Quick method
The left hand side is an upright quartic (positive coefficient of x^4) which intersects the x axis at the four zeros: -5, -1, 1, 2.
So it will be negative precisely in the intervals (-5, -1) and (1, 2).
Here's a graph of y = 1/20 (x+5)(x-2)(x-1)(x+1) ...
graph{(x+5)(x+1)(x-1)(x-2)/20 [-10, 10, -5.5, 4.5]}
color(white)()
A little slower...
Let f(x) = (x+5)(x-2)(x-1)(x+1)
f(x) is a continuous function which is zero when x in { -5, -1, 1, 2 }, so does not satisfy the inequality f(x) < 0 at those points.
These points divide the Real line into 5 intervals in which the sign of f(x) cannot change:
(-oo, -5), (-5, -1), (-1, 1), (1, 2), (2, oo)
Since the zeros are each of multiplicity 1 - which is an odd number - the sign does change at each of these points.
When x < -5, the sign of each of the linear factors is negative. When they are all multiplied, the result is positive.
That is, when x < -5:
{ ((x+5) < 0), ((x-2) < 0), ((x-1) < 0), ((x+1) < 0) :}
Hence (x+5)(x-2)(x-1)(x+1) > 0
So f(x) has the following signs in each of the 5 intervals:
+, -, +, -, +
respectively.
So f(x) < 0 in (-5, -1) uu (1, 2)
