Question

Find all the roots of the equation 3x³-x²-8x-2=0

Answer 1

$3x^3-x^2-8x-2=0$

Trying with rational factors of 2/3 - does not give us a root.  This equation has irrational roots.
We reduce the equation by the following method.

Let a = 3,  b = -1,   c = -8   ,   d = -2

Substitute y = x + b/3a       ie., y = x -1/9

You will get 
$y^3 - \frac{73}{27}y-\frac{704}{729} = 0,\ \ \ or, \ \ \ y^3-\frac{73}{3^3}y-\frac{11*2^6}{3^6} = 0 \\ \\ let\ \ p = -\frac{73}{3*27}=-0.90123,\ \ \ \ \ \ q=704/729 = 0.9657 \\ \\Solution\ for\ y^3-3p\ y-q= 0,\ \ is\ given\ by\\ \\ y_k, \ for\ k=0,1,2,\ \ is = 2\sqrt{p}\ *\ Cos[ \frac{1}{3}*\ Cos^{-1}(\frac{q}{2p^{\frac{3}{2}}} ) - k\ *\ \frac{2 \pi }{3} ] \\ \\y_0 = 1.800005105\\y_1=-0.37699\\y_2=-1.42305\\ \\ x = y +\frac{1}{9}\\ \\ x=1.911164,\ \ \ -0.26588,\ \ \ -1.31194$

We follow the above when p is positive in reduced cubic equation.  If p is negative in the reduced cubic equation then we do the following.  Otherwise, we have work with complex roots. It becomes difficult to evaluate roots.

$y^3+py-q=0\\ \\Substitute\ y=s-\frac{p}{s},\ \ to\ get\ \ s^6-q\ s^3-p^3=0\\ \\ s^3=\frac{q}{2}+-\frac{1}{2}\sqrt{q^2-4*p^3}\\ \\t^3=-\frac{q}{2}+-\frac{1}{2}\sqrt{q^2-4*p^3}\\ \\Find\ s\ and\ t\ from\ s^3\ and\ t^3.\\ \\ y = s - t\\ find \\ x = y - b/3a$