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shahfahad: Senior | Next Rank: 100 Posts; Posts: 42; Joined: Fri Oct 30, 2015 9:01 am

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by shahfahad » Tue Nov 10, 2015 12:09 pm

Please help me with this question. I am a bit confused with the factoring process in this video.

https://www.gmatprepnow.com/module/gmat- ... video/1054

For factoring, you need to factor the least common numbers/figures/variables from 2 or more equations. However, from equation on the left, 4x-2 is factored whereas 4x should have been factored. Like in equation on the right side, 4(2x) is factored from 42x + 1 - 42x. I understood this part because 42x is least common between both equations. I did not understand the first part.

Can you explain the factoring outcomes if we take 4x as a common factor rather than 4x-2.

Finally, can i get a link to a good guide on factoring? A website, document or a detailed video lecture.

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Source: Beat The GMAT — Problem Solving | Tue Nov 10, 2015 12:09 pm

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pm; Location: New York, NY; Thanked: 13060 times; Followed by:1906 members; GMAT Score:790

by GMATGuruNY » Wed Nov 11, 2015 3:42 am

If [4^x - 4^(x-2)]/5 = 4^(2x+1) - 4^(2x), then x = ?

-3.5
-3
-2.5
-2
-1.5

We can PLUG IN THE ANSWERS, which represent the value of x.
Exponents on the GMAT tend to be INTEGER VALUES.
Thus, the correct answer choice is likely B or D.

Strategy:
Factor out the term with the SMALLER EXPONENT.

D: -2
If we plug x=-2 into [4^x - 4^(x-2)]/5 = 4^(2x+1) - 4^(2x), we get:
(4Â¯Â² - 4Â¯â�´)/5 = 4Â¯Â³ - 4Â¯â�´.

On each side, factor out 4Â¯â�´:
4Â¯â�´(4Â² - 1)/5 = 4Â¯â�´(4 - 1)

Cancel out the terms in red and simplify the terms in blue:
15/5 = 3

3 = 3.

Success!

The correct answer is D.

Algebraic solution:

[4^x - 4^(x-2)]/5 = 4^(2x+1) - 4^(2x)

On the left side, factor out 4^(x-2).
On the right side, factor out 4^(2x):
4^(x-2)(4Â² - 1)/5 = 4^(2x)(4 - 1)

Simplify the terms in blue:
4^(x-2)(3) = 4^(2x)(3)

Cancel out the terms in blue:
4^(x-2) = 4^(2x)

Since the bases match, so must the exponents.
Thus:
x-2 = 2x
-2 = x.

The correct answer is D.

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shahfahad: Senior | Next Rank: 100 Posts; Posts: 42; Joined: Fri Oct 30, 2015 9:01 am

by shahfahad » Wed Nov 11, 2015 4:00 am

Ok. I think i get it.

In the algebraic approach, when you factored 4^(x-2) from the equation of the left and the answer was 4^2. The approach was to:
1) Take the smaller exponent which is: 4^(x-2)
2) When you factor 4^(x-2) from 4^x then you need to find a value that converts 4^(x-2) to 4^x which is 4^2. Because 4^x-2(+2) makes 4^x. By adding 4^+2 we converted 4^x-2 to 4x. Therefore, 4^2 remains after factoring 4^x-2 from 4^x. Is this correct?[/b]

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pm; Location: New York, NY; Thanked: 13060 times; Followed by:1906 members; GMAT Score:790

by GMATGuruNY » Wed Nov 11, 2015 4:03 am

shahfahad wrote:Ok. I think i get it.

In the algebraic approach, when you factored 4^(x-2) from the equation of the left and the answer was 4^2. The approach was to:
1) Take the smaller exponent which is: 4^(x-2)
2) When you factor 4^(x-2) from 4^x then you need to find a value that converts 4^(x-2) to 4^x which is 4^2. Because 4^x-2(+2) makes 4^x. By adding 4^+2 we converted 4^x-2 to 4x. Therefore, 4^2 remains after factoring 4^x-2 from 4^x. Is this correct?[/b]

Correct!

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shahfahad: Senior | Next Rank: 100 Posts; Posts: 42; Joined: Fri Oct 30, 2015 9:01 am

by shahfahad » Wed Nov 11, 2015 4:11 am

But in this question:

https://www.beatthegmat.com/advanced-qua ... 79853.html

Brent factored (5^y) from 5^x - 5^y and the result was (5^y)[5^(x-y) - 1]

I am having trouble understanding this part. How can you factor 5^y from 5^x. Then dont have anything in common. And even if you do, the result should have been (according to me which i know is incorrect obviously because Brent cant be incorrect) (5^y)(5^x*y-1) rather than (5^y)[5^(x-y) - 1]. Why -y? Please help.

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shahfahad: Senior | Next Rank: 100 Posts; Posts: 42; Joined: Fri Oct 30, 2015 9:01 am

by shahfahad » Wed Nov 11, 2015 4:18 am

Also, can i get a link to any article/website/video lecture or anything that explains in detail the factoring of numbers and especially variables? Any good resource? Because i am struggling with the factoring of variables.

