# TNDOT 3: The Remainders of the Day

Here’s a third entry in our Teaching New Dogs Old Trigs Tricks series: the first three questions from the 1907 Victorian Matric Algebra exam.* As well as the computational aspects, readers might be interested in contemplating what was expected for the proof parts, as well as how one might fruitfully present such results to students now, a century later.

*) TNDOT 2 is also in the process of being updated. Stay tuned.

## UPDATE (06/04/24)

OK, finally, I’ll get to updating this one, and then hopefully I’ll post a TNDOT 5 in the next day or so. I’ll do the update in bits, as I have time, first doing the substantive questions and then maybe writing a bit about the proofy bits. As usual I’ve been guided by the ideas in the comments, particularly BiB’s, and also Simon’s solutions.

First, a little background and then an answer to Q1. As BiB noted, the remainder theorem was definitely intended to be a prompt for Q1, and also “symmetry” was very much a thing at the time. I took a look at the old books, and in particular A New Algebra (Vol 2) by Barnard and Child had a very nice section on such problems.*

Just so we’re clear on the terminology, a polynomial (or any function) is symmetric if swapping any two of the letters keeps the function the same. So, for example, the polynomial is symmetric. Now consider the polynomial in Q1:

P is not symmetric, but it is cyclic, meaning that cycling the letters in some particular order, in this case , keeps the function the same. P is also alternating, meaning that swapping any two letters changes the sign of the function.

Now setting a = b gives P = 0, and thus (a – b) is a factor. Similarly, or by being cyclic, (b – c) and (c – a) are also factors. Thus Q = (a – b)(b – c)(c – a) is a factor of P. So we can write

and it remains to figure out R.

To this end, note that R is a homogenous (purely) quadratic polynomial, and that also R = P/Q is symmetric (because P and Q are alternating, and so the minus signs cancel out). That doesn’t leave a lot of options. Also, since P has no variables to higher than a cubic power, R can no squares. So, except for a constant factor, this leaves ab + bc + ca as the only option. Checking, say, the coefficient of a2b3, we finally get

*) First published in 1912, and my edition is 1955. I haven’t been able to locate an online copy.

## UPDATE (07/04/24)

On to Q2. To help with the discussion, let’s write

Now, letting F stand for the rational function to be simplified, we can naturally get a common denominator. Then

with

and

Note that we’ve taken a minus sign out and written K and N in a natural “cyclic” form. Now, the question is, how to handle the quadratic N. We’ll outline two-ish approaches.

As a first approach, Barnard and Child work through a similar but simpler example, using the Nike method: they just do it. This is not as gruesome as it might appear; since N is cyclic it is not too difficult to extract the coefficients of N, and then tidy them. (Barnard and Child also lists a string of “important identities”, the one-time basics of cyclic polynomials, which may have been considered appropriate to use in the exam). I’ll give a slightly less Nike version of this.

To begin, note that setting a = b makes N identically zero, implying (a – b) is factor. Then by the cycling or whatever it follows that

where P, Q and R are polynomials of degree 0, 1 and 2, respectively. Then, since Q and R are cyclic, it is easy to equate coefficients to obtain

Plugging into N, and then F, and cancelling the K, that finally gives

A second approach is to note, as did BiB, that

Writing

this gives us three equations for p, q, and r:

(BiB claims, without proof, that “the denominator of the simplified expression will be abc”. This amounts to claiming that K is a factor of N, which effectively gives BiB equations for P, Q and R. The subsequent solving of the equations is no simpler, however.)

As Simon noted in his solutions, there’s a Vandermonde matrix there, and one can solve using the inverse matrix. It is easier to solve directly ( = row reduction), however, as outlined by BiB: (4) = (1) – (2) (and cancel a – b), (5) = (1) – (3) (and cancel a – c), (6) = (4) – (5), etc. This leads to the expression for F as given above.

