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Current Student
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[3x^2(x-2)-x+2]/(x-2)=? [#permalink]
 24 Jul 2005, 08:57
Question Stats:
0% (00:00) correct
 0% (00:00) wrong based on 0 sessions
Please show your working....
[3x^2(x-2)-x+2]/(x-2)=?
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VP
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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
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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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Senior Manager
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Re: PS...actual gmat exam question... [#permalink]
24 Jul 2005, 10:45
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
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VP
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[3x^2(x-2)-x+2]/(x-2)=?
3x^2 (x-2) –x + 2
3x^3 – 6x^2 – x + 2
divided by
X – 2
X out of 3x^3 = 3x^2
3x^2 * x – 2 = 3x^3 + 6x^2
(3x^3 – 6x^2 – x + 2) – (3x^3 + 6x^2) =
-x + 2
-x + 2 divided by x = -1
-1 * x – 2 = -x + 2
(-x + 2) – (-x + 2) = 0
So ans is 3x^2 - 1
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Senior Manager
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Re: PS...actual gmat exam question... [#permalink]
25 Jul 2005, 11:20
mandy wrote:
hence left (3x^2-1) =0
Mandy, how did you equate the LHS to 0.
HMTG.
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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.
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SVP
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Folaa3 wrote: [3x^2(x-2)-x+2]/(x-2)=?
3x^2 (x-2) –x + 2
3x^3 – 6x^2 – x + 2
divided by
X – 2
X out of 3x^3 = 3x^2
3x^2 * x – 2 = 3x^3 + 6x^2
(3x^3 – 6x^2 – x + 2) – (3x^3 + 6x^2) =
-x + 2
-x + 2 divided by x = -1
-1 * x – 2 = -x + 2
(-x + 2) – (-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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Senior Manager
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so is 3x^2-1 the answer or is there any numerical answer
_________________
Fear Mediocrity, Respect Ignorance
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Re: PS...actual gmat exam question... [#permalink]
27 Jul 2005, 01:39
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
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SVP
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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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Manager
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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.
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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, 22:00
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.)
_________________
Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.
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RE: [3x^2(x-2)-x+2]/(x-2)=?
[#permalink]
09 Aug 2005, 22:00
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