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. 
 A: .Your method can be improved considerably.
$$
t^9-1 = (t^3)^3 - 1^3 = (t^3-1)(t^6 + t^3+1) = (t-1)(t^2 + t + 1)(t^6 + t^3 + 1)
$$
At this stage, we can see that one of the roots is $t = 1$. Two others are obtained by solving for the quadratic $t^2+t+1  = 0$, easily done via the usual formula. 
EDIT : The last term is $t^6 + t^3 + 1 = 0$. In general, there is no standard formula (a difficult result) for solving polynomial equations of degree $5$ and above. In the  light of that result, we would  not be expected to solve the above equation analytically. However,there is a simplification, that helps us do this.
Put $y = t^3$. If you notice, the equation then simplifies to $y^2+y+1 = 0$. Now, how would we solve for $t$?
Well, we will obviously first solve for $y$. Then, once we know what values $y$ will take, we can then find out what values $t$ can take, since $t^3 = y$, so we are just asking for the cube roots of $y$.
We already know what values $y$ will take, since it's just the same quadratic equation which we solved earlier, where the variable was $t$. Now, all we need to do is take the complex cube roots of all the possible values of $y$. That will give you a complete description of all the roots of the polynomial.
More specifically, let us denote the roots of $x^2+x+1$ by $\alpha$ and $\beta$(You can find what these are using the quadratic formula, but I am simplfying notation). 
Then, by what I said earlier, solving $t^3 = \alpha$ and $t^3 = \beta$ should give us all the remaining roots of the polynomial. But how do we solve these?
Well, note that $\alpha$ and $\beta$ are complex( and not real) numbers, so the solutions of $t^3 = \alpha$ and $t^3 = \beta$ are also complex. The solutions to these equations are specified using the "cube roots of unity". These are the solutions to the equation $t^3 = 1$, which are exactly $1$ and $\frac{1 \pm \sqrt{-3}}{2}$. Usually, we denote $\omega =\frac{1 + \sqrt{-3}}{2}$, and then it turns out that the cube roots of unity are just $1 ,\omega,\omega^2$. 

In similar fashion, the cube roots of $\alpha$ are given by $\sqrt[3]|\alpha|, \sqrt[3]|\alpha| \omega,\sqrt[3]|\alpha| \omega^2$, and similarly for $\beta$, where $|\alpha|$ is the absolute value of the complex number $\alpha$.

Now , we can say that the roots are precisely $1,\alpha,\beta, \sqrt[3]{|\alpha|}, \sqrt[3]{|\alpha|}\omega,\sqrt[3]{|\alpha|}{\omega^2}, \sqrt[3]{|\beta|},\sqrt[3]{|\beta|}\omega,\sqrt[3]{|\beta|}\omega^2$.

In truth, a complete description of all roots of this polynomial is afforded by the use of De Moivre's theorem. This gives the exact description $\{e^\frac{2\pi i k}{9} : 1 \leq k \leq 9\}$  for the roots.
De Moivre's theorem can be used to solve all equations of the form $t^n = 1$ for any  $n \in \mathbb N$, so it has far more diverse applications than the one mentioned here. So even though you will find $x^7=1$ hard to solve(try it!) De Moivre at least gives you an answer in polar coordinates to the same.
A: 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 $\theta$.
A: Note that if $k,n\in\Bbb Z$ then $e^{k\pi i/n}$ is a root of $X^n-1$.
A: Yes, utilizing the exponential form of complex numbers here would be a useful strategy for making progress.
If $t=\rho e^{i\varphi}=\rho(\cos\varphi + i\sin\varphi)$, then, by De Moivre's theorem: $t^n=\rho^ne^{in\varphi}=\rho^n(\cos n\varphi + i\sin n\varphi)$. Now, use $n=9$ and make it equal to $1$:
$$\rho^9(\cos 9\varphi + i\sin 9\varphi)=1=1+0i$$
which gives you $\rho=1$ and $\cos 9\varphi=1, \sin 9\varphi = 0$. It follows that $\varphi$ is of the form $\frac{2k\pi}{9}$ for some $k\in\Bbb Z$, and you can limit yourself to $k=0,1,2,\ldots,8$ because, after that, the values of $t$ get repeated.
Thus, the solutions of $t^9-1=0$ are $t_k=\cos\frac{2k\pi}{9}+i\sin\frac{2k\pi}{9}$ ($k=0,1,\ldots\,8$). There are nine of them, and they are all different, and in the complex plane they all lie on the unit circle at the angles $0, 40^\circ, 80^\circ, \ldots, 320^\circ$ with respect to the positive real axis. The solution $t=1$ is the first of those ($k=0$).
A: The polynomial can be factorized as follow
$$r(t) = t^9 − 1 =\prod_{k=0}^8(t-\omega_k)$$
where
$$\omega_k=e^{\frac{2k\pi i}{9}} \quad k=0,1,...,8$$
are the roots of unity.
