# Parentheses and the Associative and Distributive Properties

#### (A new question of the week)

This week, we’ll look at two recent questions about how parentheses (brackets) are used, how they relate to the properties we use in algebra that let us add or drop them, and the related concept of factoring a polynomial. They are examples of how student questions can touch on details teachers tend not to mention because they don’t think like a student. When we find out how a student thinks, we can correct misunderstandings that would never occur to us on our own!

## What do brackets do?

The first question is from Teegan in early August:

Hi there,

I just had a general question about brackets in algebra for Gr. 10 Math, specifically polynomials.

My question can be phrased down to: Is x(x+6) the same as (x)(x+6)?

Thank you so so much!

Teegan

It’s a very simple question, but likely hides some further uncertainty. I answered:

Hi, Teegan.

Is x(x+6) the same as (x)(x+6)?

The quick answer is, yes. Both mean the quantity x, multiplied by the quantity x+6.

Parentheses (brackets) simply mean that what they contain should be treated as a single quantity. So when they surround a single number or variable, they have no real effect. And when two quantities are written next to one another (juxtaposed), that implies that they are to be multiplied, regardless of whether there are parentheses around either.

If there is some particular reason you asked, feel free to show us that context, and we may have more to say.

So there are really two concepts here: the parentheses that are there, and the multiplication symbol that is not! Students often get confused here, thinking the parentheses somehow mean multiplication, or that their absence changes the meaning.

### Factoring, FOIL, and dropping parentheses

Thank you so much, this is really helpful.

The reason I asked was sort of for a more general question about the unit I am currently doing rather than than a particular question, but to give some context; the unit is on factoring trinomials. I sometimes get confused about whether to apply what comes before the Parentheses to everything in the Parentheses (when multiplying polynomials) or to FOIL the expression. (Or if that is the same?)

Originally I applied what was before the Parentheses to the Parentheses when there were no Parentheses on the first expression.

e.g. 2xy(4x^2+6y)

And then used FOIL when there were two sets of Parentheses.

e.g. (2xy+7)(4x^2+6).

But perhaps this changes when there is more than one term in front of the Parentheses…

I became a bit more confused when I saw a step in part of a solution in my practice questions that went from ((2x-1) – 2) to (2x-3). That original question was:

(2x-1)^2 – 5(2x-1) + 6

Let’s say u = 2x-1

u^2 – 5u + 6

[(2x-1) – 2][(2x-1) – 3]

(2x-3)(2x-4)

I guess I am just a little confused about where Parentheses are multiplied together and where they can be dropped (like the above last step) and when there are no Parentheses – what to multiply – or if it is the same.

Thanks so much,

Teegan

“FOIL” stands for “First, Outside, Inside, Last”, and is a mnemonic for how we distribute a product of two binomials. Teegan is exactly right in the supposition that the real difference in distributing is whether there are two terms (a sum) inside the parentheses, or a product. I find many students learn to distribute so well that they do it when it is inappropriate; in fact, a student I was tutoring just this week did it.

### What the properties do

I replied, starting with the last question, about dropping parentheses:

Thanks. Often, general questions are best asked in terms of examples, so this helps me see how you’re thinking.

Basically, in order to drop parentheses, you need to be applying one of the properties of operations, typically the associative properties,

(a+b)+c = a+(b+c) = a+b+c

(ab)c = a(bc) = abc

or the distributive property

a(b+c) = ab + ac

(a+b)c = ac + bc

I’ve written each of these a little differently than we usually teach them.

The difference in my presentation of the associative properties is the last part, where I dropped parentheses; and in the distributive property, it is that I showed it on both the left and the right.

(I used the plural for the associative properties, because there is one for addition and another for multiplication. I used a singular for the distributive property because it is one property, which applies to a combination of addition and multiplication; it is the same property when applied to either side, because of the commutative property.)

