How to find the roots for $x^3+2x+2$ I seem to be forgetting my college algebra. Can someone help me understand how to find the complex roots of the polynomial: $x^3+2x+2=0$
I tried synthetic devision for the possible real roots of $\pm1,\pm2$ but, alas, it doesn't reduce to a lower degree. 
Apparently, the roots are: $x\approx -0.8,0.4\pm1.6i$
What are ways I can find the complex roots of this equation? 
 A: Let $x=u+v$.
Hence, $$u^3+v^3+3uv(u+v)+2(u+v)+2=0.$$
Now, let $u^3+v^3+2=0$.
Thus, since $u+v\neq0$, we obtain $$3uv+2=0$$ or $$u^3v^3=-\frac{8}{27},$$
which says $u^3$ and $v^3$ they are roots of the following equation
$$t^2+2t-\frac{8}{27}=0.$$
$t_1=-1+\sqrt{\frac{35}{27}}$, $t_2=-1-\sqrt{\frac{35}{27}}$, which gives an unique real root
$$\sqrt[3]{-1+\sqrt{\frac{35}{27}}}+\sqrt[3]{-1-\sqrt{\frac{35}{27}}}.$$
By the Viete's theorem we can get two complex roots:
$$x_2=-\frac{u+v}{2}+\frac{(u+v)\sqrt3}{2}i$$ and
$$x_3=-\frac{u+v}{2}-\frac{(u+v)\sqrt3}{2}i.$$
A: If you got the real root $a$ by numerical means or other, you can split off the linear factor as you tried, getting
$$
f(x)=f(x)-f(a)=(x-a)((x^2+ax+a^2)+2).
$$
Solving the quadratic equation for the second factor gives the roots
$$
x=-\frac a2\pm i\sqrt{\frac{3a^2}4+2}
$$
From general root bounds we know that the real root is inside the interval $[3,3]$. Using a bracketing method can look like this in Python (or use the polynomial roots function of the scipy module, but what is the fun in that)
def regula_illinois(f,a,b,eps):
    fa, fb = f(a), f(b)
    while abs(b-a) > eps: 
        c=a-fa*(b-a)/(fb-fa); fc = f(c);

        if (fc<0)!=(fa<0) :
            b, fb = a, fa;
        else:
            fb *= 0.5;
        a, fa = c, fc;
        if abs(fc)<eps : break;
    return a


a = regula_illinois(lambda x : x**3+2*x+2, -3, 3, 1e-17);
print "real root : %.15f"%a
print "complex roots : %.15f +- i*%.15f"%(-a/2, (3*a*a+8)**0.5/2)

with the output
real root : -0.770916997059248
complex roots : 0.385458498529624 +- i*1.563884510526956

