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[3x^2(x-2)-x+2]/(x-2)=? [#permalink ]
24 Jul 2005, 07:57

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Please show your working.... [3x^2(x-2)-x+2]/(x-2)=?

Director

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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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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 ]
24 Jul 2005, 09:45

fresinha12 wrote:

from a past GMAT exam... Please show your working.... [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

Director

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[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 ]
25 Jul 2005, 10:20

mandy wrote:

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

Mandy, how did you equate the LHS to 0.

HMTG.

Director

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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.

VP

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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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so is 3x^2-1 the answer or is there any numerical answer

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

SVP

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[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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RE: [3x^2(x-2)-x+2]/(x-2)=? [#permalink ]
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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