We need to find the solution for x^3 + 5x^2 -x - 5 < 0.

x^3 + 5x^2 -x - 5 < 0

=> x^2( x + 5) - 1(x + 5) < 0

=> (x^2 - 1)(x + 5) < 0

=> (x - 1)(x + 1)(x + 5) < 0

For the expression on the left to be less than 0, either all three factors have to be less than 0 or only one of them is less than 0.

If all three are less than 0,

=> x - 1 < 0 , x + 1 < 0 and x + 5 < 0

=> x < 1 , x < -1 and x < -5

x < -5 satisfies all the three conditions. Or x lies in (-inf , -5).

If only one of them is less than 0,

x - 1 < 0 , x + 1 > 0 and x + 5 > 0

=> x < 1 , x > -1 and x > -5

This gives the values of x in the set ( -1 , 1)

x - 1 > 0 , x + 1 < 0 and x + 5 > 0

=> x > 1 , x < -1 and x > -5 , no valid solution

x - 1 > 0 , x + 1 > 0 and x + 5 < 0

=> x > 1 , x > -1 and x < -5, no valid solution

**The required values lie in x = (-inf.,-5) U (-1, 1).**