# How to Teach Kids About Factoring a Polynomial

Recently, I applied to a fellowship with Math for America, a program dedicated to improving mathematics education in U.S. public schools by recruiting, training, and retaining highly qualified secondary school math teachers. In my quest to get the fellowship, I started to fiddle around with some math ideas that made me curious. One of those is the idea of factoring a polynomial, and specifically, how we teach it.

For one, I'm not a fan of FOIL (first-outside-inside-last) for a plethora of reasons. While I think it's handy to have an acronym that reminds students of a procedure, it only works in a very special case. In this case, FOIL works only for multiplying a binomial by another binomial. Does FOIL lead students toward understanding multiplication of all types of polynomials, or understanding why the distributive property works even with variables? I'm not so sure.

We *do* know that there are alternate ways of approaching multiplication of binomials, but I'd like to focus on using the geometric method of multiplication because, well, because I can.

### Multiplication and Factoring Using Areas

Multiplying (*x* + 2) by (*x* + 3) can be represented like so (see Figure 1):

This makes the following operations look rather simple:

*(x* + 2)( *x* + 3)*x *^{2} + 2*x* + 3*x* + 6*x*^{2} + 5*x* + 6

Using the area method for multiplying binomials also makes factoring an easy task. We can visualize the squares and rectangles in this shape while thinking to ourselves, "Which two numbers have a sum of 5 (second term) and a product of 6 (third term)?" If we look carefully, students have another method for understanding why we get the terms we do after multiplying the binomials shown. How does this relate to trinomials? Let's see.

### Multiplication and Factoring of Cubes

Let's take the last example and multiply it by (*x* + 4). Like so (see Figure 2):

(x + 2)(x + 3)(x + 4)

(x2 + 5x + 6)(x + 4)

x3 + 9x2 + 26x + 24

Or geometrically (see Figure 2): This has awesome implications for finding both the surface area and volume of this figure. Since we already figured out the "face" of this cube earlier (*x*^{2}+ 5*x* + 6), we're basically multiplying that face by the length of *x* and by the length of 4. This yields:

*x*(*x*^{2} + 5*x* + 6) + 4(*x*^{2} + 5*x* + 6)

. . .*x*^{3} + 9*x*^{2} + 26*x* + 24

### Factoring The Cube

Once we find the quadrinomial, the cube gives us a hint for finding the lengths that created the quadrinomial. One would only need to figure out which three numbers give us a sum of 9 (second term) and a product of 24 (last term). These numbers are 2, 3, and 4, so we'll get (*x* + 2)(*x* + 3)(*x* + 4).

Is this much better than using the cubic formula? Absolutely, especially for our students.

What about a quadrinomial like 2*x*^{3} - 11*x*^{2} + 12*x* + 9? We can try to determine all the real roots of this polynomial, or we can take a look at the second and last terms. The simplest combination for a product of 9 is multiplying 3, 3, and 1. The first term's coefficient, 2, makes getting a -11 tricky. We can't get -11 from the set of numbers without considering the first coefficient.

Yet, as we've seen with the other cubes (see Figure 3):

We will see that the term of 2*x* will multiply with any lengths that aren't associated with it. Thus, (2*x* - 3) + (2*x* - 3) will give us 6*x* + 6*x*, or 12*x*. Since we needed to get -11*x*, this means the remaining 1*x* is positive and the 12*x* is actually -12*x*.

Therefore, our quadrinomial gets factored to (*x* - 3)( *x* - 3)(2*x* + 1) or (*x* - 3)^{2}(2*x* + 1).

### Word of Caution

I'm still exploring this for other cases (looking at *x*^{3} + 27, for example), and what imaginary numbers would look like using this model. These methods work great as a way to draw students in, but some special cases will probably require more space than I could lend it here.

Also, this was my rebellion against the cubic formula which, as many mathematicians know, makes little sense to introduce in the classroom.

Let me know what you think in the comments below.