## UPDATE (08/04/24)

Finally, modulo the proof bits, Q3. This is a weird one, since the natural approach would seem to be to get a common denominator and proceed as with Q1 and Q2. So, with p = 2, the THING2 we’re trying to simplify is given by

where K = (a – b)(b – c)(c – a), as in Q2. But then, by the standard stuff in Q1 and Q2, or directly from one of Barnard and Child‘s important identities, it is easy to see that N = -K, and so,

Similarly, and/or also by an important identity,

Very easy (once we’ve gotten into the swing of it), and clearly not what was intended, given the partial fractions preamble to the question. Looking at Barnard and Child, and also the authors’ Higher Algebra (p 102), this problem was of a standard type, with a standard and very nice method of solution. The method seems like overkill for this exam question, and should not have been required since there is no “hence” attached to the partial fractions. But the method is interesting and definitely useful for harder problems of the type. So, I’ll outline the intended approach.

For p = 2, the intention was that students split the function x2/(x-a)(x-b) into partial fractions (plus remainder), which is easily done:

Now let x = c:

Moving the fractions on the right to the left, and watching the minus signs, this gives THING2 = 1, as before.

Similarly, for p = 3, we first calculate

Setting x = c and moving things around, we get THING3 = a + b + c, as above.

## UPDATE (08/04/24)

OK, just a final quick update, a few comments on the proof bits.

First on the remainder theorem, I was a little puzzled by this being asked about, since the proof is a trivial consequence of Euclidean division, but which itself is difficult to prove. Then, Red Five remarked on a different proof of the remainder theorem, relying solely upon the factorisation of xn – hn. A version of this proof (as well as the Euclidean proof) is given in Barnard and Child, and in other texts of the time, and I wonder if this is what was intended.

To outline the proof, suppose we have a polynomial f(x) = ax4 + bx4 + cx2 + dx + e. (The argument is essentially identical whatever the degree of the polynomial.) Then,

So, f(x) = (x-h)(STUFF) + f(h). It follows that the remainder of f when divided by (x – h) is f(h).

On the partial fractions proof, I’ll just make a couple quick points. First, I wondered if the expected method of finding the values of A, B and C was by simultaneous equations or by substitution. As it happens, both methods were very standardly taught at the time, with intelligent preference given to substitution; my guess is that the substitution method was expected in the exam answer, since this is also easier to express. The second point is that the exam question implicitly assumes that a, b and c are all distinct (and the conclusion is false if they are not). That’s notable, since it’s the kind of “generic” assumption that VCAA argued (implausibly and invalidly) for their 2022 complex question. It would seem that such wording was standard, or at least common, for the time, since a 1912 exam contains an essentially identical assumption.

## 27 Replies to “TNDOT 3: The Remainders of the Day”

1. Red Five says:

Suppose (I haven’t followed this idea through to completion yet) you re-wrote the expression as and then looked for a common factor of – might this save a lot of expanding, regrouping and re-factorising?

(Not expecting an answer, just posing it as a thought).

EDIT (2 minutes later): I don’t think this helps much at all. Still convinced there is a trick somewhere though.

2. Red Five says:

OK. If anyone wants to compare answers/solutions, I can get the expression in part (1) down to a product of four factors.

The “trick” I missed was right there in the line above… and my initial idea was not completely wrong (just 90% wrong)

1. Back in Black says:

I’m limiting my comments to one per month. January (so that RF is not alone here):

The preamble:
“State and prove the remainder theorem.”
plus the obvious symmetry makes factorising the expression simple.

Clearly the expression is zero for a = b, a = c and b = c therefore factors are (a – b), (a – c) and (b – c). (In fact, you only need to spot one of these, symmetry gives the other two).

To find the remaining factor, consider the coefficient of, say, a^3 in the given expression and the coefficient of a^3 in the factorised expression (a – b)(a – c)(b – c) (coef a + ‘no a’s’):

c^2 – b^2 = (b – c)(coef) therefore coef = -(b + c).
So the fourth factor has the term -a(b + c) = -ab – ac. Since the expression is symmetric in a, b, c, the fourth factor must also include the term -bc.
So the fourth factor is -(ab + ac + bc).

Students of the era would be well aware of and practiced in such tricks, a competent student would nail something like this in short order.

(Brute force also works since terms cancel. But the resulting expression does not obviously factorise (especially for a brute) and there’s also significant time penalty attached to this method).

1. Red Five says:

Yep. Same starting point (recognising factors that in hindsight are more obvious than I am willing to admit).

Slightly different method for getting the fourth factor, but same result.