The associative properties themselves just say you can move the parentheses; but the order of operations tells us that, say, a+b+c means (a+b)+c because we do the left operation first, so the associative properties ultimately mean you can drop parentheses when the operations are the same (all multiplications, or all additions).

This is what applies to your example of (2x – 1) – 2: subtraction is really addition of the negative, so this means (2x + -1) + -2, and we can apply the associative property, moving the parentheses to make 2x + (-1 + -2), and then doing the addition: 2x + -3 = 2x – 3. With experience, you see subtraction as just a modified addition, and don’t need to write all this out.

So it is the order of operations that really permits dropping parentheses in $$(a+b)+c$$, because they say to do exactly what the order of operations already says to do for $$a+b+c$$.

I wrote the distributive property with the multiplication on either side, which is appropriate because of the commutative property, which says the multiplication and addition behave the same way in either order. The “FOIL” method is really just an extension of this idea, applying the basic property repeatedly:

(a+b)(c+d) = a(c+d) + b(c+d) = ac + ad + bc + bd

This is best thought of simply as “each times each“: to multiply two sums, we multiply each term of the first by each term of the second. The key idea is that this applies to sums of terms; and because of the order of operations, we need parentheses to write such an expression, grouping a sum together so we can then multiply it by something.

Some of the issues I’ve been discussing here are covered in

Order of Operations: Subtle Distinctions

### Distributing all at once or bit by bit

Now I dealt with the first question, about parentheses around a product. I found this a very interesting question, because it feels so automatic to me to think of the “$$2xy$$” as a unit here that I wouldn’t even consider breaking it up; but in fact that would not be wrong:

Now, if the first factor is a product, like your 2xy(4x^2 + 6y), then you could, if you wanted, treat 2, x, and y separately, distributing them one at a time (y, then x, then 2); but that would be a waste of effort. What we would all do is to insert parentheses (at least mentally) seeing it as (2xy)(4x^2 + 6y), which we can do because of the associative property (or just order of operations, really); then we just have one quantity to distribute:

(2xy)(4x^2 + 6y) = (2xy)(4x^2) + (2xy)(6y) = 2xy4x^2 + 2xy6y = 8x^3y + 12xy^2

Here, once I had only multiplications in each term, I dropped the parentheses (by the associative property) and combined factors.

There’s a lot going on here, and it’s very easy for a teacher (who knows all this too well) to go faster than you are ready for, not explaining why each step is “legal”. That’s why asking questions is good.

Breaking up the product would mean taking multiple steps:

$$2xy(4x^2+6y)=2x(4x^2y+6y^2)\\=2(4x^3y+6xy^2)=8x^3y+12xy^2$$

It is the associative property that allows this … if you had a reason to do it.

You say,

I sometimes get confused about whether to apply what comes before the Parentheses to everything in the Parentheses (when multiplying polynomials) or to FOIL the expression. (Or if that is the same?)

Yes, these are really the same: the distributive property.

The important difference in your first pair of examples is whether something in parentheses is a product like 2xy (in which case you treat it all as a single term, which it is), or a sum like (2xy+7) (in which case you multiply each term by each term in the other).

Let’s do that second one, carrying out the “FOIL” by multiplying each term by each term:

$$(2xy+7)(4x^2+6y)=(2xy)(4x^2)+(2xy)(6y)+(7)(4x^2)+(7)(6y)=8x^3y+12xy^2+28x^2+42y$$

You might think of parentheses as a tool you can use at will. Just as you might tie two objects together to make them easier to carry, you can put (or imagine) parentheses around a product when it is appropriate to treat them as a single unit (as directed by the properties and the order of operations); and you can untie them when you need to handle them separately. So parentheses are a piece of rope. Just be careful not to try to put the rope where it doesn’t fit!

The important thing is to preserve meaning. You can insert or remove parentheses when the operations inside and out are the same (associative); but when they are different, you would need to distribute.

Teegan closed:

Thank you so much – this was really helpful.

## Factoring: undistributing

Three days later, we got another question on a related topic:

Hi there!

