Last visit was: 18 Nov 2025, 22:35 It is currently 18 Nov 2025, 22:35
Home
GMAT
Messages
Tests
Schools
Events
Rewards
Chat
My Profile
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
User avatar
MeanRev
User avatar
Joined: 08 Mar 2016
Last visit: 22 Apr 2016
Posts: 2
Posts: 2
Kudos: 0
How do you solve this part of the equation? 08 Mar 2016, 11:11
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hello,

I am studying the GMAT Quant section and having not been in school for a long time my algebra is very rusty. I am struggling to get my head around something quite simple and was wondering if someone could break it down for me. In the official GMAT guide there is the following example for factoring:

(1) x^3 - 2x^2 + x = -5(x - 1)^2

(2) x^3 - 2x^2 + x + 5(x - 1)^2 = 0

(3) x(x^2 - 2x + 1) + 5(x - 1)^2 = 0

(4) x(x - 1)^2 + 5(x - 1)^2 = 0

(5) (x + 5)(x - 1)^2 = 0

(6) x + 5 = 0 OR (x - 1)^2 = 0

(7) therefore, x = -5 OR x = 1

My confusion comes from how they moved from step (3) to (4)? The rest makes sense but I cannot connect this leap.

Thanks for your time.

MR



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
avatar
Jmarcotte
avatar
Joined: 03 Mar 2016
Last visit: 25 Feb 2021
Posts: 1
Own Kudos:
1
 [1]
Given Kudos: 8
Posts: 1
Kudos: 1
 [1]
How do you solve this part of the equation? 08 Mar 2016, 11:34
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Someone can correct me if I am wrong.

It is basically reverse FOIL.
1. (x-1)^2
2. (x-1)(x-1) (FOIL to get the equation below)
3. (x^2 - 2x + 1)

In your example, they started at number 3 and went backwards up to 1.
User avatar
ENGRTOMBA2018
User avatar
Joined: 20 Mar 2014
Last visit: 01 Dec 2021
Posts: 2,325
Own Kudos:
3,837
Given Kudos: 816
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE:Engineering (Aerospace and Defense)
Products:
My Reviews
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
Posts: 2,325
Kudos: 3,837
How do you solve this part of the equation? 08 Mar 2016, 12:26
Kudos
Add Kudos
Bookmarks
Bookmark this Post
MeanRev
Hello,

I am studying the GMAT Quant section and having not been in school for a long time my algebra is very rusty. I am struggling to get my head around something quite simple and was wondering if someone could break it down for me. In the official GMAT guide there is the following example for factoring:

(1) x^3 - 2x^2 + x = -5(x - 1)^2

(2) x^3 - 2x^2 + x + 5(x - 1)^2 = 0

(3) x(x^2 - 2x + 1) + 5(x - 1)^2 = 0

(4) x(x - 1)^2 + 5(x - 1)^2 = 0

(5) (x + 5)(x - 1)^2 = 0

(6) x + 5 = 0 OR (x - 1)^2 = 0

(7) therefore, x = -5 OR x = 1

My confusion comes from how they moved from step (3) to (4)? The rest makes sense but I cannot connect this leap.

Thanks for your time.

MR
Show more

Jmarcotte has provided the correct answer to your question. Do note that you will have to use a lot

\(a^2-b^2 = (a+b)(a-b)\)

\((a+b)^2 = a^2 + 2*a*b + b^2\) and

\((a-b)^2 = a^2 - 2*a*b + b^2\)

Thus, memorize these 3 formula and understand the application of these formulae in the steps you have posted.

So once you get in (3) : (3) \(x(x^2 - 2x + 1) + 5(x - 1)^2 = 0\) ---> realize that \(x^2 - 2x + 1 = (x-1)^2\). Thus substitute this new expression and you get,

(4) \(x(x - 1)^2 + 5(x - 1)^2 = 0\)

Hope this helps.
User avatar
MeanRev
User avatar
Joined: 08 Mar 2016
Last visit: 22 Apr 2016
Posts: 2
Posts: 2
Kudos: 0
How do you solve this part of the equation? 13 Mar 2016, 21:55
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Thank you for your responses. This makes a lot more sense now and has filled the gap in logic that I was missing. Thanks again!
User avatar
ccooley
User avatar
Manhattan Prep Instructor
Joined: 04 Dec 2015
Last visit: 06 Jun 2020
Posts: 931
Own Kudos:
1,642
 [2]
Given Kudos: 115
GMAT 1: 790 Q51 V49
GRE 1: Q170 V170
Expert
Expert reply
GMAT 1: 790 Q51 V49
GRE 1: Q170 V170
Posts: 931
Kudos: 1,642
 [2]
How do you solve this part of the equation? 26 Mar 2016, 14:49
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
In general, if you need to factor a quadratic, here's the safest way to think about it.

