Last visit was: 23 Apr 2026, 19:38 It is currently 23 Apr 2026, 19:38
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
smcgrath12
Joined: 28 Jun 2004
Last visit: 11 Jun 2005
Posts: 61
Own Kudos:
Posts: 61
Kudos: 742
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
smcgrath12
Joined: 28 Jun 2004
Last visit: 11 Jun 2005
Posts: 61
Own Kudos:
Posts: 61
Kudos: 742
Kudos
Add Kudos
Bookmarks
Bookmark this Post
avatar
hardworker_indian
Joined: 20 Jul 2004
Last visit: 08 Sep 2011
Posts: 311
Own Kudos:
Posts: 311
Kudos: 407
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
cbrf3
Joined: 05 Feb 2004
Last visit: 22 Oct 2006
Posts: 187
Own Kudos:
Location: USA
Posts: 187
Kudos: 42
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Ditto hardworker_indian!!
User avatar
ian7777
Joined: 09 Mar 2003
Last visit: 24 Jan 2010
Posts: 227
Own Kudos:
Posts: 227
Kudos: 178
Kudos
Add Kudos
Bookmarks
Bookmark this Post
you can also just do straight division:

(2x^3 - 3x^2 + 7x + p)/(x+1)

I can't do it here, but if you use the regular division bar (the one that looks like a square root) you can solve this and get 2x^2 - 5x + 12. When you do, the only number for p that will cancel out perfectly and make it work is 12.
User avatar
smcgrath12
Joined: 28 Jun 2004
Last visit: 11 Jun 2005
Posts: 61
Own Kudos:
Posts: 61
Kudos: 742
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Some of you are just awesome. I was betting that nobody could answer that, because it goes to the Remainder Theorem and synthetic division concept, which many people won't remember from their algebra days. Is this concept explained in any standard GMAT guide?
User avatar
neelam
Joined: 20 Jul 2004
Last visit: 17 Sep 2004
Posts: 10
Posts: 10
Kudos: 0
Kudos
Add Kudos
Bookmarks
Bookmark this Post
hi .. could u pls tell me why does not 1 work here..if you substitute x=1, even then the number is divisible by x+1.
pls explain..
User avatar
afife76
Joined: 19 Jun 2003
Last visit: 21 Dec 2004
Posts: 75
Own Kudos:
Posts: 75
Kudos: 66
Kudos
Add Kudos
Bookmarks
Bookmark this Post
hardworker_indian
smcgrath12
Yes, its correct. Please post your steps if possible. Thanks.

For any polynomial experssion f(x), if x-a is a factor, then f(a)=0

So, f(x) = 2x^3 - 3x^2+7x+p
f(-1) = -2 - 3 -7 + p = 0
p = 12



hardworker_indian,

Does this rule hold true for (x+a) as well? Since you are taking a=-1, shouldn't we thinking as "if (x+a) factor of ...." ?

Thanks
User avatar
smcgrath12
Joined: 28 Jun 2004
Last visit: 11 Jun 2005
Posts: 61
Own Kudos:
Posts: 61
Kudos: 742
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Yes, for any function f(x), x+a is a factor if f(-a)=0 or x-a is a factor if f(a)=0.
User avatar
lastochka
Joined: 07 Oct 2003
Last visit: 15 Aug 2006
Posts: 193
Own Kudos:
Location: Manhattan
Posts: 193
Kudos: 71
Kudos
Add Kudos
Bookmarks
Bookmark this Post
hardworker_indian
smcgrath12
Yes, its correct. Please post your steps if possible. Thanks.


For any polynomial experssion f(x), if x-a is a factor, then f(a)=0

apparently my algebraic skills aren't where they should be... I couldn't follow this explanation
User avatar
ywilfred
Joined: 07 Jul 2004
Last visit: 06 Mar 2012
Posts: 1,987
Own Kudos:
Location: Singapore
Posts: 1,987
Kudos: 2,051
Kudos
Add Kudos
Bookmarks
Bookmark this Post
2x^3 - 3x^2 + 7x +p will equate to 0 if x+1 is a root.

So 2(-1)^3 - 3(-1)^2 +7(-1) + p = 0
p = 12
avatar
hardworker_indian
Joined: 20 Jul 2004
Last visit: 08 Sep 2011
Posts: 311
Own Kudos:
Posts: 311
Kudos: 407
Kudos
Add Kudos
Bookmarks
Bookmark this Post
neelam
hi .. could u pls tell me why does not 1 work here..if you substitute x=1, even then the number is divisible by x+1.
pls explain..
If you substitute x=1,
=> x=1
=> x-1=0
we are not looking to see if x-1 is a factor. We are looking to see if x+1 is a factor. As a simple rule, you can equate the factor (x+1) to zero (x+1=0) and find the value of x (x+1=0 => x=-1) and then subsitute this in the original expression.


lastochka

apparently my algebraic skills aren't where they should be... I couldn't follow this explanation

I will try to explain as much as I can - from little notes and distant memory.

