By the rational root theorem, any rational roots of F(x) = 0 must be of the form p/q in lowest terms, where p, q in ZZ, q != 0, p a divisor of the constant term 15 and q a divisor of the coefficient 8 of the term 8x^3 of highest degree.
That means that the only possible rational roots are:
+-15, +-15/2, +-5, +-15/4, +-3, +-5/2, +-15/8, +-3/2, +-5/4, +-1, +-3/4, +-5/8, +-1/2, +-3/8, +-1/4, +-1/8
F'(x) = 24x^2-6x+5 > 0 for all x in RR
So F(x) = 0 has exactly one root.
F(-1) = -8-3-5+15 = -1
F(-3/4) = -27/8-27/16-15/4+15
= (-54-27-60+240)/16 = 99/16
So the root lies strictly between -1 and -3/4
None of the possible rational roots lies in this range, so F(x) = 0 has no rational roots.
graph{8x^3-3x^2+5x+15 [-1.688, 0.812, -0.62, 0.63]}
If you still want to solve the original problem, first let t = x - 1/8. This is called a Tschirnhaus transformation, giving us:
F(x) = 8(t^3+37/64t+499/256)
Next solve t^3+37/64t+499/256 = 0 using Cardano's method
Let t = u + v
Then 0 = t^3+37/64t+499/256
=(u+v)^3+37/64(u+v)+499/256
=u^3+v^3+(3uv+37/64)(u+v)+499/256
Now let v = -37/(192u) so that 3uv+37/64 = 0
Then multiplying our equation through by u^3 we get:
(u^3)^2+499/256(u^3)-37^3/192^3 = 0
Since this derivation is symmetric in u and v we find:
t = root(3)((-499/256+sqrt((499/256)^2+4*37^3/192^3))/2) + root(3)((-499/256-sqrt((499/256)^2+4*37^3/192^3))/2)
and
x = 1/8 + root(3)((-499/256+sqrt((499/256)^2+4*37^3/192^3))/2) + root(3)((-499/256-sqrt((499/256)^2+4*37^3/192^3))/2)
So it can be done, but it's a little complex.
