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

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by vinni.k » 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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by mbawisdom » 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.

Answer is A.
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by GMATGuruNY » 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.

The correct answer is A.
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by vinni.k » 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.
Answer is A.
Thanks for your reply, but i have one or two doubts:-

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.

Please clarify.

Thanks
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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.
The correct answer is A.
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 ?

Please clarify

thanks
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abcdef

by GMATGuruNY » 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.
The correct answer is A.
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 ?

Please clarify

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.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
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by mbawisdom » 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.
Answer is A.
Thanks for your reply, but i have one or two doubts:-

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.

Please clarify.

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.
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by mbawisdom » 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.
Answer is A.
Thanks for your reply, but i have one or two doubts:-

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.

Please clarify.

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.
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by vinni.k » Wed Mar 07, 2018 8:01 am
Experts,

Thanks for your reply <i class="em em-grinning"></i>

Regards
Vinni
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by mbawisdom » Sat Mar 10, 2018 5:59 am
vinni.k wrote:Experts,

Thanks for your reply <i class="em em-grinning"></i>

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