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133 Qant Guide

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133 Qant Guide

by cpay3245 » Mon Apr 22, 2013 3:13 am
If x cannot equal 2, the 3x^2(x-2) - x +2 / x-2 =

A. 3x^2 - x + 2
B. 3x^2 + 1
C. 3x^2
D. 3x^2 - 1
E 3x^2 - 2
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by chaithu_bunny » Mon Apr 22, 2013 3:51 am
cpay3245 wrote:If x cannot equal 2, the 3x^2(x-2) - x +2 / x-2 =

A. 3x^2 - x + 2
B. 3x^2 + 1
C. 3x^2
D. 3x^2 - 1
E 3x^2 - 2
Answer is [spoiler]D[/spoiler] As simple as that...

Take (x-2) common in the numerator. You will be left with, (3x^2 - 1)

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

Hope this helps... :)
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by Anju@Gurome » Mon Apr 22, 2013 4:04 am
cpay3245 wrote:If x cannot equal 2, the [3x²(x - 2) - x + 2]/(x - 2) =
Numerator = [3x²(x - 2) - x + 2]
= [3x²(x - 2) - (x - 2)]
= (3x² - 1)(x - 2) [Take (x - 2) common from both terms]

So, given expression = (3x² - 1)(x - 2)/(x - 2)
As x ≠ 2, (x - 2) ≠ 0 and we can divide the numerator by (x - 2)

So, given expression = (3x² - 1)

The correct answer is D.
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by Brent@GMATPrepNow » Mon Apr 22, 2013 6:07 am
cpay3245 wrote:If x cannot equal 2, the [3x^2(x-2) - x + 2] / (x-2) =

A. 3x^2 - x + 2
B. 3x^2 + 1
C. 3x^2
D. 3x^2 - 1
E 3x^2 - 2
Simplifying the numberator of this expression is similar to simplifying 3x - 1x to get 2x

We get:
[3x^2(x-2) - x + 2] / (x-2) = [3x^2(x-2) - 1(x+2)] / (x-2)
= [(3x^2 - 1)(x+2)] / (x+2)
= 3x^2 - 1
= D

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Brent
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by Anju@Gurome » Mon Apr 22, 2013 6:34 am
The easiest way to solve this problem is to plug some simple numbers.

Let's plug x = 0.
Given expression = [3x²(x - 2) - x + 2] /(x - 2) = [0 - 0 + 2]/(0 - 2) = 2/(-2) = -1

So, the correct option must be equal to -1 for x = 0.

Only option D works.

The correct answer is D.
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