Remedial students have seen quadratics before but, perhaps they don't elicit positive memories. The textbook (designed for people taking the course for the first time, not for remedial students) spends a long time on "multiplication of binomials" before there is any factoring. The idea (I think) is you will see the patterns such as "difference of two squares" and the "perfect square trinomial" then, when it's time to factor, you can recognise these patterns in the new context.

Many remedial students lack confidence in their abilities, when they see patterns they don't seem to trust themselves to have recognised something significant. They have no problem learning to multiply the binomials. But, this time feels wasted because when we move to do the reverse and factor I'm met with silence. In addition, multiplying binomials isn't hard, especially for students who have done it before. What's worse, I risk boring them.

When I start factoring examples, I always start trying to nurture their intuition. But I feel like it always ends up being an algorithm.

For $a \neq 1$, our book and syllabus recommend something called the "AC Method" which I had never heard of until I started planning. It's an algorithm, my students like it. I hate it. In the end, I'm happy to see them getting correct answers under their own power and feeling successful. That's why I teach it. That's why I skip the multiplying. But I want MORE for them. I want the math to seem a bit more tame and transparent than it was before they began my course. I didn't learn math the way that I'm feeling pressured to teach math.

Are there any "methods" I'm overlooking? I do think their remedial experience should shape the course and the lessons, but I would hope it could act as a springboard rather than a wall.

  • 3
    $\begingroup$ I believe your question is closely related to an earlier question and answer here on MESE: matheducators.stackexchange.com/q/6031/267 $\endgroup$ Commented Apr 19, 2015 at 22:05
  • $\begingroup$ That helps for $a\neq0$ $\endgroup$
    – futurebird
    Commented Apr 19, 2015 at 22:07
  • $\begingroup$ I teach a similar class and in the past have found myself with some of your feelings. Perhaps evaluating an expressions (both factored and not) for some values will help. You may also try graphing quadratic equations to help give these problems more meaning. Ultimately, if your remedial class is like mine, your job as the teacher is to prepare them for the exam. I feel obligated to make sure they have learned all of the things they see on the university-created exam. May I ask why you feel "pressured to teach math" this way? $\endgroup$ Commented Apr 19, 2015 at 22:23
  • $\begingroup$ The exam, is part of it. Keeping students happy is another part. $\endgroup$
    – futurebird
    Commented Apr 19, 2015 at 22:24
  • 1
    $\begingroup$ I posed this quesiton some time ago. My colleague uses this method now and loves it; has had great results. This works for all quadratics provided that the GCF has been factored previously. $\endgroup$ Commented Apr 22, 2015 at 20:12

4 Answers 4


The AC-method is often presented as an unmotivated rule-based algorithm, e.g. here. If that is the way that you learned it, then I can understand why you hate it. Below I explain the (little-known) algebraic ideas that lie at the heart of the algorithm, e.g. I emphasize how the non-monic factorization algorithm can be viewed as a conjugation of the monic factoring algorithm.

Understanding these underlying conceptual ideas may greatly assist in constructing more motivated presentations of the method. Further, it may help you to develop a more positive view of the method after learning that there is some interesting mathematics behind the method (it is intimately connected to various refinement-based views of factorization and related lattice-theoretic ideas that play key roles in noncommutative generalizations - see below).

Let's examine the classical case, which reduces factorization of general quadratic polynomials to the simpler special case when the polynomial is $\rm\color{#c00}{monic}\,$ (lead coeff $\color{#c00}{=1})$ as follows

$$ {\begin{eqnarray} \rm\: a\:f(x)\:\! \,=\,\:\! a\:(a\:x^2 + b\:x + c)\, &\,=\,&\rm\,\ \color{#c00}{X^2}\, +\, b\:X\, +\, \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{\overbrace{\color{#0a0}{ac}}^{\rm\qquad\ \ \ \ \ {\bf\color{#0a0}{AC}-method}}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! =\, g(X),\ \ \ X = a\:x \\[.5em] \rm e.g.\qquad\qquad\ \ \ \ f \, &\,=\,&\rm \ \, 2\ x^2-\ 3\ x\ +\ 1\\ \rm \Rightarrow\ \ 2\,f\, &\,=\,&\rm\!\ (2x)^2\! -3(2x)\!+\!2\\ &\,=\,&\rm \ \ \ \color{#c00}{X^2}-\, 3\ X\ +\ 2,\,\ \ X\, =\, 2x\\ &\,=\,&\rm \ \ (X-2)\ (X-\,1)\\ &\,=\,&\rm \ (2x-2)\,(2x-1)\\ \rm \Rightarrow\ \ f\:=\: 2^{-1}\,(2\,f)\, &\,=\,&\rm \ \ \, (x- 1)\ (2x-1)\\ \end{eqnarray}}$$

