Why in the FOIL Method the terms are taken with their signs?

Question

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 $(x+2)(x-3)$?

Answer 1

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.)

Answer 2

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.)

Answer 3

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?