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I was watching a first-year high-school-algebra student struggle with factoring quadratics last night. Given a quadratic $ax^2+bx+c$ (I'll give you the exact example in a moment), her method — presumably her teacher's — was as follows: find the factors of $ac$, and see which pair add up to $b$.

It seems to me that multiplying $a$ by $c$ is needless work. True, it's $ac$ whose factors sum to $b$. But when writing out the factors as (say) $(a_1x+c_1)(a_2x+c_2)$, one's actually working with not the factors of $ac$ but rather the $c_i$ and the $a_i$, factors of, respectively, $c$ and $a$.

So my question is: Is there any advantage to working with $ac$ — finding its factors, seeing which ones sum to $b$ — and, if so, what is that advantage?

Here's the example she was working, so you get a better under standing of what I mean. The problem was (or amounted to) $9x^2-47x+60=0$. This poor girl found $9\cdot60$ and started examining its factors to see which sum to $47$. Eventually, she hit upon the answer, $20\cdot27$, and put them in her parentheses as $(9x-20)(x-3)$ (somehow divining that the $27$ was to be split up as $9\cdot3$, and the $9$ as $9\cdot1$; I'm not sure how she hit upon that).

My method would have been instead to consider $(9x-c_1)(x-c_2)$ or $(3x-c_1)(3x-c_2)$. (I'd reject the latter because $3\nmid47$, but I wouldn't expect that of my high schooler. So consider both possibilities.) Then find factors of $60$ that possibly fit in one of those pairs of parentheses, and hit upon $3\cdot20$.

Again, what if anything is the advantage to factoring $ac$? (The advantage, if any, may be pedagogic.)

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Did she multiply and then find factors of $540$, or did she continue straight to $3^2\cdot 6\cdot 10 = 2^2\cdot 3^3\cdot 5$? –  Neal Apr 30 '12 at 15:54
    
@Neal, I wish she'd have done the latter (if she had to deal with $ac$ at all)! No, she found factors of $540$. –  msh210 Apr 30 '12 at 15:55
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Maybe the same advantage as the supposed $1+2+3+\cdots +100$ assignment to Gauss's class. Keeps them busy. –  André Nicolas Apr 30 '12 at 16:04
    
I think that pedagogically, it creates uniformity with the method for factoring $1x^2 + bx + c$ (that is, find factors of $c$ that add to $1$). –  Neal Apr 30 '12 at 16:10
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@Neal: But factoring $a$ and $c$ also creates pedagogical uniformity. The special case $a=1$ is factor $1$, factor $c$. –  André Nicolas Apr 30 '12 at 16:30

4 Answers 4

This is sometimes called the AC method and it works for higher degree polynomials too. Namely, we can reduce the problem of factoring a non-monic polynomial to that of factoring a monic polynomial by scaling by a power of the leading coefficient $\rm\:a\:$ then changing variables $\rm\: X = a\:x$

$$\begin{eqnarray} \rm\: a\:f(x)\:\! \,=\,\:\! a\:(a\:x^2 + b\:x + c) &\,=\,&\rm\: X^2 + b\:X + ac\, =\, g(X),\ \ \ X = a\:x \\ \\ \rm\: a^{n-1}(a\:x^n\! + b\:x^{n-1}\!+\cdots+d\:x + c) &\,=\,&\rm\: X^n\! + b\:X^{n-1}\!+\cdots+a^{n-2}d\:X + a^{n-1}c \end{eqnarray}$$ After factoring the monic $\rm\:g(X)\, =\, a^{n-1}\:f(x),\:$ we are guaranteed that the transformation reverses to yield a factorization of $\rm\:f\:$ since the constant $\rm\:a^{n-1}\:$ must divide into the factors of $\rm\:g\:$ by Gauss' Lemma, i.e. primes in $\rm\mathbb Z$ remain prime in $\rm\mathbb Z[X],\:$ so $\rm\:p\ |\ g_1(x)\:g_2(x)\:$ $\Rightarrow\:$ $\rm\:p\:|\:g_1(x)\:$ or $\rm\:p\:|\:g_2(x).$

This technique also works for multivariate polynomial factorization, e.g. it applies to this question.

Remark $\ $ Those who know university algebra might be interested to know that this works not only for UFDs and GCD domains but also for integrally-closed domains satisfying

$\qquad\qquad$ Primal Divisor Property $\rm\ \ c\ |\ AB\ \ \Rightarrow\ \ c = ab,\ \ a\ |\: A,\ \ b\ |\ B$

Such domains are called Schreier rings by Paul Cohn (or Riesz domains, because they satisfy a divisibility form of the Riesz interpolation property). In Cohn's Bezout rings and their subrings he proved that if $\rm\:D\:$ is Shcreier then so too is $\rm\:D[x],\:$ by using a primal analogue of Nagata's Lemma: an atomic domain $\rm\:D\:$ is a UFD if some localization $\rm\:D_S\:$ is a UFD, for some monoid $\rm\:S\:$ generated by primes. These primal and Riesz interpolation viewpoints come to the fore in a refinement view of unique factorization, which proves especially fruitful in noncommutative rings (e.g. see Cohn's Monthly survey on unique factorization).

In fact Schreier domains can be characterized equivalently by a suitably formulated version of the above "factoring by conjugation" property. The connection between this elementary AC method and Schreier domains appears to be unnoticed in the literature.

