Remember solving polynomials?
(a + b)^2 = a^2 + 2ab + b^2
I have no problem working these problems out, but just what does it accomplish?
Is the equation on the right easier to understand than the equation on the left? I don’t think so.
Perhaps we need to do this to solve equations in the higher mathematical fields like integral calculus or non-linear regression, which I never took in HS or college. Can anyone enlighten me?
The skills of algebraic distribution and factoring are used all the time in the sciences; the undergraduate physics majors I teach are completely comfortable in doing this type of thing to solve physical problems, and usually have to apply this skill many times in their homework each week.
It’s important to know you can switch between the two forms, it’s not that one form is inherently “better” than the other. Often, you start with the form on the right, and switch to the form on the left. It’s a tool you can use to manipulate complicated equations.
f.coli writes:
> Is the equation on the right easier to understand than the equation on the left?
You need to get the terminology right. It’s one equation, not two of them. You can refer to the right side of the equation or the polynomial on the right. You can similarly refer to the left side of the equation or the polynomial on the right. You can’t refer to the equation on the right or the equation on the left.
I had never heard this referred to as the FOIL method before, but apparently some people do use that term. It’s not particularly useful to memorize first-outside-inside-last as a mnemonic. That’s only applicable to multiplying a polynomial with two terms by another one with two terms. There’s no similar mnemonic for multiplying a polynomial with three terms by another one with three terms, for instance. What you need to learn is the distributive law, not the term “FOIL”. The distribution law is important. The clever acronym is trivial:
You need it to factor a polynomial.
Also you’ll need it later on to pack the thing back up so it looks like the left side.
The truth?
It’s not easy teaching math. Math up until college is taught as a set of specific things, basically “how to” steps to solving fairly ridiculous artificial problems that often have little resemblance to real life.
Why?
People put up a mental block when it comes to math. It’s pretty hard to explain to kids that the skills they are learning aren’t “here’s how to distribute” but “I’m trying to show you basic problem solving skills in a limited, stilted context so you can use those lessons in logic, rational thinking, and problem solving in a larger context later”. Most kids (actually, most people) don’t get this. Hence, the more modern attempts to relate math problems to the real world.
And algebra is really where you start doing that problem solving. Arithmetic provides the basic tools. Algebra is where you start explaining “how” to use those tools while adding more tools.
And once you know how to use the basic tools, you can, if you choose to go that way in your studies, move on to more advanced tools and methods - like going from a hammer to a crane operator or from knowing how a magnifying glass works to working with electron microscopes.
ETA: And that answers that question: when will I ever use this in real life? The same answer as with a hammer or a phone or a computer. Math as taught in school is another life tool, one that not everybody may use daily but which is as ubiquitous as that hammer or phone or computer and one for which you could hire somebody to use for you but certainly useful to know how to use yourself.
I don’t think the op is asking what’s the point of math, but why we focus on this particular application as it seems to make things more complicated not less, thus being counterproductive.
The short, general answer is that the form on the right is what we sometimes need to work with when we need to combine the expression with other stuff (so we can add like terms, for example).
It comes up quite often, even in relatively basic college-level math, so that a student is at a disadvantage if they can’t look at a binomial squared (like the expression on the left) and remember that it can be expanded into the form on the right; and they’re at a major disadvantage if they don’t know how to somehow figure this out, or they assume that they can just distribute the exponent so that (a + b)[sup]2[/sup] = a[sup]2[/sup] + b[sup]2[/sup]. :smack:
Note: this is a special case of the Binomial Formula (see Wikipedia or the more elementary explanation here).
I agree, and I too have tried to speak out against the misuse of the word “equation.” Using “equation” when you mean “expression” is a little like using “sentence” when you mean “word” or “phrase.”
I disagree with this. It’s true that it’s the distributive law and being able to multiply two polynomials together that’s fundamental, and that binomial times binomial is merely a special case of this; but it’s a special case that’s so common, and that needs to be mastered so thoroughly, that it really does help to have a mnemonic to remind one how that special case works.