If we denote our factoring algorithm by $\,\cal F\ $ then the above transformation is simply

$$\rm {\cal F}\ f\ =\ a^{-1}{\cal F}\, a\ f\qquad\! $$

Thus we've transformed by conjugation $\rm\ {\cal F} = a^{-1} {\cal F}\ a\ \,$ the problem of factoring non-monic polynomials into the simpler problem of factoring monic polynomials. It works for any degree:

$$\begin{eqnarray}\phantom{1^{1^{1^1}}}\!\!\!\!\!\!\rm a\:f(x)\:\! \,=\,\:\! a\:(a\:x^2 + b\:x + c) \!&\,=\,&\!\!\rm\: X^2 + b\:X + \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\smash[t]{\overbrace{\color{#0a0}{ac}}^{\rm\qquad\ \ \ \ \ {\bf\color{#0a0}{AC}-method}}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! =\, g(X),\ \ \ X = a\:x \\[5px] \rm\: a^{n-1}(a\:x^n\! + b\:x^{n-1}\!+\cdots+d\:x + c)\! &\, =\,\,&\!\!\rm\: X^n\! + b\:X^{n-1}\!+\cdots+a^{n-2}d\:X + \color{#0a0}{a^{n-1}c} \end{eqnarray}\qquad\quad $$ After factoring the monic $\rm\,g(X)\, =\, a^{n-1}f(x),\,$ the transformation must reverse to yield a factorization of $\rm\:f,\, $ because $\rm\ a^{n-1}$ must divide into the factors of $\rm\ g\ $ by invoking Gauss' Lemma, i.e. primes $\rm\,p\in\rm\mathbb Z\,$ remain prime in $\rm\,\mathbb Z[X],\,$ so $\rm\ p\ |\ g_1(x)\:g_2(x)\,$ $\Rightarrow$ $\,\rm\:p\:|\:g_1(x)\:$ or $\rm\:p\:|\:g_2(x).$

The AC-method also works for multivariate polynomial factorization, e.g below we combine it with $\color{#0a0}{\text{completing a product,}}$ in a similar way to completing a square (from this MSE question)

$$ \begin{eqnarray} &&\rm \ \ \ a\ x\ y &+&\rm b\ x&+&\rm c\ y &=&\rm\ \ d\\ \smash{ \overset{\times\ a}\iff} &&\rm \ \ ax\,ay &+&\rm b\,ax &+&\rm c\,ay &\rm\,=&\rm ad\\ \iff &&\rm \ \ \ {X\ Y} &+&\rm {b\,X} &+&\rm {c\,Y} &=&\rm ad,\quad X = ax,\ \ Y = ay\\ \iff &&\rm \color{#0a0}{(X\!+\!c)}\!\!&&\rm\!\!\!\!\!\! \color{#0a0}{(Y\!+\!b)}&\color{#0a0}-&\rm \color{#0a0}{bc} &=\,&\rm ad,\quad {\rm by\ \ \color{#0a0}{completing\ the\ product}}\\ \iff &&\rm (ax\!+\!c)\!\!&&\rm\!\!\!\!\!\!(ay\!+\!b)\!\! &&\rm &=&\rm ad\!+\!bc \end{eqnarray}$$

Now it's easy to find all integer roots $\rm x,y$ by spliting $\rm ad\!+\!bc$ into two factors and checking if they have the form on the LHS. There are only finitely many such splittings so it is a finite process.

Readers familiar with university algebra may wonder how this relates to better known factorization properties. The AC-method works not only for UFDs and GCD domains, but additionally for any integrally-closed domain satisfying the

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

Elements $\rm\,c\,$ satisfying this are called primal. One easily checks that atoms are primal $\!\iff\!$ prime. Further, products of primes are primal. So "primal" may be viewed as a generalization of the notion "prime" from atoms (irreducibles) to composites.

Integrally closed domains whose elements are all primal 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 Schreier 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 forms of unique factorization. They prove especially fruitful in noncommutative rings (see Cohn's 1973 Monthly survey Unique factorization domains).

