3
$\begingroup$

That was the most boring title I could choose but in all honesty, it is what the question is. Here is a reminder of the FOIL method that is used for multiplying two binomials. For example, to multiply $$(x+2)(x-3),$$ the "FOIL" algorithm has students sum the following four products:

  1. First terms: $x.x$,
  2. Outer terms: $x.(-3)$,
  3. Inner terms: $2.x$,
  4. Last terms: $2.(-3)$.

Personally, I never used it as a teaching tool. But, I was writing something on the use and misuse of signs, that I came to the FOIL method. The "common" method of doing it took me by surprise, as I personally never take the terms with their signs. For me, it is going like this: $x.x$, $x.3$ (then looking at the sign of the second binomial) I write $x^2-3x$, and so on. Basically, I multiply the terms and then decide about the signs.

The question is whether there is any pedagogical advantage or disadvantage to either of these approaches. Obviously, I am biased towards one of them :) Having confessed that, let me put you in the shoes of a student who knows how to do something like $$(100+2)(100-3)$$ but still doesn't know negative numbers. Which approach would be more "natural" for the student when doing $(x+2)(x-3)$?

Improve this question
$\endgroup$
14
  • 4
    $\begingroup$ You can "define" the FOIL method as $(a+b)(c+d)=ac+ad+bc+bd$. Thus, $(x+2)(x-3)=(x+2)(x+-3)=x(x)+x(-3)+2x+2(-3)$. $\endgroup$ – Joel Reyes Noche May 1 '19 at 8:04
  • $\begingroup$ @JoelReyesNoche Am I right to say that this has a bit of both approach, providing that we have negative numbers? $\endgroup$ – Amir Asghari May 1 '19 at 8:44
  • $\begingroup$ Actually, I can't fully see the difference between the "common" method and "your" approach. Are you saying that your approach doesn't require knowledge of negative numbers? If so, then how would you explain $(x-2)(x-3)$ with your approach? If a student knows that "negative times negative is positive," then doesn't that imply that the student knows about negative numbers? $\endgroup$ – Joel Reyes Noche May 1 '19 at 9:15
  • 2
    $\begingroup$ @JoeTaxpayer, please type your answer even if it duplicates my comment. $\endgroup$ – Joel Reyes Noche May 1 '19 at 11:24
  • 2
    $\begingroup$ Just as a comment related to the last bit of the question, in the United States most children are exposed to negative numbers repeatedly long before 'FOIL' and its ilk would be broached. I don't know if that obtains universally, but I suspect so. $\endgroup$ – kcrisman May 1 '19 at 11:42
7
$\begingroup$

I'm going to answer with something of a polemical frame challenge: FOIL is evil, and probably shouldn't even be taught. Okay... that's a bit extreme. How's this: FOIL is a mnemonic that is, in my opinion, not all that useful, and should not be taught. In my own experience teaching college algebra and precalculus courses, students come to rely on FOIL far to much without understanding what they are really doing. Then they have to multiply trinomials, and it all falls apart.

Instead, I think that students should spend more time actually working with the distributive property of multiplication over addition. FOIL then becomes a special case. That is, \begin{align} (a+b)(c+d) &= (a+b)c + (a+b)d && \text{(distribute $(a+b)$)} \\ &= ac + bc + (a+b)d && \text{(distribute $c$)} \\ &= ac + bc + ad + bd. && \text{(distribute $d$)} \end{align} While I think that your concern about working with negative numbers is overblown (students who are working at this level of symbol manipulation should already be familiar with negative numbers), but even so, it can be handled fairly naturally. In your example \begin{align} (x+2)(x-3) &= (x+2)x - (x+2)(3) && \text{(distribution over subtraction)} \\ &= x^2 + 2x - (3x + 2(3)) \\ &= x^2 + 2x - 3x - 6 && \text{(distribute the negative)} \\ &= x^2 - x - 6. \end{align} Personally, I find this a little awkward, as one has to remember to distribute the negative (I would have preferred to keep the negation and the $3$ together, i.e. write $(x+2)(x-3) = (x+2)x + (x+2)(-3)$, but this is a minor point).

I prefer to put the emphasis on the distributive property of multiplication over addition (and subtraction, which is really just addition in the other direction) because it is far more generalizable. For example, if I want to multiply polynomials of degree two and three, I would write \begin{align} &\color{red}{(ax^2 + bx + c)}(dx^3 + ex^2 + fx + g) \\ &= \color{red}{(ax^2 + bx + c)}dx^3 + \color{red}{(ax^2 + bx + c)}ex^2 + \color{red}{(ax^2 + bx + c)}fx + \color{red}{(ax^2 + bx + c)}g \\ &= (ax^2 + bx + c)\color{blue}{dx^3} + (ax^2 + bx + c)\color{green}{ex^2} + (ax^2 + bx + c)\color{orange}{fx} + (ax^2 + bx + c)\color{purple}{g} \\ &= ax^2\color{blue}{dx^3} + b\color{blue}{dx^3} + c\color{blue}{dx^3} + ax^2\color{green}{ex^2} + bx\color{green}{ex^2} + c\color{green}{ex^2} + ax^2\color{orange}{fx} + bx\color{orange}{fx} + c\color{orange}{fx} + ax^2\color{purple}{g} + bx\color{purple}{g} + c\color{purple}{g} \\ &= adx^5 + bdx^4 + aex^4 + cdx^3 + bex^3 + afx^3 + cex^2 + bfx^2 + agx^2 + cfx + bgx + cg \\ &= adx^5 + (bd + ae)x^4 + (cd + be + af)x^3 + (ce + bf + ag)x^2 + (cf + bg)x + cg. \end{align} This is horribly tedious and rather difficult to convey in text (this is what would end up on the board after five minutes of lecture; the story really needs to unfold in real time). Even so, I typically very quickly get to the idea of distributing terms one at a time (i.e. first distribute the $ax^2$ to each term of the cubic, then distribute $bx$ to each term of the cubic, etc). I also tend to suggest that students line up like terms as they multiply, which makes it easier to add coefficients in the end: \begin{align} &(ax^2 + bx + c)(dx^3 + ex^2 + fx + g) \\ &\qquad \begin{matrix} =& adx^5 &+& aex^4 &+& afx^3 &+& agx^2 &&&&& \text{(distribute $ax^2$)} \\ &&+& bdx^4 &+& bex^3 &+& bfx^2 &+& bgx &&& \text{(distribute $bx$)} \\ &&& &+& cdx^3 &+& cex^2 &+& cfx &+& cg. & \text{(distribute $c$)} \\ \end{matrix} \\ \end{align} The columns may be added to get the same thing as above.

