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.

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.

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)

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).

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.

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.)

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.

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

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).

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.

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.

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.

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?

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 …

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!

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.

*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.

Thanks, BiB. I’ll try to look soon.

* (1) – (2) not (1) – (3).

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?