How to find all real and complex solutions to the polynomial $r(t) = t^9 − 1$ Thanks for the detailed answers everyone. As i understand it, this class of problem is standard and can be solved with De Moivre's theorem, even if I haven't informed myself about that yet. 
Otherwise, I would just state that my knowledge of mathematics is very low, which is unfortunately making it difficult for me to understand some of the answers so far.

I have been assigned several tasks in which i need to find the real and complex solutions of higher (than order 2) degree polynomials. 
the final question has the rather large (in my experience) polynomial.
$r(t) = t^9 − 1$
up until now my protocol has been to:
I) Substitute factors of the coefficient as potential solutions
II) Once a solution is found, Perform polynomial division until the the solution has been factored out.
III) repeat I) and II) until the polynomial is of degree 2 and then try and either factorize the expression by inspection or by the quadratic formula. 
In the following polynomial this strategy doesn't get me very far. Can anyone give me some tips on how to find the remainder of the solutions to $r(t) = t^9 − 1$ ?

Progress so far:
sub factors of coeffcient $\rightarrow t = 1$ is a solution.
$\frac{t^9 - 1}{t - 1}$ = $(t -1)(t^8+t^7+t^6+t^5+t^4+t^3+t^2+t+1)$
Here it might be worth rewriting the expression inside the second bracket as the equation
$t^8+t^7+t^6+t^5+t^4+t^3+t^2+t+1 = 0$
so that there is a series of terms raised to an exponent and a $-1$ term. 
$t^8+t^7+t^6+t^5+t^4+t^3+t^2+t = -1$
I'm not sure if utilizing the exponential form of complex numbers here would be a useful strategy for making progress. 

Hint Write $t$ in exponential form $t= R e^{i \theta}$. Then your equation becomes 
$$R^9 e^{9 i \theta}=1$$
Taking the modulus on both sides gives you $R$, and then you can figure out $\theta$.