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GMAT/MBA Expert

Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pm; Location: Vancouver, BC; Thanked: 5254 times; Followed by:1268 members; GMAT Score:770

by Brent@GMATPrepNow » Wed Nov 11, 2015 6:55 am

shahfahad wrote:Also, can i get a link to any article/website/video lecture or anything that explains in detail the factoring of numbers and especially variables? Any good resource? Because i am struggling with the factoring of variables.

We have 4 free videos on factoring:
- https://www.gmatprepnow.com/module/gmat- ... /video/954
- https://www.gmatprepnow.com/module/gmat- ... /video/955
- https://www.gmatprepnow.com/module/gmat- ... /video/956
- https://www.gmatprepnow.com/module/gmat- ... /video/957

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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shahfahad: Senior | Next Rank: 100 Posts; Posts: 42; Joined: Fri Oct 30, 2015 9:01 am

by shahfahad » Wed Nov 11, 2015 7:18 am

Hi Brent. Can you please explain the factoring of variables. I did not understand how you factored the above equation.

Quote

GMAT/MBA Expert

Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pm; Location: Vancouver, BC; Thanked: 5254 times; Followed by:1268 members; GMAT Score:770

by Brent@GMATPrepNow » Wed Nov 11, 2015 7:27 am

ASIDE: A lot of students struggle to see how we can factor the above.
Factoring, they may be okay with straightforward factoring like these examples:
k^5 - k^3 = k^3(k^2 - 1)
m^19 - m^15 = m^15(m^4 - 1)
But they have problems when the exponents are variables.

IMPORTANT: Notice that, each time, the greatest common factor of both terms is the term with the smaller exponent.
So, in the expression x^10 - x^7, we have two exponents, 10 and 7, and the smaller exponent is 7, so we can factor out x^7 to get x^7(x^3 - 1)

Likewise, in the expression 5^x - 5^(x-3), we have two exponents, x and (x-3), and the smaller exponent is (x-3), so we can factor out 5^(x-3)

Likewise, w^x + x^(x+5) = w^x(1 + w^5)
And 2^x - 2^(x-2) = 2^(x-2)[2^2 - 1]

Now let's get to your question:
In the expression 4^(2x+1) - 4^(2x), we have two exponents, (2x+1) and (2x), and the smaller exponent is 2x, so we can factor out 4^(2x) to get 4^(2x)[4^1 + 1]

Does that help?

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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shahfahad: Senior | Next Rank: 100 Posts; Posts: 42; Joined: Fri Oct 30, 2015 9:01 am

by shahfahad » Wed Nov 11, 2015 7:31 am

When you factored (5^y) from 5^x - 5^y and the result was (5^y)[5^(x-y) - 1]

I am having trouble understanding this part. How can you factor 5^y from 5^x. Then dont have anything in common. And even if you do, the result should have been (5^y)(5^x*y-1) rather than (5^y)[5^(x-y) - 1]. Why -y? Please help.

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DavidG@VeritasPrep: Legendary Member; Posts: 2663; Joined: Wed Jan 14, 2015 8:25 am; Location: Boston, MA; Thanked: 1153 times; Followed by:128 members; GMAT Score:770

by DavidG@VeritasPrep » Wed Nov 11, 2015 8:49 am

shahfahad wrote:When you factored (5^y) from 5^x - 5^y and the result was (5^y)[5^(x-y) - 1]

I am having trouble understanding this part. How can you factor 5^y from 5^x. Then dont have anything in common. And even if you do, the result should have been (5^y)(5^x*y-1) rather than (5^y)[5^(x-y) - 1]. Why -y? Please help.

Maybe think of it this way: when you're factoring something out of an expression, you're dividing each term by the thing you've factored out. So 2x - 4 becomes 2(x-2), because both the '2x' term and the '4' term have been divided by 2. Similarly, 4^5 - 4^3, becomes 4^3(4^2 - 1) because you've divided each term by the 4^3 that you've factored out.

But using this logic, we can factor out anything we want. If you wanted to, you could factor a '3' out of 2x-4 and call the expression 3 * [(2x/3) - 4/3).] Even though '3' isn't necessarily a factor of either term, we're effectively multiplying by 3/3, and so we haven't changed the expression. (And you can see that if we distribute the '3', then 3 * [(2x/3) - 4/3), will become 2x - 4. The expressions are the same.

Similarly, if we started with 5^x - 1 we can factor out whatever we want. You could factor out a 2, and call the expression 2[(5^x)/2 - (1/2). There's no real reason to do this, but it's mathematically legal. And we could start with 5^x - 1 and then factor out a 5^y, to get 5^y*[(5^x)/(5^y) - 1/(5^y).] We just divided each term by 5^y. (Note that (5^x)/(5^y) = 5^(x-y). I'm using (5^x)/(5^y) for simplicity's sake.) Note, also, that if we distribute the 5^y, then 5^y*[(5^x)/5^y) - 1/(5^y), will become 5^x - 1. So when we factor something out, we're really just dividing each term by the thing we've factored.

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