I was factoring the question:

3x(14-4y) + 15x(7-2y)

and the answer I keep getting is

27x(7-2y)

21x(7-2y)

This is what I did:

3x(14-4y) + 15x(7-2y)

First, I factored out the GCF, 3x(7-2y) and got:

3x(7-2y)(2+2) + 5

3x(7-2y)(9)

27x(7-2y)

In the solution of the correct answer, before factoring out the GCF – they simplified the polynomial, by dividing (14-4y) by 2. In order to make the quantities in the brackets match.

I see how they got 21x(7-2y) but I am just confused about why you cannot take the GCF out always – and what the difference is between simplifying first (which is technically still factoring out a factor – whether or not the greatest one)?

Thank you so much!

Shannon

This is the sort of question we love to get, clearly showing her thinking, so we don’t have to guess what went wrong! The GCF is right, but something went wrong in factoring it out.

Hi, Shannon.

You just need to be a little more careful in your work; taking more and smaller steps will help.

Here is your work, taken more slowly:

3x(14-4y) + 15x(7-2y)

Factor out the GCF, 3x(7-2y), from each term:

3x(7-2y)*2 + 3x(7-2y)*5

Factor out the GCF from the entire expression:

3x(7-2y)(2+5)

3x (7-2y)(7)

21x(7-2y)

You somehow doubled the 2. Perhaps you can explain your thinking in writing 2+2.

You also omitted parentheses in your work, at

3x(7-2y)(2+2)+5

If you expand this, you won’t be multiplying the 5 by anything, which surely is not what you intended. It appears, from the next line, that you meant

3x(7-2y)(2+2+5)

What I showed here is a good strategy for finding errors: do it again, but more slowly! That’s the same thing I do to find something I’ve lost …

### Factoring, canceling, and dividing

Shannon replied,

Thank you so much for the quick reply – this is really helpful!

I think I wrote 2+2 as a result of dividing 14-4y by 7-2y but it sounds like I am dividing this incorrectly – as if I just had 2 then I would get the correct answer.

I have been dividing it by going 14/7 -4/-2 y/y

And then getting 2 + 2

Thanks again,

It appears, as I look at it again now, that her division is like this: $$\frac{14-4y}{7-2y}=\frac{14}{7}+\frac{-4}{-2}\frac{y}{y}=2+2\cdot 1$$ I didn’t quite see this at the time; this represents a common error, in which students “cancel” a fraction by crossing out anything in the top and bottom that matches, forgetting that you can only cancel a factor of the entire numerator and denominator, not a factor of a term. Similarly, you can divide by a monomial by dividing each term in the numerator by the single term in the denominator, as in $$\frac{14-4y}{2}=\frac{14}{2}+\frac{-4y}{2}=7-2y,$$ but what Shannon did is wrong.

I responded with two better ways to have done this:

Your dividing is sort of like dividing 144 by 72 and getting 22.

It will definitely be better to think of this as a factoring step rather than a division; or at least not doing the division in your head.

As factoring, it’s 14 – 4y = ?(7 – 2y), and seeing that ? = 2 works.

As (long) division, it’s

                 2
----------
-2y + 7 ) -4y + 14
-4y + 14
--------
0

Or, to put it differently, when you divide, you should always check by multiplying. If you think that (14 – 4y)/(7 – 2y) = 2+2 = 4, you would check whether 4(7 – 2y) = 14 – 4y.

So I think of factoring as simply looking for a common factor and pulling it out (once!) from every term. Another way to express that would be to first break each term into a product, and then “undistribute” that common factor:

$$14-4y=2\cdot7-2\cdot2y=2(7-2y)$$

The long division is a considerably longer process; at the time I imagined Shannon doing this (rather than the incorrect canceling) but writing the 2 above each column.

In the check, of course, we would find that $$4(7-2y)=28-8y$$, which is not what we want, and the fix it.

Shannon closed:

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