A quick graph of this function makes it look like the roots are probably messy. For these cases, I find that using Newton's method of iterating the root works nicely.
The derivative of the function
F(x) = x^3 - 4x + 2
F'(x) = 3x^2 - 4
So the iteration function becomes
xnew = x - F(x)/F'(x) = (x*F'(x) - F(x))/F'(x)
xnew = (3x^3 - 4x - x^3 + 4x - 2)/(3x^2 - 4)
xnew = 2(x^3 - 1)/(3x^2 - 4)
If we start with x=-2.2, after 3 iterations we get x = -2.21432 If we start with x=0.5, after 3 iterations we get x = 0.5391889 If we start with x=1.7, after 3 iterations we get x = 1.6751309
The derivative of a*x^b is a*b*x^(b-1). The derivative of a sum is the sum of the derivatives.