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Manager  G
Joined: 03 Jun 2019
Posts: 79
If x ≠ -1/2, then (6x^3 + 3x^2 - 8x - 4)/(2x + 1) =  [#permalink]

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Difficulty:   15% (low)

Question Stats: 79% (01:46) correct 21% (02:17) wrong based on 191 sessions

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If $$x ≠ -\frac{1}{2}$$, then $$\frac{6x^3 + 3x^2 - 8x - 4}{2x + 1} =$$

A. $$3x^2 + \frac{3}{2}x -8$$

B. $$3x^2 + \frac{3}{2}x -4$$

C. $$3x2 – 4$$

D. $$3x – 4$$

E. $$3x + 4$$

PS16980.02
GMAT Club Legend  V
Joined: 11 Sep 2015
Posts: 4999
GMAT 1: 770 Q49 V46
Re: If x ≠ -1/2, then (6x^3 + 3x^2 - 8x - 4)/(2x + 1) =  [#permalink]

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parkhydel wrote:
If $$x ≠ -\frac{1}{2}$$, then $$\frac{6x^3 + 3x^2 - 8x - 4}{2x + 1} =$$

A. $$3x^2 + \frac{3}{2}x -8$$

B. $$3x^2 + \frac{3}{2}x -4$$

C. $$3x2 – 4$$

D. $$3x – 4$$

E. $$3x + 4$$

If you're not sure how to simplify the given expression, you can also solve this question by testing values.

Since we're looking for an equivalent expression, both the given expression and the correct answer must evaluate to have the same value for any value of x.

So, for example, if $$x = 2$$, then $$\frac{6x^3 + 3x^2 - 8x - 4}{2x + 1} =\frac{6(2)^3 + 3(2)^2 - 8(2) - 4}{2(2) + 1}$$

$$=\frac{48 + 12 - 16 - 4}{4 + 1}$$

$$=\frac{40}{5}$$

$$=8$$

This tells us that, the correct answer must also evaluate to be $$8$$ when $$x = 2$$

So let's plug $$x = 2$$ into each answer choice and see which expression evaluates to be $$8$$

A. $$3(2)^2 + \frac{3}{2}(2) -8=7$$. No good. We need the expression to evaluate to be $$8$$

B. $$3(2)^2 + \frac{3}{2}(2) -4=11$$. No good. We need the expression to evaluate to be $$8$$

C. $$3(2^2) – 4=8$$. Woo woo!!!

D. $$3(2) – 4=2$$. No good. We need the expression to evaluate to be $$8$$

E. $$3(2) + 4=10$$. No good. We need the expression to evaluate to be $$8$$

Cheers,
Brent
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Director  D
Joined: 25 Jul 2018
Posts: 732
Re: If x ≠ -1/2, then (6x^3 + 3x^2 - 8x - 4)/(2x + 1) =  [#permalink]

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If $$x ≠ -\frac{1}{2}$$, then
$$\frac{6x^3 + 3x^2 - 8x - 4}{2x + 1} =$$
= $$\frac{3x^{2}( 2x+ 1) —4(2x+1)}{2x+1}=\frac{ (3x^{2} —4)(2x+1)}{ 2x+1}= 3x^{2} —4$$

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Re: If x ≠ -1/2, then (6x^3 + 3x^2 - 8x - 4)/(2x + 1) =  [#permalink]

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$$\frac{(6x^3 + 3x^2) - (8x + 4)}{2x + 1}$$

$$=\frac{[3x^2(2x+1)]- [4(2x+1)]}{2x + 1}$$

$$=\frac{(3x^2-4)(2x+1)}{2x + 1}$$

$$= 3x^2-4$$

parkhydel wrote:
If $$x ≠ -\frac{1}{2}$$, then $$\frac{6x^3 + 3x^2 - 8x - 4}{2x + 1} =$$

A. $$3x^2 + \frac{3}{2}x -8$$

B. $$3x^2 + \frac{3}{2}x -4$$

C. $$3x2 – 4$$

D. $$3x – 4$$

E. $$3x + 4$$

PS16980.02
Math Expert V
Joined: 02 Aug 2009
Posts: 8795
Re: If x ≠ -1/2, then (6x^3 + 3x^2 - 8x - 4)/(2x + 1) =  [#permalink]

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1
parkhydel wrote:
If $$x ≠ -\frac{1}{2}$$, then $$\frac{6x^3 + 3x^2 - 8x - 4}{2x + 1} =$$

A. $$3x^2 + \frac{3}{2}x -8$$

B. $$3x^2 + \frac{3}{2}x -4$$

C. $$3x2 – 4$$

D. $$3x – 4$$

E. $$3x + 4$$

PS16980.02

Two ways..

$$\frac{6x^3 + 3x^2 - 8x - 4}{2x + 1} =$$? , and all choices are having integer coefficient of x^2, so we can say that the denominator should be dividing the numerator..
So 2x=1 is a factor of $$6x^3 + 3x^2 - 8x - 4$$ and pair of (6,3) and (8,4) should make you relate 2x+1 to both pairs as 6:3 and 8:4 are also 2:1..
\frac{3x^2(2x+1) -4(2x+1) 8x - 4}{2x + 1} =\frac{(3x^2-4)(2x+1)}{2x+1}=3x^2-4[/m]

Choices..
First point
We have just 1 power of x in denominator in 2x, where as we have power of 3 in numerator in $$3x^3$$, so our answer has to have a term in $$\frac{x^3}{x} =x^2$$.
Eliminate C and D
Second point
Take x=1, the simplest value possible ...\frac{6x^3 + 3x^2 - 8x - 4}{2x + 1} =\frac{6+3-8-4}{2+1}=\frac{-3}{3}=-1[/m]
So our answer cannot have a fraction .
Eliminate A and B as both have a fraction 3/2 when you substitute x as 1.

C
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Sloan MIT School Moderator V
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Re: If x ≠ -1/2, then (6x^3 + 3x^2 - 8x - 4)/(2x + 1) =  [#permalink]

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parkhydel wrote:
If $$x ≠ -\frac{1}{2}$$, then $$\frac{6x^3 + 3x^2 - 8x - 4}{2x + 1} =$$

A. $$3x^2 + \frac{3}{2}x -8$$

B. $$3x^2 + \frac{3}{2}x -4$$

C. $$3x2 – 4$$

D. $$3x – 4$$

E. $$3x + 4$$

PS16980.02

The best way to solve this is to do the normal division. Since picking numbers can be suicidal.
Taking x = 0 results
$$\frac{6x^3 + 3x^2 - 8x - 4}{2x + 1} = -4$$

B,C and D also give -4. So not good.

Taking x = 2, $$\frac{6x^3 + 3x^2 - 8x - 4}{2x + 1} = 8$$
Only C satisfies the condition and gives -4.

OR,
$$\frac{6x^3 + 3x^2 - 8x - 4}{2x + 1} =$$
$$= \frac{(6x^3 + 3x^2) - (8x + 4)}{2x + 1}$$

$$=\frac{[3x^2(2x+1)]- [4(2x+1)]}{2x + 1}$$

$$=\frac{(3x^2-4)(2x+1)}{2x + 1}$$

$$= 3x^2-4$$

_________________ Re: If x ≠ -1/2, then (6x^3 + 3x^2 - 8x - 4)/(2x + 1) =   [#permalink] 23 Jun 2020, 02:57

# If x ≠ -1/2, then (6x^3 + 3x^2 - 8x - 4)/(2x + 1) =  