Question #2019c

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Question #2019c

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Answer 1 · 2017-03-03T15:11:27

$x (x + 2) (x - 2) = 0$ $x(x+2)(x-2)=0$

so $x = - 2, 0, 2$ $x=-2,0,2$

Explanation:

You can take out a factor of $x$ $x$ because everything has at least one $x$ $x$ , so

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

Now you have a squared $x$ $x$ minus a square number, which is always in the form $(x + b) (x - b)$ $(x+b)(x-b)$ where $b$ $b$ is the root of the square number.

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

So you can say the whole thing is

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

which gives the solutions of $x = 0, - 2, + 2$ $x=0,-2, +2$

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Question #2019c

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