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# If x^3 - 4x^2 + 5x -5 = 0, then x(x - 2)^2/(5 - x) =

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Math Expert
Joined: 02 Sep 2009
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Kudos [?]: 139663 [0], given: 12794

If x^3 - 4x^2 + 5x -5 = 0, then x(x - 2)^2/(5 - x) = [#permalink]

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10 Sep 2017, 04:37
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35% (medium)

Question Stats:

83% (01:40) correct 17% (02:50) wrong based on 46 sessions

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If $$x^3 - 4x^2 + 5x -5 = 0$$, then $$\frac{x(x - 2)^2}{(5 - x)} =$$

A. -0.5
B. 0.5
C. 1
D. 1.5
E. 2
[Reveal] Spoiler: OA

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Kudos [?]: 139663 [0], given: 12794

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Re: If x^3 - 4x^2 + 5x -5 = 0, then x(x - 2)^2/(5 - x) = [#permalink]

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10 Sep 2017, 05:37
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The expression $$\frac{x(x - 2)^2}{(5 - x)}$$ can be re-written as $$\frac{x(x^2 - 4x + 4)}{(5 - x)}$$ = $$\frac{(x^3 - 4x^2 + 4x)}{(5 - x)}$$

It has been given $$x^3 - 4x^2 + 5x -5 = 0$$ which can re-written as $$x^3 - 4x^2 + 4x + x - 5 = 0$$
Therefore, $$x^3 - 4x^2 + 4x = 5 - x$$

Substituting this value in the simplified expression, we will get the value as $$\frac{(x^3 - 4x^2 + 4x)}{(5 - x)} = \frac{(5 - x)}{(5 - x)} = 1$$(Option C)
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Re: If x^3 - 4x^2 + 5x -5 = 0, then x(x - 2)^2/(5 - x) =   [#permalink] 10 Sep 2017, 05:37
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# If x^3 - 4x^2 + 5x -5 = 0, then x(x - 2)^2/(5 - x) =

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