it seems like it would be useful if we can express P of X in terms of Q of X because then when we evaluate these the parts that would be Q of Z 1 or Z 2 or whatever would evaluate to 0 so let's try to do that over there so the first thing I'm going to try to do I'm going to try to get Express these higher degree terms in terms of Q of X and to do that if I multiply Q of x times x squared this X to the 4th will become an X to the 6 this X to the 3rd we would come an X to the fifth it won't be exact but we might be able to adjust it a little bit to get something right over here so let's try to do that so let's think a little bit about what let's think about what x squared x squared times Q of X is going to look like that's going to be x squared times X to the fourth that's X to the sixth minus X to the fifth minus X to the fourth minus x squared just multiply it all of these terms times x squared now that isn't exactly what we have up here for P of X in order to get it exact in order to get it exact we need to what are we going to have to do let's see we don't have an X to the fourth up here so we have to get rid of this X to the fourth so let's add X to the fourth to both sides so let's add X to the fourth to both sides of this equation so that would get rid of an X to the fourth and let's see we have No to the third here we do have an X to the third there so let's subtract X to the third from both sides of this equation so I'm going to subtract X to the third from both sides of this equation and then we do have a negative x squared we do have a negative x squared over here and then we have a negative x here I don't see a negative x over here so let's subtract a negative x from both sides so negative X this is all for the left-hand side and so we are left with the left-hand side of the equation we get x squared times Q of X x squared times Q of X plus X to the fourth plus X to the fourth minus X third minus X is equal to what's this equal to it's equal to X to the sixth I didn't want to use that shade of purple let me use the blue one it's equal to X to the sixth minus X to the fifth these guys cancel out minus X to the third minus X to the third minus x squared minus x squared and then you have a minus X there minus X which is exactly this is we were actually able to completely construct P of X this is equal to P of X it's equal to all of this business on the left hand side now we could just stop there but it looks this part right here this part right here looks tantalizingly close to Q of X as well in fact it is that part right there using the exact same process would be Q of X just the green part it's Q of X what these things gotten rid of so we would have to add an x squared so it's Q of X plus x squared plus x squared plus one right if you start with Q of X you add an x squared out of one these two will go away you just have an X to the fourth minus X to the third so now we can rewrite P of X we can rewrite P of X I'll do it right over here P of X is equal to x squared times Q of X these people do know that's not the same blue it's equal to x squared times Q of X plus this business plus Q of X plus x squared plus one and then we have this last negative X or last minus X right over there now when we evaluate if I say if I say that R is a root R so if R is a root of Q or if I write Q of R is equal to zero then what is P of R P of R is going to be equal to x squared times Q of R plus Q of R plus R squared I'll switch these minus R plus 1 so this is going to be 0 obviously Q of R is 0 so this whole term is 0 this whole term is 0 so P of R is going to be equal to R squared minus R plus 1 now obviously all of these guys are roots so we can use that same logic we can use that same logic we can say P of let me do a new color here a new color is warranted P of Z 1 is equal to Z 1 squared minus Z 1 plus 1 P of Z 2 P of Z 2 is equal to Z 2 squared minus Z 2 plus 1 I could I'll go all for P of Z 3 P of Z 3 is equal to Z 3 squared minus Z 3 plus 1 and then finally we have P of Z for P of Z 4 in a slightly different shade of orange is Z sub 4 squared minus Z sub 4 plus 1 now remember our problem we just want to find the sum of these things we just want to take we just want to take the sum so this sum is going to be the sum of these expressions and you might be saying Sal we still don't know what these sums are I could sum all the ones I'll get a 4 there but not have to sum all the roots of Q of X and have to sum all the squares of the roots of Q of X and this is not a simple thing and frankly it's not taught in most algebra classes or precalculus classes this is a problem to really separate the people who study who study kind of quirky properties of polynomials from the people who don't study quirky properties polynomials but the last two videos I just recorded in the competition math playlist and actually added them to the algebra playlist as well tell you how to find the sum of the roots of a polynomial or the sum of the squares of the roots of a polynomial so let's think about this a little bit so let's add up all the ones that's easy we get four now we're also going to have to add up all of when you add up all of these guys over here let me do this in a new color when I add up all of these guys over here it's going to be negative Z 1 plus Z 2 plus Z 3 plus Z 4 now two videos ago I think it was we show that the sum of the roots of a polynomial and this is the sum of the roots of Q of X that that is equal to the negative of the coefficient on the second to highest degree term or I should say that on the degree term that's one less than the highest degree term so if we go to Q of X we go to Q of X it's a fourth degree polynomial the sum of the roots to find the sum of the roots we go one degree less than that so we look for the third degree term that's this over here the coefficient on the third degree term over here is negative one it has a negative one coefficient so the sum of the roots the sum of the roots this over here is going to be called the negative of that coefficient so negative negative one so it's the negative of negative 1 which is just +1 and of course we have a negative out front here so this whole thing is going to evaluate to negative one so that whole business evaluates to negative one now we have to take the sum of the squares we have to do all of this over here we have to do all of this over here the sum of Z 1 squared plus Z 2 squared plus Z 3 squared plus Z 4 squared now in the last video where I talked about the sum of this question I didn't prove it for the fourth degree case it's actually pretty easy to extend it from the third degree case we did in that video to the fourth degree and it's a little messy to do it generally do the in to hold induction thing that I did with the straight up so I'm not the sum of the squares but hopefully you see the the or at least you understood how to apply it and if you're interested you actually could extend it to the fourth power to the nth power but this is going to be equal to the coefficient on I should look at Q of X the coefficient on the degree 1 less than the highest degree so it's going to be so this is the negative 1 so it's going to be equal to negative 1 squared negative 1 squared minus 2 times the coefficient on the degree below that so minus 2 times the coefficient over here so over here so we're we're looking we started at this degree now we're going even one lower than that to lower than the degree of the polynomial so go to the x squared term this coefficient is negative 1 so minus 2 times negative 1 this is equal to 1 1 plus 2 which is equal to 3 so this thing over here in green when you take the sum of the squares it's equal to 3 and so when you take the total sum of P evaluated at each of those pop possibly complex numbers we get 3 minus 2 plus 4 is equal to 6 and we're done