### 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 $f \in {\Bbb R}[x]$, 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
$\null A_n x^n + A_{n-1} x^{n-1} + \dots + A_2 x^2 + A_1 + A_0$,
where “the A‘s are Real constants”—we might instead say here that “$A_i \in {\Bbb R}$, for $i \in \{0, 1 , 2, \dots n\}$“… for example if there were some need to be more concise, or more precise, or maybe because the symbols are astonishingly beautiful [and, with sufficient practice, reveal themselves to be much easier to understand than mere words…]— and $A_n \not= 0$).

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
$f(x) = (x-R)\cdot g(x)$
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)] definingx – 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…)

1. 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

2. 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).

3. 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

• ## (Partial) Contents Page

Vlorbik On Math Ed ('07—'09)
(a good place to start!)