Yeah but I was thinking of algorithms that are not the quadratic formula in disguise. Chronos pretty much has it. For my purposes as an algebra teacher, I teach it to my students only if a=1 so you are left with factoring c.
[Ignore this post]
Do you know about – and are you actually thinking of – the “AC method” for factoring trinomials?
I’ve posted it before. In fact, on googling it, I see I’ve posted it twice:
http://boards.straightdope.com/sdmb/showpost.php?p=18311715&postcount=11
http://boards.straightdope.com/sdmb/showpost.php?p=16803232&postcount=43
Notwithstanding the profusely fawning praise you got from Your Great Darsh Face for this, there remains another school of thought to the effect that the FOIL method is an unspeakable abomination, unfit to defile the pages of any algebra textbook.
The problem is, too often it’s taught without sufficient (or any) attention to the underlying Distributive Law upon which it’s based (or, alternatively, students just let it go in one ear and out the other). The FOIL mnemonic, taken without sufficient context, does nothing to illuminate WHY it works, nor does it generalize well to other polynomial product problems, like multiplying polynomials with more than the basic two term (binomials) each.
One of the threads I linked in my post just above, namely this one
discusses the FOIL rule (and some debate about it) in more detail.
One interesting point is that it’s not entirely trivial to prove that Senegoid’s “AC method” always works; there’s a subtlety to deal with (this same subtlety also must be dealt with in establishing the correctness of the factorizability criterion I gave above; I was going to pose the challenge of observing this subtlety in a follow-up post, then decided no one would care, and have now changed my mind again). Namely:
Suppose you’ve successfully re-expressed the quadratic Ax[sup]2[/sup] + Bx + C which you wish to factor as Ax[sup]2[/sup] + (P + Q)x + C where AC = PQ. [Specifically, given that B[sup]2[/sup] - 4AC is a perfect square D[sup]2[/sup], as it must be, we will have that {P, Q} is {(B + D)/2, (B - D)/2}. Thus we can avoid tediously checking through candidate factorizations altogether!].
Next, Senegoid tells us that we can turn this into x * (Ax + P) + (Qx + C), and then finish off by pulling integer factors out of Ax + P and Qx + C till they match (the “factoring by grouping” mentioned here).
The subtlety is this: Why are we guaranteed that we can pull out integer factors from Ax + P and Qx + C in such a way that they come to match?
Well, I don’t know how you prove things like that really formally. (Well, maybe I could if I dust off my rusty algebra and actually think about it for a while.)
But one can gain some insight by noting an interesting feature of this AC method: In it, you can actually see the steps of multiplying two binomials happening in reverse (providing you actually work out the steps and don’t use the FOIL method)! The usual factoring method (factoring A and C separately) doesn’t really show this.
Start with:
(ax + b) (cx + d)
Working out the steps, in detail with repeated use of the Distributive Law:
(ax + b)cx + (ax + b)d =
(acx[sup]2[/sup] + bcx) + (adx + bd) =
acx[sup]2[/sup] + bcx + adx + bd =
acx[sup]2[/sup] + (bc + ad)x + bd
If a given trinomial to be factored is formed in this way, then it is factorable, and the AC method (and the factoring by grouping step in particular) essentially just steps through the above process in reverse.
The converse is less obvious, and is (I suppose) what remains to be proved: Only if the trinomial is formed in this way is it factorable. Can a trinomial exist that is not formed as the product of two binomials in this way, and yet is factorable? On the surface, that sounds oxymoronic.
Since you know what I meant, why kick up a ruckus? It’s not necessarily obvious how to recognise when a quadratic has real integer roots, but many people know how to look for them once they’re told they exist, so my M.O. is to hand out part of the solution and let the student try the next part for themselves.
If it turns out they still don’t know what to do next then I do indeed give more help. Perhaps we should compare math degrees and pedagogy some time.
Too trivial to Pit and yet too crass to pass entirely without comment. Manners cost nothing.
The purpose of the post of mine you replied to was to clarify that what I was reacting to wasn’t fundamentally about wording. Rather, I was surprised that there seemed to be some reluctance to consider the general quadratic formula, when that would be precisely the tool that would help people in precisely this situation. It may indeed be the case that this person can cast about and find integer solutions sometimes on their own (and I never want to deny anyone the chance to explore on their own!). Then again, maybe not. I worried that saying “you’ll find you don’t need the quadratic formula - the expression factorises”, as though that should somehow make the answer obvious, was unfortunate as a thing to say to someone who could very well find that they still have no greater idea how to solve this problem or others like it.