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


This is specifically in response to the question: Are there any other "methods" I'm overlooking?

Here's how I do it.

After I teach factoring $x^2 + bx + c$, but before I teach factoring $ax^2 + bx + c$ for $a ≠ 1$, I like to teach factoring by grouping of four-term cubic polynomials such as $2x^3 + 12x^2 + 5x + 30$.

I teach students to start by factoring the GCF out of each pair of terms.

$$2x^3 + 12x^2 + 5x + 30 = 2x^2 (x+6) + 5 (x+6)$$

Then pull out the common $(x+6)$ from these two terms.

$$(x+6) (2x^2 + 5)$$

I have the students practice this with several more examples. Then I give them an example that doesn't work, such as:
$$3x^3 + 15x^2 + 4x + 8 = 3x^2 (x+5) + 4 (x+2)$$

Students recognize that we are stuck at this point because $(x+5)$ and $(x+2)$ are not the same, so we can't pull anything out.

I circle the coefficients in the original polynomial for one of the examples that worked and one of the examples that didn't work:
$$2x^3 + 12x^2 + 5x + 30\\ 3x^3 + 15x^2 + 4x + 8$$ Then I ask how we could have known by looking at those four coefficients which one was going to work and which one was not. Most classes (sometimes with a little guidance) will arrive at something like this:

In the first example, $\frac{12}{2} = 6$ and $\frac{30}{5} = 6$ as well.
In the second example, $\frac{15}{5} = 3$, but $\frac{8}{4} = 2$, so they aren't the same.

Usually then I shift to factoring $ax^2 + bx + c$ with $a ≠ 1$. I remind them that they've gotten good at factoring stuff like $x^2 + 7x + 10$, but I warn them that it's more complicated when there is a number in front of the $x^2$ (and I use my finger to point at the spot to the left of $x^2$).

I write on the board $2x^2 + 11x + 12$. I ask them if we could use the factoring by grouping method to factor it, just like we used to factor $2x^3 + 12x^2 + 5x + 30$. When they say no, because there are only three terms, I agree that we want to have four terms, and suggest that we split up the middle term. $$2x^2 + 11x + 12 = 2x^2 + 8x + 3x + 12$$ I make sure that they're on board with us being allowed to rewrite $11x$ as $8x + 3x$. Usually I draw two slanted lines downward out of the $11x$ to the $8x$ and $3x$, respectively. Then I ask them to factor this by grouping, which they are usually able to do:

$$2x^2 + 11x + 12 =\\ 2x^2 + 8x + 3x + 12 =\\ 2x (x+4) + 3 (x+4) =\\ (x+4) (2x+3)$$

Then I ask whether factoring by grouping would always work no matter how we break up the $11x$; for instance, will it work if we break up the $11x$ like this:

$$2x^2 + 11x + 12 =\\ 2x^2 + 2x + 9x + 12$$ I tell them to try it.
$$2x^2 + 11x + 12 =\\ 2x^2 + 2x + 9x + 12 =\\ 2x (x+1) + 3 (3x+4)$$ ...and we are stuck, since $(x+1)$ and $(3x+4)$ don't match each other.

I ask, how can we tell just by looking at the four coefficients that
$2x^2 + 8x + 3x + 12$ will work, but $2x^2 + 2x + 9x + 12$ will not work?

Someone will give the same sort of answer as the one given for cubic polynomials.
The first one works because $\frac{8}{2} = 4$ and $\frac{12}{3} = 4$ as well.
The second one doesn't work because $\frac{2}{2} = 1$, but $\frac{12}{9} \neq 1$.

I ask the class to consider $ax^2 + bx + c$ in general, where we break up the middle term to rewrite the quadratic as:
$$ax^2 + bx + c =\\ ax^2 + rx + sx + c$$

I ask, what must be true of $a, r, s,$ and $c$? Someone usually says that $\frac{r}{a}$ and $\frac{c}{s}$ have to be the same. So I write the equation:
$$\frac{r}{a} = \frac{c}{s}$$

I ask them how to get rid of fractions to make the equation a little nicer. Usually someone says to cross-multiply. If not, I just ask the class how to change the equation by cross-multiplying, and someone remembers how to do that. They arrive at: $$r * s = a * c$$ So when we have $2x^2 + 11x + 12$, and we are trying to break it into
$2x^2 + rx + sx + 12$ for some numbers $r$ and $s$,
we need to make sure that $r * s = a * c$.
In this case, that means $r*s = 2*12 = 24$.
I ask whether we could choose $r = 4$ and $s = 6$, since those multiply to $24$.
$$2x^2 + 11x + 12 =\\ 2x^2 + 4x + 6x + 12$$
I ask if that is allowed. Someone will say no, because $4 + 6 \neq 11$.

