We know the shape of the equation of this cubic graph. If we substitute -20 and + 20 in the equation, we get a large negative value and a large positive value, so the zeros of this cubic must be between these two points.
P(-20) = -16798, P(20) = + 14402
So if we try the midpoint we get P(0) = 2. So there must be some zeros around x = 2.
P(5) = 77 and P(-5) = -223, so the zeros must be between these two points.
P(.1) = -.028 so there is a zero between x = .1 and x = 0.
Continuing to experiment like this we can locate the approximate zeros at
x = .099, x = 3.96, and x = -2.56