We generally note algeraic expressions as functions, f(x) -- this means that f(x) is a polynomial in x and for each value of x, f(x) gives you a result.

Rules:
For any polynomial f(x):
1. To find factors, solve: f(x)=0
2. For any value a, f(a) gives the reminder when f(x) is divided by x-a.

#2 inutrn means,
For any factor x-a, f(a) gives 0
For any nonfactor x-a, f(a) gives reminder


Lets take an example: f(x) = x^2-5x+6

To find factors:
equate f(x) = 0,
x^2-5x+6=0
(x-3)(x-2)=0
x=3 or 2

To see if x-3 is a factor:
x-3=0 => x=3
f(x) = x^2-5x+6
f(3) = 3^2-5.3+6
f(3) = 0
So, x-2 is a factor of x^2-5x+6

To see if x+5 is a factor:
x+5=0 => x=-5
f(x) = x^2-5x+6
f(5) = 5^2+5.5+6
f(3) = 56
So, x+5 is NOT a factor of x^2-5x+6.
And when you divide x^2-5x+6 by x+5, you will get a reminder of 56.

I think this should be enough for this topic, but algebra in itself is a ocean. :-D
User avatar
lastochka
Joined: 07 Oct 2003
Last visit: 15 Aug 2006
Posts: 193
Own Kudos:
Location: Manhattan
Posts: 193
Kudos: 71
Kudos
Add Kudos
Bookmarks
Bookmark this Post
hardworker_indian, appreciate you taking time with explanation, it's been years since I've done f(x), thanks for refreshing the memory.

--== Message from the GMAT Club Team ==--

THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION.
This discussion does not meet community quality standards. It has been retired.


If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
User avatar
MA
Joined: 25 Nov 2004
Last visit: 09 Aug 2011
Posts: 695
Own Kudos:
Posts: 695
Kudos: 533
Kudos
Add Kudos
Bookmarks
Bookmark this Post
For what value of p is the expression 2(x*x*x)-3(x*x)+7x+p divisible by x+1 ?

--== Message from the GMAT Club Team ==--

THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION.
This discussion does not meet community quality standards. It has been retired.


If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
User avatar
tl372
Joined: 10 May 2006
Last visit: 16 Jul 2013
Posts: 100
Own Kudos:
Location: USA
Posts: 100
Kudos: 10
Kudos
Add Kudos
Bookmarks
Bookmark this Post
P = 12

I basically took 2(x^3)-3(x^2)+7x+p and divded it by x+1 (long division style).

I ended up with 2(x^3)-3(x^2)+7x+12 = [2(x^2) -5x +12] * [x+1]

P needs to be 12 in order to divide x+1 into the equation evenly.
User avatar
haas_mba07
Joined: 02 Jun 2006
Last visit: 26 Jun 2008
Posts: 662
Own Kudos:
Posts: 662
Kudos: 218
Kudos
Add Kudos
Bookmarks
Bookmark this Post
This one was a weird one for me... didn't expect cubic equation division to come in GMAT.

Well the answer (long way again... as tl mentioned) is p= 12.

The bigger question I have is should I be prepared to expect such a question on the GMAT?
User avatar
tl372
Joined: 10 May 2006
Last visit: 16 Jul 2013
Posts: 100
Own Kudos:
Location: USA
Posts: 100
Kudos: 10
Kudos
Add Kudos
Bookmarks
Bookmark this Post
In my opinion, because this question tested creative thinking more so then any specific properties of the cubic square, I think that it's fair game. I had to think for a little while before realizing that straight division would work.

The first time I took the gmats, I didn't remember seeing anything like this.....who knows though...I'll have a better gauge this Sat after I attempt the Gmats for the second time!
User avatar
haas_mba07
Joined: 02 Jun 2006
Last visit: 26 Jun 2008
Posts: 662
Own Kudos:
Posts: 662
Kudos: 218
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Given that you solved this, I am sure you are more than prepared for anything they throw at you.

Good luck.


tl372
In my opinion, because this question tested creative thinking more so then any specific properties of the cubic square, I think that it's fair game. I had to think for a little while before realizing that straight division would work.

The first time I took the gmats, I didn't remember seeing anything like this.....who knows though...I'll have a better gauge this Sat after I attempt the Gmats for the second time!
User avatar
tl372
Joined: 10 May 2006
Last visit: 16 Jul 2013
Posts: 100
Own Kudos:
Location: USA
Posts: 100
Kudos: 10
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Thanks Haas... :)
User avatar
iced_tea
Joined: 11 May 2006
Last visit: 29 Jun 2008
Posts: 146
Own Kudos:
Posts: 146
Kudos: 396
Kudos
Add Kudos
Bookmarks
Bookmark this Post
for 2(x*x*x)-3(x*x)+7x+p to be divisible by x+1,

x+1 should be a root of 2(x*x*x)-3(x*x)+7x+p = 0

subsitute x=-1 and you get p = 12
 1   2