Factoring and long division

This morning, I’ve been watching YouTube videos. I started with Tarleen Kaur’s video on Middle Term Splitting. What I find interesting about Kaur’s Chapter to Chapter videos is that, because she’s a student in India, her methods are often different from those I’m familiar with. That’s the case in this video as well. I haven’t heard of middle term splitting at all, but then, we’ve come to use a bunch of tricks for factoring that I wasn’t taught. When I was in school, we were taught to memorize the quadratic formula and not muck around with trying to find factors and such.

The strategy Kaur uses to find two numbers that multiply to \(ac\) and add to \(b\) is the opposite of what I’ve seen taught, and I’m curious if my students would have a better time with it. The method I’m used to: List all the factor pairs of \(ac\) and then see which add to \(b\). Hers is: Start with two numbers that add to \(b\). Adjust them until their product is \(ac\). For instance, if \(b = 17\) and \(ac = 30\), start with \(\{16, 1\}\). \(16 + 1 = 17\), but \(16 \times 1 = 16\), so that doesn’t work. Subtract one to the first value, add one to the second value to get \(\{15, 2\}\). \(15 + 2 = 17\) and \(15 \times 2 = 30\), so there’s the proper set. I like this because there’s much less guesswork involved. If we’re going to use a procedure, better it be a procedure that’s systematic. (Naturally, the quadratic formula is also a procedure, but one students struggle with because of the radicals and the plus/minus.)

From there, I jumped a few videos and wound up at MathBFF’s video on doing long division with polynomials. What struck me here is the language. When dealing with fractions, we speak of “numerator” and “denominator”, and I’ve been continuing that language when reviewing fractions. But Nancy (I wish she provided a last name) uses “divisor” and “dividend”. This illustrates, properly, the conceptual shift that ought to be occurring in high school, if it hasn’t already. At the same time, though, it juxtaposes the two perceptions of what numbers are in the first place.

I have a longer piece brewing on the difference between numbers-as-quantities and numbers-as-measurements, but the applicable portion here is that the numerator/denominator language represents numbers in terms of measurements, while the divisor/dividend language represents numbers in terms of quantities (even if, in practice, that’s not how each term set is used).

My biggest aha! of the morning so far, though, was tucked into this silly, embarrassing video, in which working Millennials are asked to simply do long division problems. Three out of four clearly have no idea what they’re doing (although one of them does eventually start writing in the traditional format, at least). But around 4:00, Matt Lieberman says that, if \(6 \times 8 = 48\), then \(6 \times 800 = 4800\). He’s doing this to help his partner, and those parts of the video are a great clinic in how a top student could encourage and guide a struggling student.

As Lieberman implicitly demonstrates throughout the rest of his portion of the video, what we’re doing in long division is successively subtracting multiples of \(10^k\) from our dividend until our remainder is zero. In this case, the question was \(\frac{5166}{8}\). 5166 – 4800 = 366, 366 – 320 = 46, 46 – 40 = 6, 6.0 – 5.6 = 0.4, and 0.40 – 0.40 = 0.

This is obscured in how we normally write long division because we leave off the extraneous zeros (as on the left), and talk about bringing numbers down. This is the procedure, and I learned the procedure just fine. I’m great at long division and am confused at people who struggle with it. What I apparently never did, at least not to the point of recognizing it, was internalize the concept.

Now, it’s possible that I did in fact learn it with all the zeros once upon a time (as on the right), and that at some point in elementary school we were told we could skip all those. This is the concept, after all: We’re starting with some huge pile of things (5,166 of them, to be exact), then taking away¬†groups of items in convenient chunks. We don’t have to take multiples of ten, but that’s easiest. We could take away multiples of seven if we really wanted to. It doesn’t matter, but because it doesn’t matter, we go with what’s easiest.

It’s also possible that this falls into the category of those things that adept mathematicians simply take for granted (as in Nancy’s video, where she makes the point multiple times that zero times anything is zero, but fluidly glosses over \(1 \times x = x\)). This is something to be persistently mindful of, particularly if concept takes a backseat to procedure.

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