Pre calc help x^3-4x>0 ?

Question

i'm sure its one of the easy ones i just don't remember.

Answer 1

Factor out an x: x(x^2 - 4) > 0

Reverse FOIL: x(x+2)(x-2) > 0

For this to happen either two of the three terms are negative with one positive [which happens if x < 0 but >  -2], or all three terms are positive [which happens if x >2].

So -2 < x < 0 or x > 2

Answer 2

To differentiate:

3x^2 - 4 > 0

OR

To integrate:

0.25 x^4 - 2x^2 > 0

OR

To solve:

x(x^2 - 4) > 0

x(x+2)(x-2) > 0

x = 0, -2, 2

Answer 3

In two steps. First note that x=0 is not a solution.

Consider the case x>0.

Divide by x and obtain x^2-4>0 or x^2>4 which yields x>2 (remember that x>0).

Consider now the case x<0.

Divide by x and obtain x^2-4<0 or x^2<4 which yields x>-2. Since x<0, the answer here is 0>x>-2

Thus the answer is x in (2,+ infinity) union (-2,0)