Post your question in the Comment section below, and a GRE expert will answer it as fast as humanly possible.

- Video Course
- Video Course Overview
- General GRE Info and Strategies - 7 videos (free)
- Quantitative Comparison - 7 videos (free)
- Arithmetic - 42 videos
- Powers and Roots - 43 videos
- Algebra and Equation Solving - 78 videos
- Word Problems - 54 videos
- Geometry - 48 videos
- Integer Properties - 34 videos
- Statistics - 28 videos
- Counting - 27 videos
- Probability - 25 videos
- Data Interpretation - 24 videos
- Analytical Writing - 9 videos (free)
- Sentence Equivalence - 39 videos (free)
- Text Completion - 51 videos
- Reading Comprehension - 16 videos

- Study Guide
- Philosophy
- Office Hours
- Extras
- Prices

## Comment on

Factoring – Greatest Common Factor## On the first reinforcement

Jason

## Question link: https:/

Question link: https://gre.myprepclub.com/forum/gre-math-challenge-603.html

Great question, Jason!

Sometimes dividing/multiplying both quantities works, and sometimes adding/subtracting works.

What determines the best approach?

The best approach will be the one that best allows us to compare the two quantities. So, if one action makes a quantity MORE complicated, then it's probably not the best approach.

Fortunately, as long as we follow the rules described in the Matching Operations video (https://www.greenlighttestprep.com/module/gre-quantitative-comparison/vi...), we won't get the wrong answer; we just may make things a little more difficult for ourselves.

To see what I mean, let's take things from the point where our solutions diverged...

QUANTITY A: p^4

QUANTITY B: p^3

If we subtract p^3 from both quantities, we get:

QUANTITY A: p^4 - p^3

QUANTITY B: 0

Did that last step make it easier or harder to compare the two quantities? It's hard to say.

Let's keep going.

Let's factor quantity A to get:

QUANTITY A: p^3(p - 1)

QUANTITY B: 0

From here, we can apply some number sense.

If p is POSITIVE, we know that p^3 is POSITIVE

Also, if 0 < p < 1, we know that (p - 1) is NEGATIVE

So, p^3(p - 1) = (POSITIVE)(NEGATIVE) = NEGATIVE

We we get:

QUANTITY A: NEGATIVE

QUANTITY B: 0

This means Quantity B is greater.

--------------------------------------------

Now let's examine the approach where we divide both quantities by some value. We'll start with:

QUANTITY A: p^4

QUANTITY B: p^3

Since we're told that p is POSITIVE, we know that p^3 is POSITIVE, which means we can safely divide both quantities by p^3 to get:

QUANTITY A: p

QUANTITY B: 1

Did that step make it easier or harder to compare the two quantities? I'd have to say easier.

Since 0 < p < 1, we can see that Quantity B is greater.

As you can see, both approaches will work.

So, when you're applying the Matching Operations approach, be sure to keep asking "Am I making it easier or harder to compare the two quantities?"

If you made it harder, you might want to try a different operation.

Does that help?

Cheers,

Brent

## Yes sir it helps. The answer

## My bad. I've edited my

My bad. I've edited my response.

Cheers,

Brent

## Hi Brent,

For this question @https://gre.myprepclub.com/forum/n-is-a-positive-integer-17313.html

when you factored quantity A, you got: (−1)^n [1+(−1)^1]

why the factoring isn't like this: (−1)^n [1+(+1)^1]

cuz if you multiply -ve by -ve that will be +ve and the original value of 1 is -ve.

## Question link: https:/

Question link: https://gre.myprepclub.com/forum/n-is-a-positive-integer-17313.html

Good question!

Let's start by examining a similar situation, with the variable bases:

For example: b^n + b^(n+1)

We can factor this to get: b^n + b^(n+1) = b^n(1 + b^1)

We can verify this by expanding b^n(1 + b^1) to get back to b^n + b^(n+1)

In the above factorization, notice that the base (b) remains the same each time.

Keep in mind that we need to apply the Product Law, which says: (b^x)(b^y) = b^(a+b)

To apply this law, the base (b) must be the same in b^x and b^y

The same applies with our original expression (−1)^n + (−1)^(n+1)

That is, we can't change one of the (−1) to a (1).

Does that help?

## Great explanation!

Thank you Brent.

## In the second 160-170 GRE

2^29 (2^1 - 1) can boil down to just 2^29? Don't exponent rules make it 2^30-2^29?

## Question link: https:/

Question link: https://gre.myprepclub.com/forum/2-9234.html

2^1 = 2

So, 2^29(2^1 - 1) = 2^29(2 - 1) = 2^29(1) = 2^29