Note that FOIL completely fails here. If we were to only multiply the first, outer, inner, and last terms, we would get \begin{align} (ax^2 + bx + c)(dx^3 + ex^2 + fx + g) &= adx^5 + agx^2 + cdx^3 + cg. && \text{$\leftarrow$ this is wrong!} \end{align} I hear the objection: "But Xander, we don't teach students to FOIL like that!"

You are correct, we don't teach students to do that. And yet, they show up in my class and do it anyway, because they don't understand the deeper concept, which is distribution. The FOIL mnemonic is not all that general, hides the reality of what is going on, and seems to confuse students. Don't teach it.

(Please note: I have intentionally written this to convey a strongly held opinion in an intentionally confrontational manner. However, this really is just my opinion—albeit one backed by my own experience. Do with this opinion as you like.)

Improve this answer
$\endgroup$
5
  • $\begingroup$ An interesting question is then what to do with students who show up beyond college algebra/precalc and remember very little of anything beyond the FOIL mnemonic, given that one probably doesn't have time to "unteach" such things and given that they probably will be intimidated by a jump straight to this from previous FOIL experience. I imagine you have a fair amount of experience trying to untangle that pedagogical knot. $\endgroup$ – kcrisman May 1 '19 at 17:50
  • 2
    $\begingroup$ "the distributive property of multiplication over addition. FOIL then becomes a special case." — which it is. But why distributing over subtraction if you can distribute over addition and treat each term as either negative or positive, with the polynomial being an algebraic sum? "Note that FOIL completely fails here." — FOIL assumes only four terms. Anyone who applies it blindly to cases like you shown should not be admitted to a college at the first place. $\endgroup$ – Rusty Core May 1 '19 at 19:39
  • 1
    $\begingroup$ @kcrisman I typically go into such classes assuming that they have never seen FOIL, and go over distribution as though it were being done for the first time. If a student brings up FOIL, I'll discuss it (and mention why I don't like it), then move on. This generally works pretty well. Every once in a while it does become necessary to "unteach" FOIL, but this part of the class is usually pretty smooth. $\endgroup$ – Xander Henderson May 1 '19 at 22:21
  • $\begingroup$ @RustyCore (1) With regard to distributing over subtraction, you'll note that in the paragraph right after the computation, I note that I prefer to regard subtraction as addition of a negative, and get on with life. So I wouldn't distribute over subtraction. However, the original asker seemed to be curious about this point, so I tried to give them a way in. $\endgroup$ – Xander Henderson May 1 '19 at 22:23
  • 5
    $\begingroup$ (2) With regard to your dismissal of college students with a weak mathematics background, I have to teach the students that I get. Whether or not I think that they should be in college, they are, and I have to deal with that somehow. Moreover, bemoaning the fact that they managed to get into college is both unproductive and has a significant element of victim-blaming. $\endgroup$ – Xander Henderson May 1 '19 at 22:25
4
$\begingroup$

The issue isn't unique to factoring or the FOIL process.

The roots of $(ax^2 + bx + c)$

can be found via the quadratic equation.

$$x = \frac{-b\pm\sqrt{b^2 - 4ac}}{2a}$$

Now, this prompts the issue;

$(2x^2 - 3x - 4)$

What is the value for a,b,c that we will plug into the quadratic equation? Isn't c = -4 ? And we need to accept that the minus operator transmogrifies into a negative sign when the term is isolated. Even though you see 2 'subtractions', how do you respond to the request to identify the 2nd or 3rd term?

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ In fact, I originally asked the question to find out how people respond to this transmogrification (I love the term; thank you for using it). But, I thought it is better to ask it on a concrete situation to make more sense, and hence, FOIL. Thank you for seeing the heart of the problem and adding another example. $\endgroup$ – Amir Asghari May 5 '19 at 16:02
3
$\begingroup$

Simple answer: Including the signs/negative numbers is algorithmic. Working out the sign afterwards requires thinking. Imagine writing computer code to multiply out the brackets. You would do it by keeping the minus signs with the relevant terms.

(Pedagogically: keeping the signs with the magnitude will extend to bigger brackets in a more natural way than working out the sign afterwards.)

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.