

Algebra techniques that don't work, except when they do

In Problems I Can't Fix in the Lecture Hall, Rudbeckia Hirta describes the efforts of a student to solve the equation 3x2 + 6x - 45 = 0. She describes "the usual incorrect strategy selected by students who can't do algebra": 3x2 + 6x - 45 = 0 3x2 + 6x = 45 x(3x + 6) = 45 She says "I stopped him before he factored out the x.". I was a bit surprised by this, because the work so far seemed reasonable to me. I think the only mistake was not dividing the whole thing by 3 in the first step. But it is not too late to do that, and even without it, you can still make progress. x(3x + 6) = 45, so if there are any integer solutions, x must divide 45. So try x = ±1, ±3, ±5, ±9, ±15 in roughly that order. (The "look for the wallet under the lamppost" principle.) x = 3 solves the equation, and then you can get the other root, x=-5, by further application of the same method, or by dividing the original polynomial by x-3, or whatever. If you get rid of the extra factor of 3 in the first place, the thing is even easier, because you have x(x + 2) = 15, so x = ±1, ±3, or ±5, and it is obviously solved by x=3 and x=-5. Now obviously, this is not always going to work, but it works often enough that it would have been the first thing I would have tried. It is a lot quicker than calculating b2 - 4ac when c is as big as 45. If anyone hassles you about it, you can get them off your back by pointing out that it is an application of the so-called rational root theorem. But probably the student did not have enough ingenuity or number sense to correctly carry off this technique (he didn't notice the 3), so that M. Hirta's advice to just use the damn quadratic formula already is probably good. Still, I wonder if perhaps such students would benefit from exposure to this technique. I can guess M. Hirta's answer to this question: these students will not benefit from exposure to anything. [ Addendum 20080228: Robert C. Helling points out that I could have factored the 45 in the first place, without any algebraic manipulations. Quite so; I completely botched my explanation of what I was doing. I meant to point out that once you have x(x+2) = 15 and the list [1, 3, 5, 15], the (3,5) pair jumps out at you instantly, since 3+2=5. I spent so much time talking about the unreduced polynomial x(3x+6) that I forgot to mention this effect, which is much less salient in the case of the unreduced polynomial. My apologies for any confusion caused by this omission. ] [ Addendum 20080301: There is a followup to this article. ]

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