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So you're suggesting that this one of those (many) cases where a technique in advanced math is used in a simpler form in basic math so the later math is easier to learn (even though, in this case, (a) the later math will most likely never be learned (how many students study abstract algebra?) and (b) the method is not the best for the subject at immediate hand). Possible, I suppose. +1. –  msh210 Apr 30 '12 at 16:41
    
@msh210 Re: (a), currently the majority of users here appear to be university level students, so they will likely study abstract algebra. Re: (b), as I explained, the theory behind the rote algorithm taught in high-school is as I described. This is similar to unique factorization: it is used long before the theory is taught (if ever). It is arguable if this is "best" pedagogically. The theory behind both could be taught in high-school. Or perhaps your "best" refers to something else? It's not clear what you mean by "the method is not the best". What method, and not best for what? –  Bill Dubuque Apr 30 '12 at 17:04
    
Re (a), the question was about high-school-level math. Re (b), I specified what I meant by "best": "best for the subject at immediate hand", viz high-school math. High-schoolers don't learn about UFDs (other than $\mathbb Z$ and ${\mathbb Z}[X]$). By "the method" I meant the method my high-school had learned for factoring. –  msh210 Apr 30 '12 at 17:10
    
@msh210 Generally I write my answers for the entire community, not only the OP. The first part is at high-school level. The second part requires knowledge of undergrad algebra. The OP might know undergrad algebra, since they were "watching a high school-student solve" it. If so, they may be able to use my answer to help explain at high-school level why the AC method works. Since this is rarely if ever explained, I thought it worth a little elaboration here. –  Bill Dubuque Apr 30 '12 at 17:18
    
Yes, as the OP (with a Ph.D. in math), I can assure you I'm familiar with college-level abstract algebra. (Or I took it, anyway. :-)) Thanks for the answer, which is both intrinsically interesting and may possibly even be put to use in explaining the reasons for my high-schooler's teacher's method to her (the student). –  msh210 Apr 30 '12 at 17:22

I've used this idea to help factor quadratic expressions where $a\neq 1$. Let me see if I can fill in the blanks and help you figure the idea out.

Using the example you provided, you found that the two appropriate factors of $9\cdot 60$ are $-20$ and $-27$. \begin{equation} 9x^2-47x+60=0\\ 9x^2-27x-20x+60=0\\ 9x(x-3)-20(x-3)=0\\ (9x-20)(x-3) \end{equation}

The idea is to use the factors of $ac$ to split $b$ into the two terms, then factor by grouping. It's not always the nicest if $a\cdot c$ is large as in this problem, but when $a\cdot c$ is relatively small, it makes finding the answer a bit more simple than if you were not able to immediately see the factorization (which I think is usually the case when $a\neq 1$.

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When you write -20 and -27 instead of $-20$ and $-27$, it comes out looking like a hyphen instead of a minus sign. Contrast $5$-$3$ with $5-3$. (I edited the answer accordingly.) –  Michael Hardy Apr 30 '12 at 16:08
    
Thank you Michael! I'll be more careful in the future. –  Kyle S. Apr 30 '12 at 16:08
    
But by grouping is not the method taught the high schooler I was watching struggle. (Maybe it should have been.) My question was seeking advantages of the method she was taught. (I admit, though, that the wording of my question sought merely advantages to factoring $ac$, and you answer provides one such. +1, then.) –  msh210 Apr 30 '12 at 16:32

(I'll just summarize my previous answer)

Your method is guess-and-check (or process of elimination).

When working with $(9x−c_1)(x−c_2)$ or $(3x−c_1)(3x−c_2)$, you still have to try all the factor pairs $c_1c_2 = c = 60, c_1+c_2 = b = -47$, which can be tedious.

The "AC" method gives the factorization quickly (well, once you have $pq = ac, p + q = b$).

$${\color{Red} 9}x^2−47x+60 = \dfrac{1}{{\color{Red} 9}}({\color{Red} 9}x - 20)({\color{Red} 9}x - 27) = (9x -20)(x - 3)$$


(You can also split up the linear term and factor by grouping, as Kyle demonstrates.)

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So you're saying my way is tedious because one has to try all the factors to see which sum to $b$, whereas the other method is not tedious once one's found the factors that sum to $b$. –  msh210 Apr 30 '12 at 16:44
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Alas, in summary, I omitted a key commentary (or two)! "Your" method (which was "my" method for ten years!) requires some accounting system to ensure that you haven't missed a possible combo of $c_i$. The AC method requires testing factors $1, ... \sqrt{ac}$. I also teach my students to input $y_1 = 1/(ac)$ into their calculators and check the table for integer entries... again, was trying to escape with a mere summary :) –  The Chaz 2.0 Apr 30 '12 at 16:49
    
Just as listing factors of $ac$ requires testing factors up to $\sqrt{ac}$, my method requires testing factors up to $\sqrt c$. But, yes, each needs testing twice (once for $3x+c_1)(3x+c_2)$ and once for $(9x+c_1)(x+c_2)$, and possibly eight times if $a$ is highly factorable), whereas in your method each needs testing but once (twice in case $ac<0$), which can result in fewer false negatives. Good point (and will be even better if it's edited into the answer!); +1. –  msh210 Apr 30 '12 at 16:53
    
Here are some calculations... in this case ($a = 9$), we have two possibilities $(9x...)(1x...), (3x...)(3x...)$. For each of these, we test against the $12$ possible pairs $(c_1, c_2)$ for $24$ total guesses. If we disallow theoretical preclusion of $\pm$, then we potentially have twice this many guesses. Contrast this with the... well crap, just saw your comment!... anywho, $24$ factor pairs of $540$. Then we can use some theory to hone in ... –  The Chaz 2.0 Apr 30 '12 at 16:56
    
Off to campus. I'll edit later. –  The Chaz 2.0 Apr 30 '12 at 16:56

I don't think there is another way other than actually guessing and checking. One way to help though is find the prime factors of $ac$ through short division.

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