Slide and Divide

Am I the only one who didn't know the slide and divide method?

If you've read my book, or the digital filter courses in this blog, you know that I know a thing or two about math in general, and polynomials in particular. But I hadn't heard of "Slide and Divide" until last week. The excellent math teacher at one of the places where I volunteer as a tutor, e-mailed me to Google on it and learn how to do it. I did, and have now been able to help several of her students with it.

You could also Google it, but I'll show it here for those who haven't seen it. To begin with, let's make an example that we know will be factorable:

(3x + 2)(5x - 4) = 15x^2 -2x -8. (Did I do that right?)

The first thing we do is replace c (the -8 in this case) with the product of a and c = -8 x 15 = -120.

Now, we write a new polynomial: x^2 -2x -120. Note that it is NOT equivalent to the one we're trying to factor. It's just a new polynomial that will help us with our job. So let's factor it. It's easier since a = 1.

(x  + 10)(x - 12)

Now, we have to divide the roots by the 15 by which we multiplied c above. (Repeat caveat about equivalence.)

(x + 10/15)(x - 12/15) = (x + 2/3)(x - 4/5)

Now, in the case of fractional roots such as these, we multiply each factor by its root's denominator: 3(x + 2/3) and 5(x-4/5). (Repeat caveat about equivalence.)

Finally, we simplify those: (3x + 2)(5x - 4)

Ordinarily we could FOIL it back out to check, but there's no need since we started with the same factors above.

Pretty slick right? The fact that it violates equivalence along the way was bothersome at first, but that's really just an accounting problem. One keeps track of what's equivalent and not, and uses the rest as intermediate calculations. In the end it's easy to demonstrate that the result is correct.