There was an image gaining support in the thread, it seemed to me, that the quadratic formula is somehow overkill when dealing with integer cases, as though the integer case is intrinsically obvious on its own. I think that’s false, and so I said so.
I didn’t really mean to pick on you, specifically (I believe that you would have returned to give more explanation if it turned out the OP did not understand yet); your post just happened to be a handy launching point for a discussion of a phenomenon I feel I run into with some frequency.
I imagine the Montessori approach would be to draw two squares (of indeterminate length).
Then find a rectangle that has 12 units in it, that fits evenly between them, and has the side lengths differing by one, and that would be how you solve a problem like Maria.
That is oxymoronoic. To be factorable is, by definition, to be the product of two binomials in this way.
Rather, the question is: Given A, C, P, and Q, with AC = PQ, how do we know there actually are a, b, c, and d such that ac = A, bc = P, ad = Q, and bd = C (so that we can factor Ax^2 + (P +Q)x + C as a product of two binomials as above)?
That is, if you happen to know ahead of time that a quadratic is factorizable, then it’s clear that the AC method must succeed in finding its factorization, as you demonstrate.
But what’s perhaps not clear (though it is true!) is that, even if you have no idea ahead of time whether a factorization exists, any time you can manage to split AC into two factors which sum to B, you are then assured that Ax^2 + Bx + C is indeed factorizable (and can then proceed as usual to determine its factorization). That’s the interesting but subtle further fact to be proved.
Let me also clarify, lest I be misinterpreted, that “quadratic formula”, “completing the square”, using “the AC method” with factorization AC = (B + D)/2 * (B - D)/2 for D = sqrt(B^2 - 4AC), etc., are all the same to me. I don’t care about the form in which the general method is expressed; just the idea that there is a general method which efficiently answers this common sort of problem without guess-and-checking or requiring mysterious intuition.
I haven’t worked it all the way through, but this is going to depend on Euclid’s prime number lemma, isn’t it?
Yes, that’s certainly a result closely connected to this one. (Over an integral domain (i.e., commutative ring without zero divisors), we can conclude Euclid’s prime number lemma from this fact; conversely, given Euclid’s prime number lemma, one can conclude this fact in the case where all relevant values have prime factorizations [which is assured in the integers, but not in more general contexts])
That having been said, you don’t have to think in terms of prime numbers to think about this; as the above indicates, this result is in some sense strictly stronger than Euclid’s prime number lemma, and I actually prefer to think of Euclid’s prime number lemma as a corollary to this one.
(Incidentally, it is worth noting that the proof Euclid gives of his prime number lemma in The Elements is flawed; see https://www.math.nmsu.edu/~davidp/euclid.pdf [which may also trigger insights(/spoilers) for this problem])
OK, I’m satisfied with recognizing the general form of the proof. Being a physicist rather than a mathematician, I don’t feel the need to go through all the details.
And I’m not at all surprised that Euclid’s proof was flawed-- Off the top of my head, I know of at least two of his proofs that are so flawed that they actually reach incorrect conclusions.
Fair enough. I’ll write it out for completeness’s sake:
Recall that the original question was, given that AC = PQ, how do we know that the binomials (Ax + P) and (Qx + C) are both scalings-up (by a whole number factor) of some common binomial?
The answer is to note that the ratios A/P and Q/C are equal [in that cross-multiplication, AC = PQ sense], and such equal ratios are always both scalings-up of some common reduced form.
Ah, but why are equal ratios always both scalings-up of some common reduced form? Well, we can take the common reduced form to be the lowest-terms form of said ratio [i.e., the one with minimal denominator].
Ah, but why is a ratio always a scaling-up of its lowest-terms form?
Now here’s a simple property I suppose most of us are convinced of by familiarity (and perhaps blind authority), but have never actually paused to see a reason for. It’s not as trivial an observation as that familiarity makes it seem! (Indeed, trying to show this is where Euclid screwed up, probably because he too was blinded by familiarity!)
