Lesson: Factoring – Greatest Common Factor

Comment on Factoring – Greatest Common Factor

On the first reinforcement

On the first reinforcement activity I am confused on when to add or subtract vs when to divide/multiply. When I got p^4 in column A and p^3 in column B I subtracted p^3 in both sides. Why did we divide here? In certain problems it seems like we subtract and others we divide and I’m not clear as to when to use which. Is there a rule or something?

Jason

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.

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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

Yes sir it helps. The answer says Quantity B is greater, your response says Quantity A is greater...It is B right?

My bad. I've edited my response.

Cheers,
Brent

Hi 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.

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!

Great explanation!
Thank you Brent.

In the second 160-170 GRE

In the second 160-170 GRE Prep Club quesiton from the list, I do not get how

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