I’ll also point out that the OP is not an example of FOIL per se. FOIL may be used to figure out that the left side is equivalent to the right, but the OP hasn’t explicitly done so.
One of the reasons you need to be able to do this is to show that (a - b)[sup]2[/sup] + 2ab = a[sup]2[/sup] + b[sup]2[/sup]. And when you can do this you can get to a nice simple diagrammatic proof of Pythagoras’ Theorem: Put four identical right-angled triangles together so that their hypotenuses are on the outside forming the sides of a square and the left-over space where the triangles meet forms another small square inside it. Label the short sides a and b (b is the shortest) and the hypotenuse c and after a little algebra you can see that in this case a[sup]2[/sup] + b[sup]2[/sup] = c[sup]2[/sup].
In essence what you are doing is converting an expression between two alternative forms. That is, between two alternative calculation methods. And you would choose the form that is most convenient for future processing.
For example in the following:
(x+2)(x+5) = x[sup]2[/sup]+7x+14
You are converting a multiplication problem to an addition problem.
As a practical application - what is 65x74?
(60+5)(70+4) =
F — 60x70 —plus
O— 60x4 —plus
I— 5x70 —plus
L— 5x4
If you have a moderately good head for mathematics and decent memory and your times tables, you can do that in your head -
4200+240+350+20=4810 (If my mental arithmetic brain cells work)
It’s just a way of ensuring you know how to simplify or expand a math expression.
FOIL has served me well for literally 35 years. Learned it in 9th grade algebra when I was 14. Its one of those things I rarely use anymore but can remember as needed.
I’m a math person, but… why would I want to be able to make a “nice simple diagrammatic proof of Pythagoras’ Theorem”?
If you were really a math person, you wouldn’t ask that. 
A simple diagrammatic proof of Pythagoras’s Theorem is its own reward.
To enjoy its ineffable quality of ‘elegance.’
there are computers to fix such kind of problems
What’s (a + b)^2 + (2a +b)(a+3b)?
It’s pretty much impossible to figure out in that form, but if you distribute and add up like terms, it’s trivial:
a^2 + 2ab + b^2
+ 2a^2 + 7ab + 3b^2
__________________
3a^2 + 9ab + 4b^2
Then you could factor back into the original form if you wanted, or wait until you have a good reason to.
Suppose the equation was
(a+4)^2 -(a+2)^2 = 84
In this form it isn’t easy to see what value “a” should have but if we multiply it out with FOIL:
a^2 +8a +16 - (a^2+4a+4) = 84
a^2 +8a +16 -a^2 -4a - 4 = 84
4a +12=84
4a= 84 -12
4a = 72
a = 18
To the OP, it’s useful to recognize that the right side of your equation is a perfect square which can therefore be expressed as it is on the left side of the equation. Once you’re comfortable with that, you can take expressions that are not perfect squares and manipulate them so that they are a perfect square plus whatever term you need to add to balance out your manipulation. This process is called ‘‘Completing the Square’’. An example:
x[sup]2[/sup] + 6x + 13
is not a perfect square, but it would be if that 13 was a 9 instead.
So move the 13 off to the side for a sec, put a 9 in there to make your perfect square and just remember to subtract the 9 back out to keep everything equal like this:
(x[sup]2[/sup] + 6x + 9) + 13 - 9
The stuff in parantheses is now a perfect square which can be expressed like the left side of your equation with the additional term needed to make it balance like this:
(x+3)[sup]2[/sup] + 13 - 9
Combine the +13 and -9 to get:
(x+3)[sup]2[/sup] + 4
Now that you know how to complete the square, you can apply that process to the general form of a quadratic equation:
ax[sup]2[/sup] + bx + c = 0
which, if you do it correctly and solve for x, you get the familiar quadratic formula:
x = [-b±(b[sup]2[/sup] - 4ac)[sup]1/2[/sup]]/(2a)
which is a reasonably useful extension of what you asked about. Of course there are other cases in which such algebraic manipulations are useful too. But the bottom line to your question is that knowing what you posted in the OP helps to solve problems as well as being a building block to learn even more math.