## Comments (15) Sign in or register to comment Follow Subscribe to comments via RSS

I guess I'm getting just a tad obsessed here. I hadn't realized until just now that @CCSSI Mathematics is the same person(s) behind Five Triangles Math. Seeing a link on the latter site to the former was a rather large hint, given that there are no other links offered other than internal ones. And of course, Five Triangles is also anonymous. My, but there seems to be quite a lot to hide going on.

At any rate, all three problems linked to are, of course, NOT trinomials. And they're not in one variable. Coincidence? Hardly. They're hand picked to NOT be relevant to the issues Jose is exploring and that James Tanton discusses in his free course and that countless students have to struggle with every year, often to ill effect (if you consider it an ill effect when something in mathematics that could be intriguing and beautiful - again, see those Tanton videos - is made tedious, pointless, and frustrating to the point where it becomes yet another hurdle that eliminates a sizeable number of, ahem, unfit students. Because clearly, if a kid can't factor quadratic equations given the classic treatment they are given in most US mathematics classrooms, s/he must be intellectually defective, fit only for a job a McDonalds, as long as the electronic cash registers are working. Otherwise, it's flipping burgers forever.

So here are the gems we were asked to consider:

First, x^2 - yz + xy - xz

Well, that's a little wicked. Of course, don't put the terms in a convenient order. Too easy. But factoring by grouping does seem to rear its head here. So let's try grouping them this way:

x^2 + xy | - yz - xz

The left group factors into x (x + y). The right factors into -z(x + y) and the distributive property gives us (x + y)(x - z).

Of course, if you have 3 variables rather than one, students are likely to freeze up, so this is a great problem to make a lot of them feel paralyzed and inferior. Always a goal for a certain breed of mathematics teacher.

2) a^2 - ac - b^2 + bc

Again, make sure not to put the terms in the most pliable order, because that would cut the frustration level a bit.

So we'll write instead a^2 - b^2 | + bc - ac

Neatly, we get the difference of two perfect squares on the left, suggesting

(a + b)(a - b). With a little thought, we can factor out a negative c on the right to get -c(a - b)

[It never hurts to think about what result MIGHT be desirable and then see if there's a way to get it]. So now we can use the distributive property to get (a-b)(a + b - c). Check to see that this multiplication of a binomial times a trinomial gives the original expression. Oh, so tricky-san is our Five Triangles host. Why, you'd almost think that these problems were created to "prove" that modeling something that is very different, namely quadratics in one variable, just won't yield to methods that of course we already KNOW they won't yield to. Hmm. What would the goal be in bringing them up in a conversation about something different?

Finally, we get this gem: x^2 - 2xy + y^2 - 1

This one is neatly designed to lead you down various primrose paths to failure.

For example, seeing that it is four terms, and given the previous two problems, we are likely to want to factor by grouping into pairs of binomials. And it would seem that a promising approach would be to group, say, x^2 - 1 | + y^2 - 2xy

But not only does this fail (you can factor both expressions, but not so you get a common factor), but so will other grouping into pairs (with grouping x^2 + y^2 obviously doomed before we start). Despair begins to set in, and our oh-so-clever problem-poser has once again made sure to list the terms in an unhelpful order.

However, we aren't going to be daunted. We try writing x^2 - 2xy + y^2| - 1

Since the left side is a perfect square trinomial, namely (x - y)^2, we have the difference of two perfect squares after all, but not as we first thought.

(x - y)^2 - 1 then factors as (x - y + 1)(x - y - 1). Bingo. But of course this is the product of two trinomials, not two binomials. We've gotten awfully far from ax^2 + bx + c = 0, haven't we? But then, that's the point. So it's not that the model being used to explore quadratic equations of the form most often dealt with in school algebra fails, but that it doesn't address problems that it was never claimed to explain. In a now archaic phrase that somehow still resonates, No Duh!

