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.)
 A: 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. 
A: 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 $a\neq 1$.
A: (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.)
