Problem Solving

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

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

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

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

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