# Why do we teach the Rational Root Theorem? (high school algebra 2)

Main question: Does anyone have any good/interesting applications of the rational root theorem or ways to teach it that don't involve conveniently ignoring computer-based tools in order to avoid rote checking of long root candidates lists?

Longer rationale: The rational root theorem feels like a redundant waste of time.

For example, the very first guided problem in my school's textbook (Holt, Algebra 2) demonstrates how to generate a candidate list, but then it immediately advises the student to use a graphing calculator to narrow down that list to the most likely roots they see on the graph in order to not have to plug and chug 10+ candidates. Why go through the trouble of generating that list if you're just going to look at the graph to identify the root? I get the point about verification, but why bother with the list at all if you trust the computer enough to believe its graph?

My students and I go through a few examples and practice problems and then check them on wolframalpha, but my students respond: "Why did just do all that work at the beginning?" I'm at a bit of a loss at what to say. This isn't like the quadratic formula, which not only is quick, but also there's clear motivation for not totally relying on computer-generated solutions (e.g., works for all quad polys; gives exact form of irrational and complex roots, not rounded). Moreover, it seems like an artificial chore to construct example polynomials with crazy enough rational roots (e.g., 12345/654321) to call into question the precision of computer-generated solutions. In other words, problems with numbers difficult enough to make us suspicious of calculator rounding are highly artificial and time-consuming to produce, as well as likely to require a computer to solve in a reasonable amount of time anyway without extreme drudgework.

I do see some drill-based skill-value in the repetitive process (factoring the constant term and leading coefficients and listing p/q for each, practice with evaluating a polynomial a various values), and I certainly think that discussions of what the theorem can and cannot tell you are important (e.g., if no rational roots, then the list doesn't apply). But more than many other topics of Algebra 2, this topic feels like its problems are far more artificially constructed, its solution process and methods are far less broadly illuminating, and the students are just going through the motions to get information that is more easily, more quickly, and more practically obtained in other ways.

Thoughts?

[Update: 2/24] I generally agree with the responses saying that the RRT is a useful logical tool to help step the student up towards Calculus-level processes that require multiple systematic steps and the observation/juggling of multiple pieces of info in a given math expression.

I guess I'm just frustrated with the textbook presentation and style of problem I've seen up to now, which not only focuses on obtaining root values (rather than logically ruling things out), but also seems blandly circular in that you're often specifically constructing polynomials to have or not have specific rational roots (and for any appreciably high-degree polynomial example (10+), using a computer program to multiply it out). As noted in some response below, many polynomials that actually come up in practice are unlikely to have nice rational roots -- hence my concern about wasting time plugging and chugging after the candidate list has been generated.

I think I'm going to spend a bit of time writing problems more like the algebra puzzle in a comment below, i.e., provide the student with a finite set of options along with ancillary info which, when combined with the RRT, will rule out all but 1 or 2 of the options. This "tool for logic puzzle" approach seems much more fruitful and useful for stimulating abstract thinking than the "generate a list and try them all" presentation that practically begs you to use a computer to grind through rote calculations.

[Update 2: 2/25]
Related thread: Should Eisenstein's Criterion be taught in high school? Should Eisenstein’s criterion be taught in high-school?

• Even as far back as the late 1990s I mostly only covered this mainly as a way of showing irrationality of certain simple explicit algebraic numbers. The theorem was on the syllabus I had to follow, but I taught it based on a handout I had that used to be on the internet, but I can't find it now. However, I've used the theorem many times in various irrationality proofs in Stack Exchange answers. I don't believe I covered it in the precalculus courses I taught in the early 2000s (last taught in 2005). This is for the USA. Other countries often have more stringent math curriculum. – Dave L Renfro Feb 23 at 19:28
• The handout was titled Proving Irrationality by the Rational Roots Theorem, and it appears now to be at only one place on the internet (at one time it was in google scholar, of all places), but at a possibly unsafe web page. For those interested, here's a nontrivial irrationality proof using the rational root theorem (the examples I had on the handout were nowhere near this difficult). – Dave L Renfro Feb 23 at 19:54
• I learned it in high school and used it fairly often afterwards. When I got to teaching college, I discovered most (US) students didn’t know it. Consequently I avoid it in my teaching thereby making it useless for H.S. students to learn it. – user615 Feb 23 at 22:30
• I have a math degree. This is the first time I've ever heard of the rational root theorem. – Ben Crowell Feb 24 at 14:26
• I have a math degree. I have studied the rational root theorem and used it many times in my algebra courses. (Just to counterbalance that other comment and provide a useful data point.) – Federico Poloni Feb 24 at 19:15

I would say the standard implementations of the Rational Root Theorem (make a huge list for the sake of making the huge list) indeed feel like a complete waste of time. However, the theorem can sometimes combine nicely with technology tools. I'll provide an example of this.

