$(α|0⟩ + β|1⟩)|0⟩$ in matrix/vector form I am currently working through superdense coding with bell states and have a question regarding this value:
$$(α|0⟩ \ + \ β|1⟩) \ |0⟩$$
I understand that $α|0⟩ \  + \  β|1⟩$ can be represented in vector format as $(α \ \  β)$ (vertically)
However, I'm not sure how to apply the outer $|0⟩$ ket? Would it be $(α \ \ β)×(1 \ \  0)$ which would give $α$?
 A: This operation is what's called the tensor product. More formally it is denoted by
$$|\psi\rangle\otimes|\phi\rangle$$
but often the $\otimes$ is left out. When calculated numerically it is called the Kronecker product which can be calculated as follows for a 2 state system
$$\pmatrix{a\\b}\otimes\pmatrix{c\\d}=\pmatrix{ac\\ad\\bc\\bd}.$$
See also the wiki page on the Kronecker product for a more general computation. To get a little more intuition you can calculate all the combinations of $\{|0\rangle,|1\rangle\}\otimes\{|0\rangle,|1\rangle\}$ where $|0\rangle=\pmatrix{1\\0}$ and $|1\rangle=\pmatrix{0\\1}$. When you do this you will see that the result is a unit vector with the $1$ at all 4 different positions: the result gives all basis vectors of a 4 state system.
Edit: It is also distrubitive so $$\big(|\alpha\rangle+|\beta\rangle\big)\otimes|\gamma\rangle=|\alpha\rangle\otimes|\gamma\rangle+|\beta\rangle\otimes|\gamma\rangle\\
|\alpha\rangle\otimes\big(|\beta\rangle)+|\gamma\rangle\big)=|\alpha\rangle\otimes|\beta\rangle+|\alpha\rangle\otimes|\gamma\rangle$$
and this should allow you to calculate the tensor product product in question.