Can't speak for anyone else, but I prefer actually understanding how the results of the multiplication of the binomial and the trinomial show you precisely why the signs have to be that way. Examining the flow between factored and unfactored forms of the expressions are supposed to eventually sink in as to what's going on. If a mnemonic helps kids get through the exam, that's fine, if the goal is getting through an exam. But there's a downside. I like mnemonics best when they are used for arbitrary information, e.g., order of operations or the order of the cranial nerves or the names of the five Great Lakes. There's just no logical way to deduce the order or the names. So having aids to memory makes sense, if for nothing else than as a checklist to make sure you've got them all in the last case, and for the ordering in the second one. Order of operations does have a kind of logic to it, but it's still got an arbitrariness, too.

When there's something conceptual at play, however, I prefer to draw students' attention to that. Algebra is at least in part about structures, and I like to point back to how two-digit by two-digit multiplication can be broken down into (10a + b)(10c +d) followed by using the distributive property or the lattice method, or any reasonable approach that stresses the underlying structure, using specific numerical examples, and then returning to the algebraic setting. Doesn't work for every student (or elementary teacher, for that matter), but it's worth looking at. And I think we've got things like that going on with sums and differences of perfect cubes. I love that Jose is playing with this visually. Of course, as someone will surely want to say, that model breaks down when we go to one higher dimension, so obviously it's a complete waste of time to even consider. :^)

I saw the cubic model as a Montessori device in my new classroom (leftover from previous teacher). Thought it was interesting.

What I'm wondering is how we decide the "size" of the x length. And how things would look if we varied that - as it is a variable after all.

Also, I have a problem teaching factoring of 3rd degree polynomials this way without saying out loud and multiple times that there are VERY FEW of these that have nice factorizations. There are also ridiculously few quadratics with factorizations, but we have the quadratic formula to (hopefully) show that the "non-nice" factorizations can also be irrational or even imaginary.

Consider x^2 - 10x +34. It "factors" into (x-5-3i)(x-5+3i) and you'd only get that mess by setting the polynomial to zero, using the quadratic formula to find the zeros/roots and apply the idea that (x-c) is a factor iff c is a zero/root.

Thank you for your comment. I might suggest that I would *have* to make the connection from the geometric to the abstract because it's hard to define a fourth dimension, unless we're talking about time or something. In any case, I guess if you MUST ... but personally, I think FOIL is a gateway drug whereas if we focus on the idea of distributive property, there's a better fluidity so people don't *have* to rely on first-outside-inside-last.

I feel similarly about SOAP a) because I've never heard of it and b) because I'm not sure if that's going to help students progress upwards ...

Thank you, Bon, and you raise legit questions here. Here's the thing, Bon. At some point, every "model" breaks down for special cases. That's why I'm playing with it right in front of you. At some point, you *must* transition into the abstract to deal with those pesky non-nice trinomials or any other polynomial. Really, it's a way to push our thinking just a bit to examine why we do what we do, or even what made us bring up polynomials at all.

Amen, amen, amen. I stopped using FOIL my 6th-7th year.

A. Men. I was taught not to use FOIL and instead use the area/volume model to teach multiplication and factoring, and I never looked back. I have worked with lower achieving students and I say that this method not only teaches the concept, but allows them to think visually (and kinesthetically, if you use algebra tiles), where their number sense trips them up.

In my mind, you want models to be too cumbersome to serve at some point, to provide impetus for abstraction. Also, doesn't understanding of what is and what is not possible open up whole new worlds of ideas? Teaching with models does not forbid abstraction, or even impede it. Rather, it deepens connections so students know, understand, and remember concepts without (sometimes silly) acronyms.

I think this a really good way to explain factoring. Geometrical representation helps the student understand the concept much better.

Have you heard of Algebra Tiles? They are just physical manipulatives for teaching with a geometric model. I've used them in my 9th grade algebra class to teach multiplying all types of polynomials, factoring binomials, and completing the square. It's helpful for kids to have something to physically mess around with, and they think they are toys so it keeps their attention. ;)

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