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## If (x^3 - 2x^2 + x - 1)^2 - 1=(x - a) (x - b) (x - c) (x - d

tagged by: vinni.k

This topic has 6 expert replies and 3 member replies
vinni.k Master | Next Rank: 500 Posts
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#### If (x^3 - 2x^2 + x - 1)^2 - 1=(x - a) (x - b) (x - c) (x - d

Tue Mar 06, 2018 7:42 am
If (x^3 - 2x^2 + x - 1)^2 - 1=(x - a) (x - b) (x - c) (x - d) (x - e) (x - f), then abcdef=?

A. 0
B. 25
C. -25
D. 24
E. -24

OA is A

How to approach this question ?

Thanks

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mbawisdom Senior | Next Rank: 100 Posts
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Tue Mar 06, 2018 2:56 pm
vinni.k wrote:
mbawisdom wrote:
(x^3 - 2x^2 + x - 2)
(x-2)(x^2 + 1)
(x)(x-1)(x-1)((x-2)(x^2 + 1)
one of a, b,c,d,e, or f = 0. therefore abcdef = 0.

1. How you factor out (x-2) from (x^3 - 2x^2 + x - 2)
2. (x)(x-1)(x-1)((x-2)(x^2 + 1). This gives a,b,c,d,e. "f" or any alphabet is still missing, but here x = 0. So, abcdef = 0 , even if there is no f.

Thanks
Hi,

Thanks for the follow up questions

(1) how did I factor out (x-2) from (x^3 - 2x^2 + x - 2)
x^3 - 2x^2 + x - 2
x^2(x-2) + (x-2)
(x-2)(x^2 +1)

(2) so the factors are 0, 1, 1, 2, i and -i (but don't worry about the complex roots i and -i, its out of the scope of the GMAT)

You are right, though, that once we have a root of zero we know abcdef =0, so we can stop at that point.

### GMAT/MBA Expert

mbawisdom Senior | Next Rank: 100 Posts
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Tue Mar 06, 2018 2:56 pm
vinni.k wrote:
mbawisdom wrote:
(x^3 - 2x^2 + x - 2)
(x-2)(x^2 + 1)
(x)(x-1)(x-1)((x-2)(x^2 + 1)
one of a, b,c,d,e, or f = 0. therefore abcdef = 0.

1. How you factor out (x-2) from (x^3 - 2x^2 + x - 2)
2. (x)(x-1)(x-1)((x-2)(x^2 + 1). This gives a,b,c,d,e. "f" or any alphabet is still missing, but here x = 0. So, abcdef = 0 , even if there is no f.

Thanks
Hi,

Thanks for the follow up questions

(1) how did I factor out (x-2) from (x^3 - 2x^2 + x - 2)
x^3 - 2x^2 + x - 2
x^2(x-2) + (x-2)
(x-2)(x^2 +1)

(2) so the factors are 0, 1, 1, 2, i and -i (but don't worry about the complex roots i and -i, its out of the scope of the GMAT)

You are right, though, that once we have a root of zero we know abcdef =0, so we can stop at that point.

### GMAT/MBA Expert

mbawisdom Senior | Next Rank: 100 Posts
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Tue Mar 06, 2018 8:01 am
vinni.k wrote:
If (x^3 - 2x^2 + x - 1)^2 - 1=(x - a) (x - b) (x - c) (x - d) (x - e) (x - f), then abcdef=?

A. 0
B. 25
C. -25
D. 24
E. -24

OA is A

How to approach this question ?

Thanks
Use difference of two squares to find the roots: a^2 - b^2 = (a+b)(a-b)
(x^3 - 2x^2 + x - 1)^2 - 1
a = x^3 - 2x^2 + x - 1
b = 1

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

one of a, b,c,d,e, or f = 0. therefore abcdef = 0.

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GMATGuruNY GMAT Instructor
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Tue Mar 06, 2018 8:11 am
vinni.k wrote:
If (x^3 - 2x^2 + x - 1)^2 - 1=(x - a) (x - b) (x - c) (x - d) (x - e) (x - f), then abcdef=?

