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

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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
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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.

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

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

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

$$= 3x^2-4$$

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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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