The factorization of a qbit product state is not unique; how should the different factorizations be interpreted? Consider two arbitrary qbits; their product state is as follows:
$ \begin{bmatrix} a \\ b \end{bmatrix} ⊗ \begin{bmatrix} c \\ d \end{bmatrix} = \begin{bmatrix} ac \\ ad \\ bc \\ bd \end{bmatrix} $
We can factor this product state into four different tensor products of two qbits:
$ \begin{bmatrix} ac \\ ad \\ bc \\ bd \end{bmatrix} =
\begin{bmatrix} a \\ b \end{bmatrix} ⊗ \begin{bmatrix} c \\ d \end{bmatrix} $,
$ \begin{bmatrix} ac \\ ad \\ bc \\ bd \end{bmatrix} =
\begin{bmatrix} -a \\ -b \end{bmatrix} ⊗ \begin{bmatrix} -c \\ -d \end{bmatrix} $,
$ \begin{bmatrix} ac \\ ad \\ bc \\ bd \end{bmatrix} =
\begin{bmatrix} ia \\ ib \end{bmatrix} ⊗ \begin{bmatrix} -ic \\ -id \end{bmatrix} $,
$ \begin{bmatrix} ac \\ ad \\ bc \\ bd \end{bmatrix} =
\begin{bmatrix} -ia \\ -ib \end{bmatrix} ⊗ \begin{bmatrix} ic \\ id \end{bmatrix} $
What is the relationship between these four factorizations? Do they all represent the same state?
 A: There aren't just four different tensor products; there are an uncountably infinite number of tensor products that give the same state, because
$$\begin{bmatrix}e^{i\phi}a\\e^{i\phi}b\end{bmatrix}\otimes\begin{bmatrix}e^{-i\phi}c\\e^{-i\phi}d\end{bmatrix}=\begin{bmatrix}ac\\ad\\bc\\bd\\\end{bmatrix}$$
for any real number $\phi$. (In fact, there are even more, if you don't require the two input vectors to be normalized; multiplying one vector by some arbitrary nonzero complex number $c$ and the other by $\frac{c^*}{|c|^2}$ will also give you the same result.)
This isn't some sort of special property; it's just due to the bilinearity of the tensor product, which states that for vectors $A$ and $B$ and complex scalars $c$ and $d$,
$$(cA)\otimes (dB)=cd(A\otimes B)$$
In this case, we take $d=c^{-1}$. Bilinearity is a property that also appears in a lot of other operators, like the dot product:
$$(a\vec{v})\cdot (b\vec{w})=ab(\vec{v}\cdot\vec{w})$$
And the cross product:
$$(a\vec{v})\times(b\vec{w})=ab(\vec{v}\times\vec{w})$$
It's also trivially shared with ordinary scalar multiplication, though in that case, bilinearity is equivalent to commutativity.
So, not only is there nothing special about the four examples you provided, there is also nothing special about any particular decomposition of a vector into a tensor product of two other vectors.
