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Difficulty: 505-555 Level,    Algebra,             
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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. \(3x + 4\)


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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



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

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

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

\(= 3x^2-4\)

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