Last visit was: 24 Jul 2024, 23:57 It is currently 24 Jul 2024, 23:57
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
SORT BY:
Date
Tags:
Difficulty: 505-555 Level,   Algebra,                     
Show Tags
Hide Tags
GMAT Club Legend
GMAT Club Legend
Joined: 12 Sep 2015
Posts: 6804
Own Kudos [?]: 30864 [1]
Given Kudos: 799
Location: Canada
Send PM
GMAT Club Legend
GMAT Club Legend
Joined: 03 Oct 2013
Affiliations: CrackVerbal
Posts: 4915
Own Kudos [?]: 7820 [1]
Given Kudos: 221
Location: India
Send PM
Manager
Manager
Joined: 11 Aug 2021
Posts: 58
Own Kudos [?]: 17 [0]
Given Kudos: 83
Send PM
Math Expert
Joined: 02 Sep 2009
Posts: 94611
Own Kudos [?]: 643703 [0]
Given Kudos: 86740
Send PM
Re: If a, b, and c are constants, a > b > c, and x^3 - x = (x - a)(x - b) [#permalink]
Expert Reply
sa800 wrote:
my question is why or where are people getting -1,0,1 from? I assume they factor …. but then Why do people automatically set the quadratic equation equal to zero after factoring?

nowhere on the question does it say that….(x)(x+1)(x-1) = 0…..

is there a fundamental theorem im missing?


You are missing the crucial piece of information given in the stem: \(x^3-x=(x-a)(x-b)(x-c)\) for ALL numbers \(x\).

Since \(x^3-x=(x-a)(x-b)(x-c)\) is true for ALL \(x-es\) than it must also be true for \(x=0\), \(x=1\) and \(x=-1\). Setting x to 0 1 and -1 allows us to determine the values of a, b, and c.

Hope it helps.
Tutor
Joined: 10 Jul 2015
Status:Expert GMAT, GRE, and LSAT Tutor / Coach
Affiliations: Harvard University, A.B. with honors in Government, 2002
Posts: 1182
Own Kudos [?]: 2439 [0]
Given Kudos: 274
Location: United States (CO)
Age: 44
GMAT 1: 770 Q47 V48
GMAT 2: 730 Q44 V47
GMAT 3: 750 Q50 V42
GMAT 4: 730 Q48 V42 (Online)
GRE 1: Q168 V169

GRE 2: Q170 V170
Send PM
If a, b, and c are constants, a > b > c, and x^3 - x = (x - a)(x - b) [#permalink]
Expert Reply
Attached is a visual that should help.­  First, factor out x—then use your quadratic identities + logic to solve.  

You must also be familiar with the specific quadratic identity (x+y)(x-y) = x^2 - y^2, or in this case (x+1)(x-1) = x^2 - 1.

­
GMAT Club Bot
If a, b, and c are constants, a > b > c, and x^3 - x = (x - a)(x - b) [#permalink]
   1   2 
Moderator:
Math Expert
94609 posts