I basically divided the entire expression by , and in turn using the method of displaced coefficients (which I saw a reference to in another exam from a similar time, so assumed it was fair game.

Brute force gets you down to a sum/difference of six terms pretty easily, but from there it is far from obvious.

Mostly because I suspect the modern question focuses (almost exclusively) its scope to factors with only two terms in each. Meh.

2. marty says:

Thank you both. A very nice approach. It’s not obvious to me, though, that this would have been the approach expected in 1907.

Playing with symmetric polynomials was much more of a thing back then, and I wonder if a more formulaic approach might have been standard. There were some hints of this in the O’Mara book, even from just a quick browse, and I think the classic texts of the time had a lot of this stuff.

Do either of you have thoughts about proving (or “proving”) the remainder theorem, either then or now? (I’m snowed, and have thought nothing about this post other than it seemed worth doing.)

1. Back in Black says:

OK, I’m breaking my rule out of guilt. But it’s close enough to February …

I can easily imagine that the division algorithm was part of the syllabus back in the day (and it’s still implicit in the Specialist Units 1&2 syllabus):

Dividend = (Divisor * Quotient) + Remainder.

I’m guessing an acceptable proof would be something like the following:

Assume that q(x) and ‘r’ are the quotient and the remainder respectively when a polynomial p(x) is divided by a linear polynomial (x – a).

From the division algorithm: p(x) = (x – a) * q(x) + r.

Substitute x = a: p(a) = (a – a) * q(a) + r = 0 * q(a) + r = r

therefore the remainder is p(a).

I’m certain the “State and prove the remainder theorem” is there to prompt finding the factors (a – b), (a – c) and (b – c). Then I suppose it’s a matter of style (or the mathematical mores of the day) how the fourth factor (and there has to be a fourth factor to get the cubic terms) is found. Using symmetry seems an obvious route.

Yes, symmetric polynomials were a thing back in the day (and still seems a small thing in NSW, with the Vietta formulas etc). Then again, an entire subject called Algebra at the Year 12 level needs content and that sort of stuff fits the bill very nicely.

1. marty says:

How would you prove/justify/assume the division algorithm?

1. Back in Black says:

Yes, that had crossed my mind (in particular, the existence of q(x) and r). I’ve assumed students could use the division algorithm without having to prove it. Because I think it’s a tall order to require proof of the division algorithm:

https://math.oxford.emory.edu/site/math125/proofDivAlgorithm/

(and there are some things in this proof that have been assumed).

Also, if it’s there only as a prompt (again, I’m assuming) for the factorising question, I wonder how detailed the proof needs to be.

The broader question is what ‘theorems’, ‘axioms’ etc. can be reasonably assumed when proving something in an exam. Unless a marking scheme is available, we won’t know for sure in this case.

(That’s March blown).

1. marty says:

Yes, that was the reason for my question. It’s not obvious what a “proof” of the remainder theorem meant for a 1907 exam question, and it’s not obvious, at least to me, what one would want to offer as a proof or semi-proof now.

1. Red Five says:

Reading other exams you have posted on, I suspect Euclidean division was part of the Algebra syllabus and so maybe the Euclidean division lemma(?) theorem(?) would be expected to assume.

I have seen a proof that does not involve Euclidean division, but does require the factorisation which I also assume would have been commonplace in 1907 Algebra courses.

2. Red Five says:

I suspect a sketch of a proof was anticipated, something basically to test had the students been shown a proof that they could mostly reproduce.

Nowadays… I highly doubt more than a handful of Methods teachers (including those who teach Specialist) would bother with a proof. Many may not know one themselves.

3. Simon says:

Hi Marty – some more fun!

As you mentioned above, symmetric polynomials used to be a more core unit of study – I’ve never studied them properly, just learnt bits I needed to solve specific problems I’ve hit in other areas. And that will show in my solutions. Q1 [similar method to BiB] and Q3 are not too bad, got stuck on Q2. There is probably some standard knowledge and tricks that will make this more accessible, but that’s enough for me now! Maybe I need O’Mara’s book!

My notes are in the same document as last time
https://www.mathcha.io/editor/0EJ5dI0VtjPidDCNm0MnjHM3WgEuNYXlJH21mz57

By the way, all of these TNDOT posts could make nice cores for building a SAC…
Maybe I need to get a hold of these old exams and books… what’s your source?