So I say, oh, you're saying that $r + s$ has to equal which letter? $a$? $b$? $c$?

Someone says $r + s$ has to equal $b$.

I write on the board: $$r + s = b\\ r * s = a * c$$

I say let's see if those rules work in another example:
Factor $3x^2 + 8x + 4$.

I write on the board:
$$r + s =$$ $$r * s=$$

I ask them how to finish out the equations using the formulas, with no numbers yet. They tell me what to write:

$$r + s = b$$ $$r * s = a * c$$

Then I write another equals sign after each line and ask for what numbers belong there. Students tell me what to write.

$$r + s = b = 8\\ r * s = a * c = 3 * 4 = 12$$

By this point, they have already been factoring expressions like $x^2 + 8x + 12$, so they proceed to figure out the two "mystery numbers" $r$ and $s$ that add up to $8$ and multiply to $12$. They get $2$ and $6$.

I ask, so, does that mean the answer is this:
$$3x^2 + 8x + 4 =\\ (x + 2)(x + 6)$$

Some students may say yes. If so, I ask them to check by FOIL-ing out the $(x+2)(x+6)$. When we arrive at $x^2 + 8x + 12$, I ask if this is what we were hoping to get. Someone says, no, we wanted $3x^2 + 8x + 4$.

I let them know that writing $(x+2)(x+6)$ at this point is a very common mistake. I remind them that the r and s in this case -- where we have a number besides 1 in front of the $x^2$ -- were intended to help us split up the middle term so that we could factor by grouping.

$$3x^2 + 8x + 4 =\\ 3x^2 + 2x + 6x + 4$$

I ask students how we should continue, and usually they are able to tell me:

$$3x^2 + 8x + 4 =\\ 3x^2 + 2x + 6x + 4 =\\ x (3x + 2) + 2 (3x + 2) =\\ (3x + 2)(x + 2)$$

If we have time, I ask them to FOIL that out in order to check, and students are pleased when we get $3x^2 + 8x + 4$.

Note: It is intentional that I don't include any negative coefficients in this part of the lesson. In general, when I'm teaching something that is complicated enough on its own, I try to remove other complications until they gain some confidence in the new topic. By the end of the lesson, or by the next day's lesson, I would start including $ax^2 + bx + c (a \neq 1)$ factoring problems with negatives. By this point, they already are comfortable handling negatives in factoring $x^2 + bx + c$, so the transition isn't too bad.


My answer was just a little too long for a comment.

I'd start with "X puzzles" (which I put in quotes because I'm not sure they have a name). These are just big X's where you have 1 number on the top and 1 on the bottom. Then they write 1 number on the left and 1 on the right that will add to the bottom and multiply to the top (or vice versa if you want).

After that I'd go to the lattice style (which is basically algebra tiles but without the tiles themselves) and have them do the multiplication (give them the outside and work in). Then show them a few that have the middle and have them find the outsides. Then gradually pull back so you have just a trinomial and see what they can do.

If you want to do the difference of two squares you can do a page of just those types of problems and ask for any commonalities they find in the factors versus what the problems look like.

Here's an example (I am not affiliated with this site, just did a quick search): http://emathlab.com/Algebra/GenericRectangles/genericRectangleFactoring.php

  • $\begingroup$ Here is a CPM document on this "lattice style": "Using the Area Model to Teach Multiplying, Factoring and Division of Polynomials" PDF download. $\endgroup$ Commented Apr 20, 2015 at 0:44

Manipulatives work well for exploring quadratic forms. Algetiles are the standard, but you could use base 10 blocks if algetiles aren't available. You can easily make a set of algetiles if don't have one available.

One of the issues with manipulatives is that students sadly don't encounter them enough, so you will probably have to run over the basics with them before you do algebra: positive and negative tiles, zero pairs, and so on.

A fast introduction to the use of algetiles for quadratics is at http://mathbits.com/MathBits/AlgebraTiles/AlgebraTiles/AlgebraTiles.html. Factoring amounts to taking a given set of algetiles and forming them into a rectangle, so it is analogous to factoring integers by taking a collection of unit squares and forming it into a rectangle.


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