### Next Stop: Multiplicities

So somehow it finally dawned on me. There’s some major sticking point whenever the subject of the connection between the *roots* of a polynomial function (*f*, say… it’s always handy to have *names* for things, after all…) and its *factors* comes up: so this connection should be mentioned right up front as clearly as possible.

This last bit is what somehow doesn’t seem to have occurred to me until, to be embarassingly specific about it, right in the middle of the *third time* through the material this week (I’ve got three different 148 classes this quarter): there I was, finally boxing off the display I’d wanted all along (as I’ll always do *eventually*; I’m more or less convincing myself here it should be one of the first displays of its lecture).I refer, if you must insist on some *math* with the navel-gazing, to the proposition that “*R* is a **root** of *f* if and only if *(x-R)* is a **factor** of *f*“. This rather innocuous looking fact gets pretty close to the Heart of the Matter (which, for us here now, is, what else, the Fundamental Theorem of Algebra). Unfortunately, however, it’s not stated precisely enough for me to relax just yet.

First of all, in the proposition I’ve just caused to appear between quote marks, one has implicitly assumed that *f* is not only a polynomial, but a polynomial in the variable *x*. Such assumptions are quite often harmless, but sometimes (for instance, if *x* is already being used with another meaning in the problem we’re working on) they can lead to confusion.

Suppose , then. That was painless enough, right? The code can be pronounced “*f* is a polynomial, in the variable *x*, with Real coefficients”. That was the only handwave that actually *bothered* me, but *I* certainly seem to be having a good time and we can sure be more precise about what’s going on if we *want* to…

Let *R* be a root of *f*, then. This means that *f(R)=0* (and it means this *first of all*: we are invoking the very *definition* of “root” here; the reason I’ve digressed to say so is that the importance *of* definitions seems to be very ill-understood by quite a few of the math laity and so I’m seizing an opportunity to stress it). Let’s assume further that *f* has degree *n* (one has, ideally, already defined a *polynomial* function in *x*, of *degree* *n*, as equal [for all values of *x*] to

,

where “the *A*‘s are Real constants”—we might instead say here that “, for “… for example if there were some need to be more concise, or more *pre*cise, or maybe because the symbols are *astonishingly beautiful* [and, with sufficient practice, reveal themselves to be *much easier to understand* than mere words…]— and ).

Where was I? Oh yes. Something about a polynomial, *f*, of degree *n*, having the root *R*. I claim that there is then a polynomial, *g*, having the property that

for all values of *x*. And that’s (almost) it: “if *R* is a root, then *x – R* is a factor”.

This claim isn’t *obviously* true. What’s much easier to see is the converse statement: “if *x-R* is a factor, then *R* is a root” (“plug in” *x = R* on the equation [*f(x) = (x-R)g(x)*] *defining* “*x – R* is a factor”; done). We’ll look at the harder direction soon (in 148… *maybe* in the blog). For now, it’ll content me—and serve you well!—if you take both directions for granted (though of course if any of this is unfamiliar, one will wish to look at a few *examples*…)

February 16, 2009 at 4:30 pm

I’m not sure you’ve identified the hard direction correctly. I think that depends on the (group of) student(s).

But I hope you continue. I enjoy the precision of the discussion.

Jonathan

February 16, 2009 at 4:42 pm

there’s a blog post in there somewhere:

“hard and easy”. i’ve mentioned in classes

any number of times that i quite often

find myself pretty close to 180-degrees

away from ordinary citizens on questions

of “what’s hard” and “what’s easy”

(seeing a point to unfamiliar definitions: easy.

seeing a point to professional sports: hard.

etcetera).

are we closer to agreement if i say

“hard to *show*” in place of “hard to *see*”?

because “factor implies root”

is proved by simply *examining definitions*,

but “root implies factor” depends on

the so-called remainder theorem

(as i complained in my lengthy follow-up

to this post).

February 16, 2009 at 6:03 pm

harder to show, absolutely.

I am much better than I used to be at predicting what kids will get stuck on… but then I get it wrong. They waltz through what I thought would be killer, while an easy topic can grind them to a halt.

But fractions. Fractions are always hard.

Jonathan

February 18, 2009 at 1:39 pm

fractions are hard!