Consider the following algebra puzzle:

Solve for $$x$$: $$7x^3 -39x^2+52x+30 = 0$$.

If you are allowing a graphing calculator but not a computer algebra system, then the rational root theorem is pretty helpful.

Start by realizing that it is a cubic polynomial, so it has at least one real root. We can go look at the graph, hoping to find a nice root: Uhoh.. 0.429. My graphing calculator doesn't know the exact value of that root. It could be any number of things. But because of the rational root theorem, I know that if it is a rational root, it is some number of sevenths. This drastically decreases the amount of effort needed to find that indeed our real solution is $$x = -\frac37$$. If you need all three values of x, you can then proceed to divide by $$7x+3$$ to factor the polynomial and find the two complex zeros.

In summary, the rational root theorem can be useful as a trial-and-error shortener: it severely restricts the number of "trials" you have to make, and in a situation with a graphing calculator but no computer algebra system, that can still have some value.

I think an important learning outcome of this part of high school Algebra 2 (theory of polynomials) is developing mathematical capacity to juggle several disparate abstract tools all at once, similar to how in calculus, one uses first derivative test, second derivative test, concavity, asymptotes, intercepts, end behavior, etc., all at once, in some mysterious mash up in the brain, to understand the graph a function. The analogous abstract tools juggled in high school Algebra 2 are rational zero test, Descartes' rule of signs, degree and parity of degree, sign of leading coefficient, factor theorem for intercepts, synthetic division, bound theorem for roots, conjugate pair theorem, etc. So these are like "training wheels" for what is done in calculus.

The point is not to develop fluency with one specific tool such as the rational zero test. Rather, the point it to train the mind to use many tools simultaneously in a mathematical context.

• I agree with this because really the whole idea of solving polynomial equations is contrived. Any polynomial found "in the wild" is unlikely to have "nice" roots. Yet we spend gobs of time on these methods. Why? Because the actual skill is beside the point. The real benefit comes in the form of synthesizing multiple pieces of abstract information. – Aeryk Feb 24 at 17:57
• @Aeryk yes, but there are plenty of topics that would practice this kind of thinking without being useless by themselves. – leftaroundabout Feb 24 at 19:58
• My training is predominantly in physics. Every topic you list from first-semester calculus is used frequently in science and engineering. Not so for the topics you describe as being parts of "high school algebra 2." Of the 8 algebra topics you list by name, I could only tell you what one of them was (synthetic division). (Possibly some of the others refer to something that I know but not by any particular name.) – Ben Crowell Feb 25 at 2:33
• I’m very skeptical of this justification. Why not just learn calculus twice? The results would surely be strictly better if that’s the only goal. – Kevin Arlin Mar 1 at 13:49
• @user52817 Sure, the chess comparison was overstated. The point remains that there are a lot of things that involve algebra, verbal statements, and checking mathematical conditions, so if that's the entire argument for studying root-finding then it's as weak as it could possibly be without immediately implying we shouldn't teach this stuff. – Kevin Arlin Mar 1 at 21:58

Agreed...it's one of the less useful parts of high school algebra. But not because "you could use a computer"--you could say that about almost everything. And then we get some of the same people who push the "use a computer" who are surprised when their kids flounder because of lack of manipulational ability in calculus. ;)

The reason this topic is less useful is because you really don't end up using it to any reasonable extent (I can't ever recall using it!) in a conventional calculus, physics, engineering, or chemistry course. Full stop. Regardless if you "check answers with WA" (I wouldn't) or not. So, if you're time-limited, just don't bother with it, or skimp on the amount of time, don't test it much, etc. And do more with power laws or the like.

