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
If (x^3  2x^2 + x  1)^2  1=(x  a) (x  b) (x  c) (x  d
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Use difference of two squares to find the roots: a^2  b^2 = (a+b)(ab)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
(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)(x2)(x^2 + 1)
(x)(x1)(x1)((x2)(x^2 + 1)
one of a, b,c,d,e, or f = 0. therefore abcdef = 0.
Answer is A.
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Let x=0.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
Plugging x=0 into (xÂ³  2xÂ² + x  1)Â²  1=(x  a)(x  b)(x  c)(x  d)(x  e)(x  f), we get:
(0  1)Â²  1 = (a)(b)(c)(d)(e)(f)
1  1 = abcdef
0 = abcdef.
The correct answer is A.
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 vinni.k
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Thanks for your reply, but i have one or two doubts:mbawisdom wrote: (x^3  2x^2 + x  2)
(x2)(x^2 + 1)
(x)(x1)(x1)((x2)(x^2 + 1)
one of a, b,c,d,e, or f = 0. therefore abcdef = 0.
Answer is A.
1. How you factor out (x2) from (x^3  2x^2 + x  2)
2. (x)(x1)(x1)((x2)(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
 vinni.k
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Mitch,GMATGuruNY wrote: Let x=0.
Plugging x=0 into (xÂ³  2xÂ² + x  1)Â²  1=(x  a)(x  b)(x  c)(x  d)(x  e)(x  f), we get:
(0  1)Â²  1 = (a)(b)(c)(d)(e)(f)
1  1 = abcdef
0 = abcdef.
The correct answer is A.
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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We are asked to determine the value of abcdef.vinni.k wrote:Mitch,GMATGuruNY wrote: Let x=0.
Plugging x=0 into (xÂ³  2xÂ² + x  1)Â²  1=(x  a)(x  b)(x  c)(x  d)(x  e)(x  f), we get:
(0  1)Â²  1 = (a)(b)(c)(d)(e)(f)
1  1 = abcdef
0 = abcdef.
The correct answer is A.
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
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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Hi,vinni.k wrote:Thanks for your reply, but i have one or two doubts:mbawisdom wrote: (x^3  2x^2 + x  2)
(x2)(x^2 + 1)
(x)(x1)(x1)((x2)(x^2 + 1)
one of a, b,c,d,e, or f = 0. therefore abcdef = 0.
Answer is A.
1. How you factor out (x2) from (x^3  2x^2 + x  2)
2. (x)(x1)(x1)((x2)(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
Thanks for the follow up questions
(1) how did I factor out (x2) from (x^3  2x^2 + x  2)
x^3  2x^2 + x  2
x^2(x2) + (x2)
(x2)(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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Hi,vinni.k wrote:Thanks for your reply, but i have one or two doubts:mbawisdom wrote: (x^3  2x^2 + x  2)
(x2)(x^2 + 1)
(x)(x1)(x1)((x2)(x^2 + 1)
one of a, b,c,d,e, or f = 0. therefore abcdef = 0.
Answer is A.
1. How you factor out (x2) from (x^3  2x^2 + x  2)
2. (x)(x1)(x1)((x2)(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
Thanks for the follow up questions
(1) how did I factor out (x2) from (x^3  2x^2 + x  2)
x^3  2x^2 + x  2
x^2(x2) + (x2)
(x2)(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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No problems Vinni, thanks for engaging with us  hope we could be of some help!vinni.k wrote:Experts,
Thanks for your reply <i class="em emgrinning"></i>
Regards
Vinni