[3x^2(x-2)-x+2]/(x-2)=? : PS Archive
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# [3x^2(x-2)-x+2]/(x-2)=?

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24 Jul 2005, 07:57
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[3x^2(x-2)-x+2]/(x-2)=?
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24 Jul 2005, 08:23
Freshina, do you have the answer choices to this question?

I got 3x^2 - 1 (using polynomial division)

3x^3 - 6x^2 - x + 2 /x-2 = 3x^2 - 1
Current Student
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24 Jul 2005, 09:18
I do...but I dont want to give em..cause..people start guessing...I want to see the workings...
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Re: PS...actual gmat exam question... [#permalink]

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24 Jul 2005, 09:45
fresinha12 wrote:
from a past GMAT exam...

[3x^2(x-2)-x+2]/(x-2)=?

[3x^2(x-2)-x+2]/(x-2)=
factor x-2

3x^2(x-2)-(x-2)/(x-2)=

(x-2)(3x^2-1)/(x-2)
CANCEL OUT ( x-2)

hence left (3x^2-1) =0

3x^2=1
x^2=1/3
X = +OR -SQ ROOT 1/3 hope it is right
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24 Jul 2005, 09:50
[3x^2(x-2)-x+2]/(x-2)=?

3x^2 (x-2) â€“x + 2
3x^3 â€“ 6x^2 â€“ x + 2
divided by
X â€“ 2

X out of 3x^3 = 3x^2
3x^2 * x â€“ 2 = 3x^3 + 6x^2
(3x^3 â€“ 6x^2 â€“ x + 2) â€“ (3x^3 + 6x^2) =
-x + 2
-x + 2 divided by x = -1
-1 * x â€“ 2 = -x + 2
(-x + 2) â€“ (-x + 2) = 0

So ans is 3x^2 - 1
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Re: PS...actual gmat exam question... [#permalink]

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25 Jul 2005, 10:20
mandy wrote:

hence left (3x^2-1) =0

Mandy, how did you equate the LHS to 0.

HMTG.
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26 Jul 2005, 11:26
Folaa3 wrote:
Freshina, do you have the answer choices to this question?

I got 3x^2 - 1 (using polynomial division)

3x^3 - 6x^2 - x + 2 /x-2 = 3x^2 - 1

I agree with fola.
Mandy: It's not an eqn. but a polynomial, so you can't assume 3X^2 - 1 to be equal to zero.
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26 Jul 2005, 11:46
Folaa3 wrote:
[3x^2(x-2)-x+2]/(x-2)=?

3x^2 (x-2) â€“x + 2
3x^3 â€“ 6x^2 â€“ x + 2
divided by
X â€“ 2

X out of 3x^3 = 3x^2
3x^2 * x â€“ 2 = 3x^3 + 6x^2
(3x^3 â€“ 6x^2 â€“ x + 2) â€“ (3x^3 + 6x^2) =
-x + 2
-x + 2 divided by x = -1
-1 * x â€“ 2 = -x + 2
(-x + 2) â€“ (-x + 2) = 0

So ans is 3x^2 - 1

Very long method. Mandy's approach is best except that LHS shudn't be equated to 0 and the ans shud be left as 3x^2-1
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26 Jul 2005, 19:34
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Fear Mediocrity, Respect Ignorance

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Re: PS...actual gmat exam question... [#permalink]

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27 Jul 2005, 00:39
HowManyToGo wrote:
mandy wrote:

hence left (3x^2-1) =0

Mandy, how did you equate the LHS to 0.

HMTG.

PLZ what is the LHS

my logic was get the x
I had left this : 3x^2 -1
since there is only one unknow i can still solve it
so 3x^2 -1=O
3x^2=1
SO x^2=1/3
Hence + OR - SQUARE root of 1/3
i try to get the most simplified form of x
i might be wrong PLZ Let me know

thanks
VP
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27 Jul 2005, 15:00
Mandy: That's wrong, there can't be any numerical ans to this. Ans shud be left as 3x^2 -1. You can't just equate 3x^2 -1 = 0 unless otherwise indicated.
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06 Aug 2005, 21:52
What if x=2 and the denominator is 0 !

So it is only (3x^2-1) if x <> 2

otherwise you have to keep the (x-2)s in.
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07 Aug 2005, 04:40
[3x^2(x-2)-x+2]/(x-2)=?

Steps...
= [3x^2(x-2)-(x-2)]/(x-2)

= [(x-2) (3x^2-1)]/(x-2)

= 3x^2-1.

(This is true only if 'x' is not equal to '2')
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09 Aug 2005, 21:00
This is basic algebra, you should make sure you feel it to be very easy.

[3x^2(x-2)-x+2]/(x-2)
=[3x^2(x-2)-(x-2)]/(x-2)
=[(x-2)(3x^2-1)]/(x-2)
=3x^2-1

(x would not be equal to 2 by definition, because the original function already have (x-2) in the denominator.)
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RE: [3x^2(x-2)-x+2]/(x-2)=?   [#permalink] 09 Aug 2005, 21:00
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