If you do have time and student capacity (e.g. upper track kids in an enriched course), I would go ahead and cover it. There is something about being able to move from guessing how to factor to some sort of systematic method, that is appealing after all the time spent factoring, especially for stronger kids. [And do cubics and quartics also--IF you have top kids. Something kind of beautiful about these discoveries.] And the work done is useful practice to be able to handle longer chains of reasoning and manipulation.

Maybe a tenous application would be that there is something similar in trying the different possibilities as compared to combinatrics, truth tables, full factorials, issue/logic trees, etc. Not saying THE SAME, but saying some similar "muscles" as for example Turing's bombe trying combinations of rotors and wiring. Again, very vague and loose. But a connection.

Also, of course, you're only spending PART of the year on this stuff. So it's not the end of world to do a small amount. But of course, priorities. Don't push lower utility, harder topics if you aren't getting more core topics mastered by the students.

Let $$x = \sqrt 2 + \sqrt 7$$ prove that $$x$$ is irrational.

\begin{align} x - \sqrt 2 &= \sqrt 7 \\ x^2 - 2\sqrt 2x + 2 &= 7 \\ x^2 -5 &= 2\sqrt 2x \\ x^4 -10x^2 +25 &= 8x^2 \\ x^4 -18x^2 + 25 &= 0 \end{align}

So $$\sqrt 2 + \sqrt 7$$ is a root of $$x^4 -18x^2 + 25$$.

According to RRT, if $$x$$ is rational, then $$x = \pm 1$$ or $$x=\pm 5$$ or $$x=\pm 25$$.

$$x = \pm 1 \implies x^4 -18x^2 + 25 = 8 \ne 0$$

$$x = \pm 5 \implies x^4 -18x^2 + 25 = 200 \ne 0$$

$$x = \pm 25 \implies x^4 -18x^2 + 25 = 379400 \ne 0$$

So $$\sqrt 2 + \sqrt 7$$ must be irrational.

The RRT is not taught in isolation. It is taught as a collection of tools for (partially) factoring polynomials. It should be taught with Descartes' rule of signs and some form of polynomial division.

It is direct preparation for understanding the proof of Eisenstein's criterion for irreducibility. The "shadow" of that application is via Gauss's lemma: factorization over the rationals and factorization over the integers are essentially equivalent. (At the cited link, the statement that is directly to this point is "A non-constant polynomial in $$\Bbb{Z}[x]$$ is irreducible in $$\Bbb{Z}[x]$$ if and only if it is both irreducible in $$\Bbb{Q}[x]$$ and primitive in $$\Bbb{Z}[x]$$.")

So why do we have these three tools?

• Descartes's rule of signs: are there any roots in the ray $$[a,\infty)$$? (Obtained by looking at the signs of the coefficients of $$p(x-a)$$.) Likewise, in the ray $$(-\infty, a)$$. This can be used to find intervals for each of which there is either an isolated root, a multiple root, or one or more complex pairs of roots.
• For each such interval, can there be a rational root? This frequently collapses the list of rational candidates by discarding those outside of the intervals. These first two steps can also be done in the reverse order or iteratively: the RRT suggests a set of partitions to binary search through the list of RRT candidates, letting lack of change in the sign of coefficients indicate that the set of roots in the ray has not changed.
• Synthetic (or long) division of found rational roots gives an entirely different set of coefficients. So any yet not-yet-found rational roots have to be on the lists for both the dividend and the quotient. One can also take the intersections of the rule of signs intervals for both the dividend and the quotient.

What does this practice? You have a bunch of tools, some will give you partial information and some will not. Learn how to select the correct tool, how to apply that tool, and how to update the information you have about the roots from the result of that application. This is the same pattern one needs for statistical hypothesis testing and (very directly) solving differential equations. It is a form of Bayesian inference (with all the equivalences that go with that).

Students always imagine they will have immediate access to technology. This is false. In my own direct experience, I was standing in a test minefield, outside of range of cell towers, and unable to use a phone or other poorly RF-shielded devices for the obvious reason. And I still had to make useful calculations in that setting. What was I calculating? The zero-crossing of an inverse function inferred from gathered discrete data. So as recently as a couple of years ago, real, live people had to find real, live roots of real, live polynomial models of functions without any access to technology.

