Factoring $ac$ to factor $ax^2+bx+c$

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.)

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.

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 $a\neq 1$.

(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.)