In other words, why does every way of expressing a given ratio have, as denominator, some multiple of the lowest possible one? [For example, the fractions equal to 8/6 are 4/3, 8/6, 12/9, 16/12, etc. The lowest-terms form is 4/3, and all the rest are scalings-up of this; there’s nothing in there with a denominator of 7 or 8 or 10 or anything not divisible by 3. Why couldn’t there be?]
I say, everyone should think about why it works out this way at least once in their life. So much so that I’ll put the final explanation in a spoiler box:
[spoiler]Consider what the possible denominators for a fixed ratio could be like. These will be the whole numbers which multiply by our ratio to give other whole numbers. So let’s take our ratio r and keep adding it to itself, to get the sequence 1r, 2r, 3r, 4r, … . We’re interested in when we get whole numbers, so let’s ignore the whole number components of these terms and focus only on the fractional leftovers (i.e., what comes after the decimal point).
Since we do the same thing to the leftovers at every step (just adding r), if the sequence of leftovers ever hits 0, it’ll begin repeating from the beginning again, in cyclic fashion. Thus, if we first hit 0 after d steps, we’ll next hit 0 after another d steps, and next after another d steps, and so on. In other words, if we find the lowest possible denominator is d, we’ll find the next possible denominator at 2d, and the next at 3d, and so on. Thus, the possible denominators are all multiples of the lowest terms denominator.[/spoiler]
There are contexts of note where this fails, even though the quadratic formula and much of the rest of familiar arithmetic continues to work. For example, suppose we take our whole numbers to be those of the form a + b * sqrt(-6), where a and b are ordinary integers. [For convenience, let us write h for sqrt(-6)].
These can be added, subtracted, multiplied, etc., and the familiar quadratic formula is still true so far as these are concerned. But note that, allowing these as coefficients, we cannot determine factorizability in quite the same way any more: for example, consider the quadratic 2x^2 + 3. Here B = 0 and AC = 6, and we can indeed find two possible coefficients which sum to B and multiply to AC; namely, h and -h. But try as we might, we won’t be able to find a whole number factorization of 2x^2 + 3 in this context; we can get the rational factorization 2(x - h/2)(x + h/2), using the fractional h/2, but there will be no way to distribute the leading 2 across both factors to make both whole; indeed, 2 remains prime in this context in the sense of having no proper factors, and thus cannot be split in twain (proof: enumerate the finitely many whole values of magnitude < 2 and see that none work).
Accordingly, this is also a context where the familiar fact about lowest-terms expressions goes awry: if we consider the ratio -h/3, it has lowest-terms form 2/h [note that |h| = sqrt(6), on the standard account of size for complex numbers, and we will not be able to do better than that for our denominator (proof: enumerate the finitely many whole values of magnitude < sqrt(6), and see that none work)]. However, -h/3 is not a scaling-up of 2/h by a whole number amount; rather, the inflation factor is the fractional h/2.
Of course, no one need care about unfamiliar arithmetic contexts such as this unless it is of interest to them, but it does serve to illustrate the point that the reasoning undergirding the “AC method” applicability/reduction to lowest-terms by cancelling of common factors/uniqueness of prime factorization in ordinary contexts is not actually so trivial as we have may have been inclined to glibly say.
[The connection to Euclid’s lemma, incidentally, is like so:
Our theorem, written out in full, is that, if AC = PQ, then there are a, b, c, and d such that ac = A, bc = P, ad = Q, and bd = C.
In other words, if P divides AC, then P can be split into complementary factors which divide A and C respectively.
In the particular case where P is prime, one of these factors must be P itself, and thus we find that if a prime divides a product, it divides one of the factors; i.e., Euclid’s lemma.]
Trivial
See if A/P = Q/C. If not, then a, b, c, d cannot exist.
A/P reduces to m/n and A/Q reduces to u/v
(mu)/(nv) give us a fraction equivalent to A/C. Choose x such that x(mu) = A and x(nv) = C. If x factors into y and z then
a=ym; b=yn; c=zu; d=zv works.
How do you know you can find such an x? Just because m/n and u/v are in lowest terms doesn’t mean mu/nv is in lowest terms. And if mu/nv is not in lowest terms, how do you know A/C is a simple scaling-up of mu/nv?