9
$\begingroup$

I am a graduate teaching assistant at a larger state university teaching a college algebra class. Today we begin our decent into polynomials and one of the facts we will soon get to is the connection between roots of a polynomial and linear factors. We do not teach polynomial long division (explicitly avoid it in fact), which is normally how I could convince students that $x - a$ is a factor of a polynomial $p(x)$ if and only if $p(a) = 0$.

My question is how can I convince my students of this fact without appealing to division? Keep in mind that these are largely college freshman who do not necessarily care for explanations. While they would take my word on something and memorize it if I told them it was necessary, I still like to convince them this is indeed mathematical fact and not something that has come out of thin air.

$\endgroup$
10
  • 1
    $\begingroup$ Without polynomial division? That means also without Euclides algorithm...that doesn't look too algebrish to me...and I really have no idea how you're going to teach that stuff and that way. To me, that sounds like trying to teach about extreme points of functions without seeing derivatives... $\endgroup$ Commented Feb 26, 2018 at 18:11
  • $\begingroup$ Write $\,p(x)\,$ in terms of $\,(x-a)\,$ and prove that the constant term is in fact $\,p(a)\,$. $\endgroup$ Commented Feb 26, 2018 at 18:18
  • 1
    $\begingroup$ Could you maybe assert a form of unique factorization? $\endgroup$ Commented Feb 26, 2018 at 18:19
  • 6
    $\begingroup$ For any $m\in\mathbb{N}$ we have $(x-a)\mid (x^m-a^m)$. By linearity it follows that $(x-a)\mid (p(x)-p(a))$, hence $p(x)=q(x)(x-a)+p(a)$. $\endgroup$ Commented Feb 26, 2018 at 20:29
  • 1
    $\begingroup$ Given the students don't require a proof, I think this is not a strong need pedagogically. $\endgroup$ Commented May 9, 2018 at 17:58

7 Answers 7

Reset to default
6
$\begingroup$

This answer presumes that the students know polynomial multiplication. In particular, $$(x-a)(x^{n-1}+ax^{n-2}+\cdots+a^{n-1})=x^{n}-a^{n}.$$ Since $$p(x) = c_{n}x^{n}+c_{n-1}x^{n-1}+\cdots+c_0=(x-a)q(x),$$ and $$p(a) = c_{n}a^{n}+c_{n-1}a^{n-1}+\cdots+c_0,$$ we have $$p(x)-p(a) = c_n (x^{n}-a^{n})+c_{n-1}(x^{n-1}-a^{n-1})+\cdots+c_1(x-a)=(x-a)r(x).$$ Consequently, $$p(a) = p(x)-(x-a)r(x)=(x-a)(q(x)-r(x)).$$ The L.H.S. is a constant, whereas the R.H.S. contains $x$. So $q(x)-r(x)=0$, and $p(a)=0$.

$\endgroup$
1
  • $\begingroup$ Said structurally it's true when $p(x)$ is a monomial $x^n$ so by innate linearity it's true for all polynomials, being linear combinations of monomials. $\endgroup$ Commented May 10, 2018 at 0:07
6
$\begingroup$

One way to avoid (explicit) division is to make a change of change variables $\, X = x-a\,$ which reduces it to the following simpler special case $$ X\mid P(X) \iff \color{#c00}{P(0)} = 0$$ Explicitly we have $$\qquad\qquad\begin{eqnarray} & x-a\,&\mid\,\ p(x) \\[.21em] \iff\ & X\!\!\!\!\!\!\!\! &\mid\ \ p(X+a)\\[.2em] \iff\ & &\ \color{#c00}{\bigl[p(X+a)\bigr]_{X=0}} = 0\quad\text{by above special case}\\[.2em] \iff\ & &\ \ p(a) = 0 \end{eqnarray}$$

Remark $ $ This approach implicitly uses the fact that the above shift map is a ring automorphism - something that is implicitly known since high school but not usually proven rigorously until a course in abstract algebra. See this MSE answer for further discussion on such pedagogical subtleties.

$\endgroup$
2
  • $\begingroup$ Reminds me of this, which I recently cited here. $\endgroup$ Commented May 9, 2018 at 9:18
  • $\begingroup$ @Dave Indeed, simplifications by way of shifts (automorphisms) are helpful not only in algebra, but also in analysis, e.g. for proving inequalities (as in those explicit $\epsilon$-$\delta$ proofs of limits). $\endgroup$ Commented May 9, 2018 at 14:52
4
$\begingroup$

