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Thu, 22 Mar 2007
The most difficult question on the Algebra III exam presented the examinee with some intractable third degree polynomial—say x3 + 4x2 - 2x + 6—and asked for the sum of the cubes of its roots.
You might like to match your wits against the Algebra III students before reading the solution below.
In the three summers I taught, only about two students were able to solve this problem, which is rather tricky. Usually they would start by trying to find the roots. This is doomed, because the Algebra III course only covers how to find the roots when they are rational, and the roots here are totally bizarre.
Even clever students didn't solve the problem, which required several inspired tactics. First you must decide to let the roots be p, q, and r, and, using Descartes' theorem, say that
x3 + bx2 + cx + d = (x - p)(x - q)(x - r)This isn't a hard thing to do, and a lot of the kids probably did try it, but it's not immediately clear what the point is, or that it will get you anywhere useful, so I think a lot of them never took it any farther.
But expanding the right-hand side of the equation above yields:
x3 + bx2 + cx + d = x3 - (p + q + r)x2 + (pq + pr + qr)x - pqrAnd so, equating coefficients, you have:
Or, to take an example that we can actually check, consider x3 - 6x2 + 11x - 6, whose roots are 1, 2, and 3. The sum of the cubes is 1 + 8 + 27 = 36, and indeed -b3 + 3bc - 3d = 63 + 3·(-6)·11 + 18 = 216 - 198 + 18 = 36.
This was a lot of algebra III, but once you have seen this example, it's not hard to solve a lot of similar problems. For instance, what is the sum of the squares of the roots of x2 + bx + c? Well, proceeding as before, we let the roots be p and q, so x2 + bx + c = (x - p)(x - q) = x2 - (p + q)x + pq, so that b = -(p + q) and c = pq. Then b2 = p2 + 2pq+ q2, and b2 - 2c = p2 + q2.
In general, if F is any symmetric function of the roots of a polynomial, then F can be calculated from the coefficients of the polynomial without too much difficulty.
Anyway, I was tinkering around with this at breakfast a couple of days ago, and I got to thinking about b2 - 2c = p2 + q2. If roots p and q are both integers, then b2 - 2c is the sum of two squares. (The sum-of-two-squares theorem is one of my favorites.) And the roots are integers only when the discriminant of the original polynomial is itself a square. But the discriminant in this case is b2 - 4c. So we have the somewhat odd-seeming statement that when b2 - 4c is a square, then b2 - 2c is a sum of two squares.
I found this surprising because it seemed so underconstrained: it says that you can add some random even number to a fairly large class of squares and the result must be a sum of two squares, even if the even number you added wasn't a square itself. But after I tried a few examples to convince myself I hadn't made a mistake, I was sure there had to be a very simple, direct way to get to the same place.
It took some fiddling, but eventually I did find it. Say that b2 - 4c = a2. Then b and a must have the same parity, so p = (b + a)/2 is an integer, and we can write b = p + q and a = p - q where p and q are both integers.
Then c = (b2 - a2)/4 is just pq, and b2 - 2c = p2 + q2.
So that's where that comes from.
It seems like there ought to be an interesting relationship between the symmetric functions of roots of a polynomial and their expression in terms of the coefficients of the polynomial. The symmetric functions of degree N are all linear combinations of a finite set of symmetric functions. For example, any second-degree symmetric function of two variables has the form a(p2 + q2) + 2bpq. We can denote these basic symmetric functions of two variables as Fi,j(p, q) = Σpiqj. Then we have identities like (F1,0)2 = F2,0 + F1,1 and (F1,0)3 = F3,0 + 3F2,1.
Maybe I'll do an article about this in a week or two.