Plz help Inequality?

Let f(x) = x³ - x² - x - 2

f(2) = 2³ - 2² - 2 - 2 = 0
Hence, (x - 2) is a factor of x³ - x² - x - 2

x³ - x² - x - 2 > 0
x³ - 2x² + x² - 2x + x - 2 > 0
(x³ - 2x²) + (x² - 2x) + (x - 2) > 0
x²(x - 2) + x(x - 2) + (x - 2) > 0
(x - 2){x² + x + 1} > 0
(x - 2){[x² + 2x(1/2) + (1/2)²] - (1/2)² + 1] > 0
(x - 2){[x + (1/2)]² + (3/4)]} > 0

For all real x, [x + (1/2)]² ≥ 0 and thus {[x + (1/2)]² + (3/4)]} ≥ 3/4
Hence, x - 2 > 0
x > 2

Ok so here we go :)
We need to find the x - value (s) for which the y - values of the function
are > 0.
This occurs when x > 2.
Hope this helps !!!!!!!!!!!!!!!!!!!

Follow up :
You could try long division to get the other factor :

x^2 + x + 1
x - 2 √ (x^3 - x^2 - x - 1)
- (x^3 - 2x^2)
----------------------------
x^2 - x
- (x^2 - 2x)
------------------------------
x - 2
- (x - 2)
------------------
0

I hope this will help :)

x^3-x^2-x-2>0=>
(x-2)(x^2+x+1)>0=>
x>2
or x is in [2, inf.)