How do you factor $2 x^{3} - 4 x$ $2x^3 - 4x$ ?

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How do you factor $2 x^{3} - 4 x$ $2x^3 - 4x$ ?

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Answer 1 · 2016-04-21T06:52:18

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

Explanation:

The difference of squares identity can be written:

$a^{2} - b^{2} = (a - b) (a + b)$ $a^2-b^2 = (a-b)(a+b)$

We use this with $a = x$ $a=x$ and $b = \sqrt{2}$ $b=sqrt(2)$ later.

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First note that both terms are divisible by $2$ $2$ and by $x$ $x$ , so they are both divisible by $2 x$ $2x$ , so separate that out as a factor first...

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

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How do you factor $2 x^{3} - 4 x$ $2x^3 - 4x$ ?

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