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How do you solve $2 x^{3} - 8 x = 0$ $2x^3-8x=0$ by factoring?

1 Answer

Oct 6, 2015

$x = 0 X X X$ $x=0color(white)(XXX")$ or
$x = - 2 XX$ $x=-2color(white)("XX")$ or
$x = + 2$ $x=+2$

Explanation:

$2 x^{3} - 8 x = 0$ $2x^3-8x=0$

$\to XXX (2) (x) (x^{2} - 4) = 0$ $rarrcolor(white)("XXX")(2)(x)(x^2-4)=0$

$\to XXX (2) (x) (x + 2) (x - 2) = 0$ $rarrcolor(white)("XXX")(2)(x)(x+2)(x-2)=0$

$\to XXX x = 0 XX or XX x + 2 = 0 XXX or XX x - 2 = 0$ $rarrcolor(white)("XXX")x=0color(white)("XX")or color(white)("XX")x+2=0color(white)("XXX")orcolor(white)("XX")x-2=0$

$\to XXX x = 0 XX or XX x = - 2 XXXX or XX x = + 2$ $rarrcolor(white)("XXX")x=0color(white)("XX")orcolor(white)("XX")x=-2color(white)("XXXX")orcolor(white)("XX")x=+2$

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