1. marty says:

Thanks, Simon. I still haven’t had a chance to look at these questions. Hopefully I transcribed 2 correctly …

As for sources, the exams are here and there, in the big libraries. I’m desperately trying to find time to sort what I have, and see exactly what I have. I also might investigate copyright/permission at some point.

In terms of books, these turn of the (other) century books have a ton of stuff. O’Mara is not a particularly good one. Perhaps something on that is worth a post as well. So much to do …

1. Simon says:

Thanks Marty,

The question was transcribed fine – I assume, as it does work out – just me working too late at night and giving up too quickly!

When the new SM study design came out, I did think of going to the state library to look at the old exams with some of the reintroduced topics, but never got to it. I’m glad you took the time to have a look! As for copyright, I think (but I am not a lawyer) it would be “Crown Copyright”, which lasts 50 years. So all of this TNDOT stuff should be public domain.

No hurry on the books – my “want-to-learn” list is growing much faster then I can keep up with!

1. marty says:

Thanks, Simon. I’m sure publishing a question or two is fine, both under “Fair use” and “You gotta be kidding”. But it would be good to put up entire exams, and for that I’d want to carefully check the ground rules and/or VCAA’s view.

2. Back in Black says:

*Sigh* April is blown but Simon goaded me into giving my thoughts. A brutal approach (no symmetry):

When x = a, x = b and x = c the expression = 1/a, 1/b and 1/c so the denominator of the simplified expression will be abc.

The numerator is a quadratic in x. Let it be .

From the above observations f(x) must satisfy:

…. (1)

…. (2)

…. (3)

(1) – (3) and (2) – (3) and readily simplify:

…. (4)

…. (5)

(4) – (5): A = 1.

Sub A = 1 into (1) and (2) and readily solve for B and C:

and .

The final form of answer is a matter of style – there is an evident symmetry.

1. marty says:

Thanks, BiB. I’ll try to look soon.

2. Back in Black says:

* (1) – (2) not (1) – (3).

3. Simon says:

Thanks BiB! I appreciate you breaking your rules for me 🙂

That solution works and made me push my approach to get the same result. [Updated in my notes now]

The thing that tripped me up when I first read it was that I thought you were saying the denominator was only , but your , , could have been rational functions of it just happens that things cancel out nicely. Is it obvious that they should have cancelled?

4. marty says:

BiB, maybe I’m being obtuse but I can’t make sense of your argument that the simplified denominator is abc. Of course it’s true, and I believe you are thinking something valid to get there. But what you hve written makes no sense to me.

1. Back in Black says:

I’ve probably used a physicist’s argument.

My thoughts were that whatever (2) simplifies to, the simplified expression has to be true for any value of x. So I chose some convenient values of x to try and get insight:

When x = a, the simplified expression has to equal 1/a.

Similarly when x = b the simplified expression has to equal 1/b, and when x = c the simplified expression has to equal 1/c.

I took these results to suggest that in general, the denominator of the simplified expression would be (or at least contain the factor) abc.

I’ll be the first to concede it’s not the strongest mathematical argument (although I think ‘intuition’ plays a role in solving a problem). Maybe someone can make the above more formal or rigorous.

PS – Nice solution for (1).

1. marty says:

What is (2) here? The rational function in Q2? (Also don’t use “true” for “equal”.) Then, sure, R(a) = 1/a etc is a natural thing to try. But I don’t get what follows. Anyway, thanks, and I’ll ponder.

1. Back in Black says:

Yes, (2) is Question 2: “Simplify …”

4. marty says:

I’ve finally started updating this post. I’ve added a little background and an answer to Q1. I’ll continue adding as quickly as I can.

5. marty says:

Getting there. I’ve updated the post with a couple of solutions to Q2.

6. marty says:

I’ve updated the post with solutions to Q3. This included, at least for me, an unexpected method for solving such questions.

I’ll probably post a few quick comments on the proofy bets, and then this one is done. I’ll try to post a TNDOT very soon.

7. marty says:

OK, I’ve added a quick update on the proof bits, and I’m outta here. TNDOT 5 shortly.