• I teach it with Synthetic, but I skip the rule of signs, which feels like excess baggage to me. – Sue VanHattum Feb 24 at 21:58
• That's great! I remember now that you mention it that I used Descartes's Rule of Signs around the same time as RRT back in High School. I like the idea of putting together a bunch of divisibility, root-testing techniques into a whole unit as a toolbox to narrow down the list of candidates WITHOUT using a graphing calculator as a crutch. I could even imagine teaching regular track Algebra 2 students (i.e., not calculus-bound) the algorithm for 1st and 2nd derivatives just to throw a couple more tools in that box and make graphing by hand that much more detailed and meaningful. – Cassius12 Feb 24 at 23:01
• @SueVanHattum I would much rather go the other way around. Descartes’ rule is the first of a chain of results leading to modern real root estimation algorithms. The rational root theorem is a result of number theory, much less significant for applications. It’s good to do both if only to give students problems they can actually progress through by reducing the degree using RRT. – Kevin Arlin Mar 1 at 15:57
• My goal is to help them learn to associate equations and graphs. RRT feels like a useful tool for that. – Sue VanHattum Mar 1 at 18:35
• @SueVanHattum Well, that's double difficult to respond to, since you neither tagged me nor made a precise point. Ah, well. – Kevin Arlin Mar 1 at 20:39

I would reluctantly agree that it's not a particularly powerful tool if you have electronics at your disposal. But I might double down and say that you should be teaching synthetic substitution as well so that students can factor cubic and quartic polynomials without their calculators. About a year ago, I went down a YouTube rabbit hole of watching videos about factoring cubics and quartics by hand in Hindi (a language I do not speak). I am given to believe that they do not teach RRT in India, and it is clear from the large number of "intuitive" hacks that they showcase to factor cubic and quartic polynomials that would have fallen almost instantly to RRT.

One demonstration I like for giving a nice survey of pre-calculus algebra and trig tools is proving that $$x:=\sin 18^\circ=\frac{-1+\sqrt5}4$$. Let's take $$\theta=18^\circ$$ for the sake of clarity. Then it is obvious from the cofunction identity that $$\cos2\theta=\sin3\theta$$. Using the double and triple angle formulas, we can turn that into $$1-2\sin^2\theta=3\sin\theta-4\sin^3\theta$$, or more simply as $$1-2x^2=3x-4x^3$$. So $$\sin 18^\circ$$ is a root of $$4x^3-2x^2-3x+1$$. Here is where we can use the rational root theorem to quickly observe that $$1$$ is a root of that polynomial (and $$x\neq 1$$), obviously, so by the Remainder Theorem we can then divide the polynomial by $$x-1$$ and use the Quadratic formula to get the desired answer (again, since we can foresee that the root we want satisfies $$0 by the definition of sine).

I'm not sure if it goes without saying or not, but the Remainder theorem is essential for helping students to understand why the difference quotient $$\frac{f(x)-f(a)}{x-a}$$ always has a removable discontinuity at $$x=a$$ when $$f$$ is a polynomial function.

• This is really cool, but does it use the rational root theorem? Plugging $x=1$ into a polynomial doesn't seem to use the rational root theorem. – Chris Cunningham Feb 24 at 17:34
• @ChrisCunningham It's a fair point. But I would say that RRT has two purposes. It can rule out the existence of rational roots. But I suggest it also deserves credit for providing a strategy for testing a finite number of candidates with synthetic substitution or the like even if one them turns out to be a root. You know that it's a root because you plugged it into the equation, but how did you know to check for it? Admittedly, we could make a rule to always check 1 and -1 because they are super-easy, but at some point perhaps you're recreating RRT to avoid teaching it. – Matthew Daly Feb 24 at 18:36
• @Chris Cunningham: For those interested, this answer gives 3 ways of obtaining "square root expressed" values of $\sin18^\circ$ and $\cos18^\circ.$ – Dave L Renfro Feb 24 at 19:18

Theorem: For every integer $$m$$, the polynomial $$x^3 - mx^2 - (m+1)x - 1$$ is irreducible among polynomials with rational coefficients.