1. Put the quadratic in order, making sure that the coefficient of x^2 is 1: x^2 + b x + c = 0.

2. Look at the middle coefficient (b) and the last coefficient (c).

3. Find two numbers that multiply together to create c, and add together to create b.

4. Those two numbers represent the two question marks in the factored quadratic: (x + ?)(x + ?) = 0.

So, if we use the example you gave:

(x^2 - 2x + 1)

You'll look for two numbers that multiply to 1, and add to -2. Those numbers are -1 and -1. So, the factored version is (x - 1)(x - 1).

For another example, try the same with the following quadratic:

x^2 - 5x + 6

The two numbers should multiply to 6, and add to -5. The numbers are -2 and -3. So, the factored version is (x - 2)(x - 3).

If you have time, finish by double-checking your work with FOIL: multiply out the quadratic and make sure it matches what you started with.
User avatar
broilerc
User avatar
Joined: 17 Jan 2016
Last visit: 17 Jul 2016
Posts: 8
Own Kudos:
552
Given Kudos: 46
Posts: 8
Kudos: 552
How do you solve this part of the equation? 04 Apr 2016, 10:13
Kudos
Add Kudos
Bookmarks
Bookmark this Post
ccooley
In general, if you need to factor a quadratic, here's the safest way to think about it.

1. Put the quadratic in order, making sure that the coefficient of x^2 is 1: x^2 + b x + c = 0.

2. Look at the middle coefficient (b) and the last coefficient (c).

3. Find two numbers that multiply together to create c, and add together to create b.
Show more

I am often struggling with the highlighted steps in particular ...especially when it gets to really high numbers.

Does anybody have some tips / shortcuts to quickly find "two numbers that multiply together to create c, and add together to create b" ? Or do I just need to test numbers in my head until they work? These steps always consume most of my time for these questions.. :?

Appreciate any help!
User avatar
ccooley
User avatar
Manhattan Prep Instructor
Joined: 04 Dec 2015
Last visit: 06 Jun 2020
Posts: 931
Own Kudos:
1,642
 [2]
Given Kudos: 115
GMAT 1: 790 Q51 V49
GRE 1: Q170 V170
Expert
Expert reply
GMAT 1: 790 Q51 V49
GRE 1: Q170 V170
Posts: 931
Kudos: 1,642
 [2]
How do you solve this part of the equation? 04 Apr 2016, 14:08
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
broilerc
ccooley
In general, if you need to factor a quadratic, here's the safest way to think about it.

1. Put the quadratic in order, making sure that the coefficient of x^2 is 1: x^2 + b x + c = 0.

2. Look at the middle coefficient (b) and the last coefficient (c).

3. Find two numbers that multiply together to create c, and add together to create b.

I am often struggling with the highlighted steps in particular ...especially when it gets to really high numbers.

Does anybody have some tips / shortcuts to quickly find "two numbers that multiply together to create c, and add together to create b" ? Or do I just need to test numbers in my head until they work? These steps always consume most of my time for these questions.. :?

Appreciate any help!
Show more

Not everybody finds this intuitive! The reason I usually teach it that way is because most people find that strategy more intuitive than the other way of doing it. But in your case, it might be the opposite, and that's completely fine.

The other option, which I encourage you to try, is to memorize and use the quadratic formula: https://www.purplemath.com/modules/quadform.htm Try some problems both ways, and stick with the one that's faster.

If you find that you need to just keep practicing the method I described in my last post, instead, my advice is to try factoring C before you do anything else. Once you have its prime factorization, it should be pretty obvious which pairs of numbers (or at least, which integers) could work - then check them to see if they add to B. It helps to have a sense of about how large the numbers ought to be, too... if B is 1 (for instance) you're probably looking for a positive and negative that are very close to each other, so you'll try to factor C into two very close values.



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!