Let $\,p(x) = c_n x^n + c_{n-1} x^{n-1} + \ldots +c_1 x + c_0\,$, and expand it in powers of $\,(x-a)\,$:

$$ \begin{align} p(x) &= c_n\big((x-a)+a\big)^n+c_{n-1}\big((x-a)+a\big)^{n-1}+\ldots+c_1\big((x-a)+a\big)+c_0 \\ &= c_n\left((x-a)^n+\binom{n}{1}(x-a)^{n-1}\cdot a+ \ldots+\binom{n}{n-1}(x-a)\cdot a^{n-1} + \color{red}{a^n}\right) \\ &\quad + c_{n-1}\left((x-a)^{n-1}+\binom{n-1}{1}(x-a)^{n-2}\cdot a+ \ldots+\binom{n-1}{n-2}(x-a)\cdot a^{n-2} + \color{red}{a^{n-1}}\right) \\ &\quad \cdots \\[5px] &\quad + c_1\big((x-a) + \color{red}{a^1}\big) \\ &\quad + c_0 \cdot \color{red}{a^0} \\[5px] &= (x-a)q(x) + \color{red}{c_n a^n}+\color{red}{c_{n-1}a^{n-1}}+ \ldots + \color{red}{c_1 a} + \color{red}{c_0} \\[5px] &= (x-a) q(x) + p(a) \end{align} $$

It then follows that $\,(x-a) \mid p(x) \iff p(a) = 0\,$.

$\endgroup$
1
$\begingroup$

If you could first convince them of the fundamental theorem of algebra (and if they know of complex numbers) then one writes $p(x)=c\prod_i(x-\alpha_i)$, substituting $a$ will have to convince them from that point that there is $i$ such that $\alpha_i=a$.

$\endgroup$
0
$\begingroup$

I'm not sure if there's a great way of explaining it without division (I feel like any good explanation will strongly hint at division) but maybe you could "disguise" the division by talking about repeated subtraction.

You can subtract any multiple of $x-17$ from a polynomial. That includes multiples like $10x^{82}(x-17)=10x^{83}-170x^{82}$.

In fact, given any polynomial of degree $\ge1$, say its leading term is $ax^n$, you can subtract $ax^{n-1}(x-17)$ and get a polynomial of strictly smaller degree. And as long as you keep getting polynomials of degree $\ge1$, this process can be repeated. But it can't continue forever. So eventually we get a polynomial of degree $<1$, i.e., a constant. Call that constant $c$.

So, if the original polynomial was $p(x)$, we get $p(x)-(\mbox{some multiple of ($x-17$)})=c$. Then evaluate at $x=17$.

Now that I look at what I've typed, it's not very different from the standard argument. But it's possible that phrasing it in terms of repeated subtraction, rather than polynomial division, may be helpful to some of your students.

$\endgroup$
1
  • $\begingroup$ This is precisely the standard inductive proof of the Polynomial Division Algorithm, i.e. decrease the degree of the dividend by subtracting a multiple of the divisor (scaled so they have equal lead terms), then proceed inductively on the remaining smaller degree dividend. Interpreted constructively this induction yields the recursion for the standard Polynomial Division Algorithm. See also here. This extends easily to non-monic divisors. $\endgroup$ Commented May 9, 2018 at 18:21
0
$\begingroup$

To prove the assertion, you need polynomial division. But to demonstrate with enough examples that they develop an intuition (you're probably going to help them discover it on their own, right?) all they need is the ability to factor polynomials. I'd think you could start simple with something like this:

$f(x) = (x+1)(x-5)$

What is $f(-1)$?

So the class calculates $f(-1)$ and finds it is zero. (Will you need to discuss what a root means in terms of a polynomial?) Do this three or four times with different functions, asking for a single root. Then ask for all roots, not just one. Ask the class what pattern they are seeing and if they think that pattern holds for all polynomials. See if they can guess the other root of $f(x)$, above.

This may not get the class to be able to construct a proof that goes both ways, but it should give them an understanding of how to apply the proof -- which is really the aim (assuming the class is not composed exclusively of math majors).

$\endgroup$
1
  • $\begingroup$ That's the trivial direction, i.e. for some polynomial $q(x):\ p(x) = (x-a)q(x)\,\Longrightarrow\, p(a) = 0.\,$ Proving the reverse direction is where the difficulty lies, and that has little to do with factoring polynomials. Also one doesn't "need" the Polynomial Division Algorithm to prove it, e.g. see my answer. $\endgroup$ Commented May 9, 2018 at 23:54
-3
$\begingroup$

We have by division algorithm that $$f(x)=(x-a)Q(x)+R$$where $R$ is the real residue of division therefore $$f(x)=R$$which means that $$R=0\leftarrow\rightarrow f(a)=0$$

$\endgroup$
4
  • 7
    $\begingroup$ You must have missed the important point that the OP wants to avoid completely polynomial division... $\endgroup$ Commented Feb 26, 2018 at 18:11
  • $\begingroup$ Indeed. This is exactly what I would normally do, but my students do not know polynomial division (or at least it is not covered in the class and is therefore a technique that I can rely on using). $\endgroup$ Commented Feb 26, 2018 at 18:14
  • 1
    $\begingroup$ Well! This is the most essential definition and intuition about division. How would you replace it with something better? $\endgroup$ Commented Feb 26, 2018 at 18:17
  • $\begingroup$ @MostafaAyaz It is beyond my pay grade. Like I said, this is indeed how I would do it. $\endgroup$ Commented Feb 26, 2018 at 18:18

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.