Proof: This polynomial has degree $$3$$, so if it is a product of lower-degree polynomials, the decomposition is (linear)(quadratic) or (linear)(linear)(linear). Either way, there is a linear factor and thus a rational root ($$ax + b$$ with rational $$a$$ and $$b$$ where $$a \not= 0$$ has root $$-b/a$$). By the rational roots theorem, if $$x^3 - mx^2 - (m+1)x - 1$$ has a rational root that root must be of the form $$a/b$$ where $$b$$ is a factor of $$1$$ and $$a$$ is a factor of $$-1$$. So $$a/b$$ is $$\pm 1$$. Let's try each of them:

$$f(x) = x^3 - mx^2 - (m+1)x - 1 \Rightarrow f(1) = 1 - m - (m+1) - 1 = -2m -1 \not= 0$$ and $$f(x) = x^3 - mx^2 - (m+1)x - 1 \Rightarrow f(-1) = -1 - m + m + 1 - 1= -1 \not= 0.$$ Therefore $$f(x)$$ doesn't have $$1$$ or $$-1$$ as a root, so it is irreducible as a polynomial with rational coefficients. QED

Good luck trying to prove every polynomial in an infinite family of examples is irreducible by computer. Do you only ever illustrate results in math in your class with individual examples, instead of an infinite family of examples?

Homework problem: each student needs to come up with their own infinite family of cubic polynomials and use the rational root theorem to prove every polynomial in the family is irreducible (in $$\mathbf Q[x]$$). In contrast, there is no simple nonconstant polynomial formula whose values are all prime numbers: $$5n+2$$ is prime for infinitely many $$n$$, but it's also composite for infinitely many $$n$$. We think $$n^2 + 1$$ is prime for infinitely many integers $$n$$, but this is still an unsolved problem, and in any case it is composite for sure infinitely often (let $$n$$ be an odd number bigger than $$1$$).

I really think this questions gets to a bigger issue. This is part of a general problem with our approach to math education.

The point of high school math class, from any practical standpoint, is not to teach students things that they will constantly be using all the time (or even ever). Empirically this is verified in myriad examples, not just the rational root theorem.

So I think it's better to not even pretend like we're trying to teach people things that are immediately useful. Absolutely 100% of word problems that students encounter before reaching vector calculus or maybe even differential equations are completely and utterly contrived, and usually made trivial with computer assistance. Students see through this, and by my observation they are always complaining about how non-useful the math they're learning is. It's interesting to observe that they don't make that same complaint nearly as often about other classes, yet I can honestly say (as far as I'm aware) I have never used a single thing I learned in high school outside of math, foreign-language, and web design classes. I certainly share your frustration with textbooks for not realizing this; but you are the teacher and you even if you have to follow the curriculum of the book you do not have to adopt its attitudes.

So then what is the point of teaching math in high school?

There's a couple of reasons I can think of that are not related to direct applicability:

1. It forces students to think abstractly and to grasp new concepts. The rational root theorem is an example here. Simply put, it forces you to think about polynomials in a different way than simply finding the roots with WolframAlpha or by the quadratic formula. From my experience as both a mathematician and as a math educator, it is impossible to be good at math unless you can understand every concept in multiple ways. High school math, if nothing else, is an exercise in being able to think abstractly and think in new ways. If done correctly, it should help to keep student's minds flexible and indirectly fosters creativity. High school math is to the mind what PE is to the body.

2. It is foundational for more advanced mathematics. Analogies to phonics classes in elementary school, you need to learn the basics of English orthography if you want to be able to read and write effectively. It's not directly useful for the purposes of reading comprehension to need to be able to read words out loud, but this is a necessary prerequisite step in learning English. In the same way, all the basic concepts you learn in high school aren't broadly applicable, but it's necessary to go through them in order to get to the parts of math that are applicable. Perhaps not 100% of the things you learn are necessary foundations, so this probably doesn't directly relate to the rational root theorem (it'd be hard to argue that's foundational knowledge for any higher branch of math). But it is another reason to teach the basic, non-applicable concepts.

3. Math is something that we as a society believe to be value and it's culturally and historically enriching to learn. To me this is a weak argument but it is the entire reason we teach history and literature in high school. Thus learning theorems that have historical significance is beneficial just like reading Shakespeare or learning about the War of 1812 is beneficial.

