Question

How do you solve the inequality `x^3-x^2-6x>0`?

Answer 1

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

`x^3 - x^2 - 6x > 0`
`rarr` `(x) (x-3) (x+2) > 0`

Note that `x != 0` since the left side must be `> 0`

If `x >0`
then `(x-3) (x+2) > 0`
`rarr` `x > 3`

if `x < 0`
then `(x-3)` will be negative
`rarr` `(x+2)` must be `>0`
(so the product `(x) (x-3)_neg (x+2)` will be `> 0`
i.e (neg) `xx` (neg) `xx` (pos) )
`rarr` # (-2) < x < 0

Therefore
`x^3 - x^2 - 6x > 0`
for `x > 3` or `(-2) < x < 0`