Sat, 01 Mar 2008
More rational roots of polynomials
Suppose you have a polynomial P(x) = xn + ...+ p = 0. If it has a rational root r, this must be an integer that divides p = P(0). So far so good.
But consider P(x-1). This is a different polynomial, and if r is a root of P(x), then r+1 is a root of P(x-1). So, just as r must divide P(0), r+1 must divide P(-1). And similarly, r-1 must divide P+1.
So we have an extension of the rational root theorem: instead of guessing that some factor r of P(0) is a root, and checking it to see, we first check to see if r+1 is a factor of P(-1), and if r-1 is a factor of P(1), and proceed with the full check only if these two auxiliary tests pass.
My notes conclude with:
Is this really less work than just trying all the divisors of P(0) directly?Let's find out.
As in the previous article, say P(x) = 3x2 + 6x - 45. The method only works for monic polynomials, so divide everything by 3. (It can be extended to work for non-monic polynomials, but the result is just that you have to divide everything by 3, so it comes to the same thing.) So we consider x2 + 2x - 15 instead. Say r is a rational root of P(x). Then:
So we need to find three consecutive integers that respectively divide 12, 15, and 16. The Britannica has no specific technique for this; it suggests doing it by eyeball. In this case, 2–3–4 jumps out pretty quickly, giving the root 3, and so does 6–5–4, which is the root -5. But the method also yields a false root: 4–3–2 suggests that -3 might be a root, and it is not.
Let's see how this goes for a harder example. I wrote a little Haskell program that generated the random polynomial x4 - 26x3 + 240 x2 - 918x + 1215.
That required a fair amount of mental arithmetic, and I screwed up and got 502 instead of 512, which I only noticed because 502 is not composite enough; but had I been doing a non-contrived example, I would not have noticed the error. (Then again, I would have done the addition on paper instead of in my head.) Clearly this example was not hard enough because 2–3–4 and 4–5–6 are obviously solutions, and it will not always be this easy. I increased the range on my random number generator and tried again.
The next time, it came up with the very delightful polynomial x4 - 2735x3 + 2712101 x2 - 1144435245x + 170960860950, and I decidedd not going to go any farther with it. The table values are easy to calculate, but they will be on the order of 170960860950, and I did not really care to try to factor that.
I decided to try one more example, of intermediate difficulty. The program first gave me x4 - 25x3 + 107 x2 - 143x + 60, which is a lucky fluke, since it has a root at 1. The next example it produced had a root at 3. At that point I changed the program to generate polynomials that had integer roots between 10 and 20, and got x4 - 61x3 + 1364 x2 - 13220x + 46800.
This is just past my mental arithmetic ability; I got 34884 instead of 34864 in the first row, and balked at factoring 61446 in my head. But going ahead (having used the computer to finish the arithmetic), the 17 and 19 in the first and last rows are suggestive, and there is indeed a 17–18–19 to be found. Following up on the 19 in the first row suggests that we look for 19–20–21, which there is, and following up on the 11 in the last row, hoping for a 9–10–11, finds one of those too. All of these are roots, and I do have to admit that I don't know any better way of discovering that. So perhaps the method does have some value in some cases. But I had to work hard to find examples for which it made sense. I think it may be more reasonable with 18th-century technology than it is with 21st-century technology.