How do you solve the inequality $x^{3} - x^{2} - 6 x > 0$ $x^3-x^2-6x>0$ ?

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How do you solve the inequality $x^{3} - x^{2} - 6 x > 0$ $x^3-x^2-6x>0$ ?

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Answer 1 · 2015-02-15T01:57:33

Factor the expression $x^{3} - x^{2} - 6 x$ $x^3 - x^2 - 6x$ on the left side of the inequality and then evaluate for each term:

$x^{3} - x^{2} - 6 x > 0$ $x^3 - x^2 - 6x > 0$
$\to$ $rarr$ $(x) (x - 3) (x + 2) > 0$ $(x) (x-3) (x+2) > 0$

Note that $x \neq 0$ $x != 0$ since the left side must be $> 0$ $> 0$

If $x > 0$ $x >0$
then $(x - 3) (x + 2) > 0$ $(x-3) (x+2) > 0$
$\to$ $rarr$ $x > 3$ $x > 3$

if $x < 0$ $x < 0$
then $(x - 3)$ $(x-3)$ will be negative
$\to$ $rarr$ $(x + 2)$ $(x+2)$ must be $> 0$ $>0$
(so the product $(x) {(x - 3)}_{\neg} (x + 2)$ $(x) (x-3)_neg (x+2)$ will be $> 0$ $> 0$
i.e (neg) $\times$ $xx$ (neg) $\times$ $xx$ (pos) )
$\to$ $rarr$ # (-2) < x < 0

Therefore
$x^{3} - x^{2} - 6 x > 0$ $x^3 - x^2 - 6x > 0$
for $x > 3$ $x > 3$ or $(- 2) < x < 0$ $(-2) < x < 0$

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