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# If x ≠ –1, then 2x^2 + 4x + 2/(x + 1)^2 =

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Math Expert
Joined: 02 Sep 2009
Posts: 58396
If x ≠ –1, then 2x^2 + 4x + 2/(x + 1)^2 =  [#permalink]

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09 Apr 2019, 03:47
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Difficulty:

15% (low)

Question Stats:

95% (00:51) correct 5% (02:12) wrong based on 21 sessions

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If x ≠ –1, then $$\frac{2x^2 + 4x + 2}{(x + 1)^2} =$$

(A) 0
(B) 1
(C) 2
(D) 4
(E) 6

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

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09 Apr 2019, 03:51
Bunuel wrote:
If x ≠ –1, then $$\frac{2x^2 + 4x + 2}{(x + 1)^2} =$$

(A) 0
(B) 1
(C) 2
(D) 4
(E) 6

$$\frac{2x^2 + 4x + 2}{(x + 1)^2} =$$
2* (x+1)^2/(x+1)^2 = 2
IMO C
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Re: If x ≠ –1, then 2x^2 + 4x + 2/(x + 1)^2 =  [#permalink]

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09 Apr 2019, 04:07
1
Bunuel wrote:
If x ≠ –1, then $$\frac{2x^2 + 4x + 2}{(x + 1)^2} =$$

(A) 0
(B) 1
(C) 2
(D) 4
(E) 6

$$\frac{2x^2 + 4x + 2}{(x + 1)^2} = \frac{2(x^2 + 2x + 1)}{(x + 1)^2} = \frac{2(x + 1)^2}{(x + 1)^2}$$ because $$(x + 1)^2 = x^2 + 2x + 1$$ = 2(Option C)
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Re: If x ≠ –1, then 2x^2 + 4x + 2/(x + 1)^2 =  [#permalink]

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10 Apr 2019, 18:02
Bunuel wrote:
If x ≠ –1, then $$\frac{2x^2 + 4x + 2}{(x + 1)^2} =$$

(A) 0
(B) 1
(C) 2
(D) 4
(E) 6

Simplifying, we have:

2(x^2 + 2x + 1)/(x^2 + 2x + 1) = 2

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Re: If x ≠ –1, then 2x^2 + 4x + 2/(x + 1)^2 =   [#permalink] 10 Apr 2019, 18:02
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