Last visit was: 25 Apr 2026, 02:44 It is currently 25 Apr 2026, 02:44
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.
Close
Request Expert Reply
Confirm Cancel
User avatar
hardaway7
User avatar
Current Student
Joined: 03 Dec 2007
Last visit: 05 Nov 2010
Posts: 72
Own Kudos:
Schools:NYU STERN
Posts: 72
Kudos: 8
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
ALD
Joined: 15 Apr 2008
Last visit: 01 Oct 2012
Posts: 90
Own Kudos:
Given Kudos: 1
Posts: 90
Kudos: 24
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
hardaway7
User avatar
Current Student
Joined: 03 Dec 2007
Last visit: 05 Nov 2010
Posts: 72
Own Kudos:
Schools:NYU STERN
Posts: 72
Kudos: 8
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
zonk
User avatar
Current Student
Joined: 12 Jul 2008
Last visit: 10 Nov 2013
Posts: 366
Own Kudos:
354
 [1]
Concentration: Finance, Entrepreneurship, Health Care
Schools:Wharton
Posts: 366
Kudos: 354
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
hardaway7
Thanks ALD-

How did you go from: 3x(x-3) + 1(x-3)=0

to:
(3x+1)(x-3)=0

Sorry I am slow!!! I don't know what is wrong with me!

now the two components of the sum -- 3x(x-3) and 1(x-3) -- both have (x-3) in common. So you can factor out (x-3) from each term.
User avatar
buffdaddy
Joined: 14 Oct 2007
Last visit: 13 Jan 2011
Posts: 546
Own Kudos:
303
 [1]
Given Kudos: 8
Location: Oxford
Schools:Oxford'10
Posts: 546
Kudos: 303
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
hardaway7
Thanks ALD-

How did you go from: 3x(x-3) + 1(x-3)=0

to:
(3x+1)(x-3)=0

Sorry I am slow!!! I don't know what is wrong with me!

Hi Hardaway7,

if you look at 3x(x-3) + 1(x-3),
(x-3) is common to the parts being added together. So one can factor this out.

its the same as 3ab + 1b can be written as b(3a + 1)

good luck

-buff
User avatar
lumone
Joined: 25 Nov 2006
Last visit: 26 Oct 2010
Posts: 275
Own Kudos:
Schools:St Gallen, Cambridge, HEC Montreal
Posts: 275
Kudos: 298
Kudos
Add Kudos
Bookmarks
Bookmark this Post
What are doing here, Buff? :P
Are you re-taking the GMAT like me?
User avatar
hardaway7
User avatar
Current Student
Joined: 03 Dec 2007
Last visit: 05 Nov 2010
Posts: 72
Own Kudos:
Schools:NYU STERN
Posts: 72
Kudos: 8
Kudos
Add Kudos
Bookmarks
Bookmark this Post
ahhh got it!

THANKS to both of you!
User avatar
buffdaddy
Joined: 14 Oct 2007
Last visit: 13 Jan 2011
Posts: 546
Own Kudos:
Given Kudos: 8
Location: Oxford
Schools:Oxford'10
Posts: 546
Kudos: 303
Kudos
Add Kudos
Bookmarks
Bookmark this Post
lumone
What are doing here, Buff? :P
Are you re-taking the GMAT like me?

Hi Lumone,

truth be told, being a fellow martian, I was compelled to help Hardaway7.

back to the application forums!

-buff
User avatar
tusharvk
Joined: 04 Jan 2009
Last visit: 24 May 2011
Posts: 119
Own Kudos:
Concentration: finance & strategy
Posts: 119
Kudos: 21
Kudos
Add Kudos
Bookmarks
Bookmark this Post
hardaway7
I am having trouble factoring problems that having something in front of the x^2

For example- how would I do this?

3x^2 - 8x -3 = 0

OG says (3x+1)(x-3)

But how do you figure that out quickly?!
a fail proof way to solve a problem of the type
ax^2+bx+c=0 is
x=-b+/-sqrt(b^2-4ac)

A simpler approach is to look at the actual values of ac and b (with the signs).
Then find two integers k1 and k2 such that k1+k2 = ac and k1k2=b.
In the present situation,
ac=-9 and b=-8. Thus, k1=-9 and k2 = 1
and we get
3x^2-9x+x-3=0 and problem solved.



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!