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# The Power of Parentheses!

I find oftentimes that one of the biggest stumbling blocks for algebra students is that beginners have difficulty seeing the "chunks" in an expression. Instead, they see a big jumbled mess of symbols.

An analogy is an orchestra. A person who has never played a musical instrument, or doesn't have much experience with listening to music, hears the orchestra as one big sound. The trumpets, flutes, strings, percussion all happening at once.

An experienced musician can isolate each instrument, and let the rest of the orchestra fade, focusing on the single melody or harmony line.

Likewise, an experienced mathematician can isolate the sections of an expression, focusing on the single term or operation that needs to be dealt with at the moment, allowing the rest of the expression to fade away for the time being, until the term or operation has been dealt with.

Consider the following problem; can you see the four operations required to solve the problem?

Find the zeros of the function
7
f(x) = 1 + -----------
x2 - 8

If you haven't seen the concept of "zeros" before, don't worry. just set the whole expression equal to 0 and solve for x. Or put another way, think about what value(s) x would have that makes the whole expression be zero.

This post is not about zeros, but about parentheses, your powerful tool for breaking down a big problem into manageable chunks.

Let's add the first set of parentheses around the biggest chunk, to begin, which would be the fraction:

7      ⌉
0 = 1 + | -----------|
x2 - 8   ⌋

Adding the parentheses here make it clear that this function is about adding one to the content of the parentheses. The inverse operation for addition is subtraction. Now we can subtract 1 from both sides to isolate the parentheses:

⌈      7       ⌉
0 - 1 = 1 + | -----------| - 1
⌊   x2 - 8   ⌋

And we don't need the big parentheses anymore.

7
-1 = -----------
x2 - 8

Time for the next set of parentheses, this time to make clear that the fraction has a numerator (7) and a denominator (x2 - 8), that we can treat as chunks.

7
-1 = -----------
(x2 - 8)

The 7 is divided by the chunk, and the inverse operation for division is multiplication. Multiply both sides by (x2-8) and divide both sides by -1 to get the chunk isolated:

(x2 - 8)                  7              (x2 - 8)
----------- * -1 = -----------  *  ------------
-1                   (x2 - 8)              -1

Simplified, removing the parentheses, now that they've done their job:

x- 8 = -7

The next chunk, is the x-squared term:

(x2) - 8 = -7

This lets us see that the 8 is subtracted from the term were interested in, and the inverse operation for subtraction is addition, so move it to the other side:

(x2) - 8 + 8 = -7 + 8

Simplifying and removing parentheses:

x2 = 1

The next set of parentheses is for the x itself, and shows us that the last operation to work with is the square:

(x)2 = 1

Taking the inverse on both sides, that is the square root, isolates the parentheses:

√((x)2) = √(1)

Simplifying:

x = √(1)

Now that we have the x all by itself on one side of the equation, there are no more parentheses required, and we perform the last arithmetic operation to get our answer:

x = ±1

By using the parentheses to chunk the expression and isolate the operation in question at the moment, you can work with one at a time, letting the contents inside the parentheses fade for a short time, like tuning out all of the orchestra except the flutes to clearly hear their melody.

Work from "outside" to "inside", getting closer and closer to x each time, and you'll overcome any obstacle, one operation at a time!

I love this! Definitely helped me figure out new ways as a tutor to explain more complicated expressions and equations to my Algebra students.

Thank you!

\$40p/h

Nathan C.

Math and Science (K through College)

50+ hours

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