How do you convert  -2 - 2sqrt3i to polar form?

You can simplify this as $- 2 \left(1 + \sqrt{3} i\right)$. This is not needed, but I like to work with easier terms. Next, you remember that $z = r \left(\cos \left(\theta\right) + i \sin \left(\theta\right)\right)$. $r$ is called the modulus of $z$ and it is defined as $\sqrt{{x}^{2} + {y}^{2}} = | z |$. Yes, that's the same symbol for the absolute value, but in complex numbers, it defines the modulus. (This has to do with metric spaces)
Hence, $r = 2$
Now, to find $\theta$ which is also called the argument of z, you use ${\tan}^{-} 1 \left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3}$
Now, not forgetting the $- 2$ at the beginning, we get:
$- 2 \left(1 + \sqrt{3} i\right) = - 4 \left(\cos \left(\frac{\pi}{3}\right) + i \sin \left(\frac{\pi}{3}\right)\right)$