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# 2003 AIME II problem 9

Video transcript

- [Voiceover] Consider the polynomials, P of X is equal to X to the
sixth 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 one. Given that Z one, Z
two, Z three, and Z four are the roots of Q of X equals zero, find P of Z one plus P of Z two, plus P of Z three, plus P of Z four. So a good place to start, and you might have already realized this, that if these guys are the
roots of Q of X equals zero, that means that Q of Z one is equal to Q of Z two, which is equal to Q of Z three, which is equal to Q of Z four, which is equal to zero. They're the roots, they are
solutions to this equation here. Now given that, and we want to evaluate P of Z one plus P of Z two, 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 one or Z two or whatever would evaluate to zero. So let's try to do that over there. So the first thing I'm gonna try to do, I'm gonna try to express
these higher degree terms in terms of Q of X. And to do that, if I multiple
Q of X times X squared, this X to the fourth will
become an X to the sixth, this X to the third will
become an X to the fifth. It won't be exact, but we might be able to adjust 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 multiplied 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're gonna 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 X 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 gonna 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're 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 to the 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 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 with 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, you add a 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, it's equal to, no,
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. 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 one. So this is going to be zero, obviously. Q of R is zero, so this
whole term is zero. This whole term is zero. So P of R is going to be equal to R squared minus R, plus one. 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 one is equal to Z one squared, minus Z one plus one. P of Z two, P of Z two is equal to Z two squared, minus Z two plus one. I could, I'll go all four. P of Z three. P of Z three is equal to Z three squared, minus Z three plus one. And then finally we have P of Z four. P of Z four, in that slightly
different shade of orange, is Z sub four squared
minus Z sub four plus one. Now remember our problem. We just want to find
the sum of these things. 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 four there, but now I have to sum
all the roots of Q of X and I 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 kind of quirky
properties of polynomials from the people who don't study quirky properties of polynomials, but the last two videos I just recorded in the competition math playlist, and I 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 gonna
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 one plus Z two plus Z three plus Z four. Now, two videos ago, I think it was, we show that the sum of
the roots of a polynomial, and this 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 equal the negative of that coefficient. So negative negative one, so it's the negative of negative one, which is just plus one. 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 one squared
plus Z two squared plus Z three squared plus Z four squared. Now in the last video where I talk about the sum of the squares, I didn't prove it for
the fourth degree case. It's actually pretty
easy to extend it from the third degree case
that we did in that video to the fourth degree and it's little messy to do it generally do
the whole induction thing that I did with the straight up sum, not the sum of the squares. But hopefully you see, or at least you understood how to apply it and if you're interested,
you actually could extend it to the fourth power or 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 one less than the highest degree,
so it's going to be, so this is a negative one. So it's going to be equal
to negative one squared, negative one squared minus two times the coefficient on the degree below that. So minus two times the
coefficient over here. So over here, so we're looking, we
started at this degree, now we're going even one lower than that. Two lower than the
degree of the polynomial. So we go to the X squared term. This coefficient is negative one. So minus two times negative one. This is equal to one, one plus two, which is equal to three. So this thing over here in green when you take the sum of the squares, it's equal to three. And so when you take the total sum of P evaluated at each of
those possibly complex numbers, we get three minus two
plus four is equal to six. And we're done.