How do you solve $3 x^{3} - 48 x = 0$ $3x^3-48x=0$ ?

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Answer 1 · 2016-03-14T12:57:13

$x = \pm 4$ $x=±4$

$3 x^{3} - 48 x = 0$ $3x^3-48x=0$
$3cancel(x)^3=48cancel(x)$
$cancel(3)x^2=cancel(48)$
$x^2=16$
$sqrt(x^2)=sqrt16$
$x=±4$

How do you solve 3x^3-48x=03x3−48x=03x^3-48x=0?