• I agree with a lot of what you say. I've found that I have MUCH more engagement/rapport with students if I'm up-front with them about the contrived/impractical nature of many of the ostensibly "relevant" math problems we cover and instead talk to them regularly about how the point is often mental training/habits of mind rather than "you will definitely use this". That said, there is a lot of "practical math" that should be taught but isn't the curricular focus anywhere I know of. I'd advocate bringing back serious HomeEc and Financial literacy classes as separate from a calculus track. – Cassius12 Feb 25 at 16:16
• I 100% agree with that. But I think that the people who design curricula don't really have any idea what they're doing tbh – Dark Malthorp Feb 25 at 16:21
• This is the best answer. Unfortunately, most math teachers and book writers don't seem to be aware of this. – Martin Argerami Feb 27 at 0:06
• The problem is that this is a fully general argument: it does nothing to distinguish RRT from literally any other piece of math. – Kevin Arlin Mar 1 at 16:00
• Indeed, I have solved the general problem of "why is X taught to highschoolers who will never need to know it?" The application to this question is left as an exercise to the reader. – Dark Malthorp Mar 2 at 18:13

This sort of thing has no application in science or engineering. It can't have any such application, because real-world polynomials don't have nice coefficients or nice roots. Forcing it down the throats of kids at age 17 is absurd. If it's in a high school textbook, it's there because publishers have an economic incentive to puff up their books with random doodads. No teacher or textbook selection committee will ever reject a book because it has a topic they don't intend to teach, but books do get rejected because they lack someone's pet topic.

The issue is not that this topic is made useless by the existence of computer algebra systems. The issue is that the topic is simply not useful to dentists, physicists, biologists, mechanical engineers, etc. We're talking about something that is going to be forced on a kid who is thinking of being a pharmacist. That kid doesn't take Galois theory. That kid will never prove or want to prove that $$7^{1/3}$$ is irrational. Our educational system would be doing well if that kid could estimate that $$7^{1/3}$$ is slightly less than 2 -- but in reality, I doubt that very many dentists and pharmacists can do so, nor does it bother me that they can't. It would bother me if my pharmacist couldn't convert grams to milligrams without getting the conversion factor flipped half the time.

• The rational root theorem and (to a lesser extent) Descartes rules of signs used to be a standard topic in precalculus courses (and strong algebra 2 courses, and most every college algebra course) in the U.S. until it started getting phased out bit-by-bit beginning in the early 1990s, mostly due to the appearance of graphing calculators. For example, it appears before the chapters on exponential functions, logarithms, the distance formula in the plane, and graphs of quadratic equations in the 1968 edition of Dolciani's Algebra and Trigonometry 2 used when I was in high school. – Dave L Renfro Feb 25 at 14:54
• It was also in every college algebra text and every precalculus text I taught from up until maybe the early 2000s, and probably even those, although the topic was by then not taught much anymore. I recall that it was taught in precalculus courses at NCSU in the early 1990s (and the departmental "lesson plan booklet" I have confirms this), but by the time I was leaving, and definitely in the mid 1990s many math departments began having discussions about omitting the topic, or at least not requiring it to be covered. – Dave L Renfro Feb 25 at 15:01
• The main application (by a long shot) of this topic was to allow students to factor nice cubics and quartics in calculus courses, such as when using a sign chart for a factored first derivative. However, in actual fact, only a very few calculus textbook problems ever needed the rational root theorem. Often in cubics and quartics you'd try $x=\pm 1$ first, then maybe $x = \pm 2,$ and perhaps $x = \pm \frac{1}{2}$ or $x = \pm \frac{1}{3}$ if the leading coefficient was $2$ or $3,$ and you'd have a root, after which you can get a quadratic by reduction of degree (divide by "$x$ minus root"). – Dave L Renfro Feb 25 at 15:07
• Science and engineering with "messy" numbers are far from the only real-world application for algebra. – R.. GitHub STOP HELPING ICE Feb 25 at 18:37
• This is a terrible answer. The point of the school curriculum is not to teach useful things, but to expand the student's minds and prepare them to learn whatever they will need in the future. – Martin Argerami Feb 27 at 0:10