A. 0
B. 25
C. -25
D. 24
E. -24
Let x=0.
Plugging x=0 into (x³ - 2x² + x - 1)² - 1=(x - a)(x - b)(x - c)(x - d)(x - e)(x - f), we get:
(0 - 1)² - 1 = (-a)(-b)(-c)(-d)(-e)(-f)
1 - 1 = abcdef
0 = abcdef.

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vinni.k Master | Next Rank: 500 Posts
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GMAT Score:
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Tue Mar 06, 2018 8:48 am
mbawisdom wrote:
(x^3 - 2x^2 + x - 2)
(x-2)(x^2 + 1)
(x)(x-1)(x-1)((x-2)(x^2 + 1)
one of a, b,c,d,e, or f = 0. therefore abcdef = 0.

1. How you factor out (x-2) from (x^3 - 2x^2 + x - 2)
2. (x)(x-1)(x-1)((x-2)(x^2 + 1). This gives a,b,c,d,e. "f" or any alphabet is still missing, but here x = 0. So, abcdef = 0 , even if there is no f.

Thanks

vinni.k Master | Next Rank: 500 Posts
Joined
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Posted:
389 messages
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2 members
6
Test Date:
2016
Target GMAT Score:
700+
GMAT Score:
620
Tue Mar 06, 2018 8:57 am
GMATGuruNY wrote:
Let x=0.
Plugging x=0 into (x³ - 2x² + x - 1)² - 1=(x - a)(x - b)(x - c)(x - d)(x - e)(x - f), we get:
(0 - 1)² - 1 = (-a)(-b)(-c)(-d)(-e)(-f)
1 - 1 = abcdef
0 = abcdef.
Mitch,

Thanks for your reply. I also tried x = 0. In fact, i tried x = 1, x = 2, and x = 3. I got the answer from x = 1, and x =2, but if i took bigger values of x , then i am not getting the answer. Is there a reason that i should stick with the smaller values of x ?

thanks

### GMAT/MBA Expert

GMATGuruNY GMAT Instructor
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Tue Mar 06, 2018 9:25 am
vinni.k wrote:
GMATGuruNY wrote:
Let x=0.
Plugging x=0 into (x³ - 2x² + x - 1)² - 1=(x - a)(x - b)(x - c)(x - d)(x - e)(x - f), we get:
(0 - 1)² - 1 = (-a)(-b)(-c)(-d)(-e)(-f)
1 - 1 = abcdef
0 = abcdef.
Mitch,

Thanks for your reply. I also tried x = 0. In fact, i tried x = 1, x = 2, and x = 3. I got the answer from x = 1, and x =2, but if i took bigger values of x , then i am not getting the answer. Is there a reason that i should stick with the smaller values of x ?

thanks
We are asked to determine the value of abcdef.
If x=0, then x disappears from the right side of the question, leaving only the value of abcdef -- the very value that we seek to determine.
For this reason, it makes sense to test x=0.
As shown in my solution above:
When x=0, abcdef = 0.
Since only one of the answer choices can be correct, there is no need to test any other values for x.

_________________
Mitch Hunt
GMAT Private Tutor
GMATGuruNY@gmail.com

If you find one of my posts helpful, please take a moment to click on the "UPVOTE" icon.

Available for tutoring in NYC and long-distance.
Student Review #1
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Student Review #3

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vinni.k Master | Next Rank: 500 Posts
Joined
17 Apr 2011
Posted:
389 messages
Followed by:
2 members
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Test Date:
2016
Target GMAT Score:
700+
GMAT Score:
620
Wed Mar 07, 2018 8:01 am
Experts,

Regards
Vinni

### GMAT/MBA Expert

mbawisdom Senior | Next Rank: 100 Posts
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Sat Mar 10, 2018 5:59 am
vinni.k wrote:
Experts,

Regards
Vinni
No problems Vinni, thanks for engaging with us - hope we could be of some help!

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