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

by vinni.k » 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.
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.
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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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