How to revert a carryless polynomial multiplication?

Question

In the carryless multiplication of two polynomials in a 8-bit environment, is it possible to obtain the original 8-bit values from the result?

As an example:

$$\begin{align} (x^{11} + x^{10} + x^9 ) + (x^7 + x^5 + x^4 + x^3 + x^1) &= (x^6 + x^5 + x^1) \times (x^5 + x^3 + x^2 + x^0)\\ (x^{11} + x^{10}) + (x^6 + x^4 + x^2 + x^1) &= \text?\times\text?\quad \hbox{(each value must fit in 8 bits)} \end{align}$$

I know that I can try every possible value (it's only two 8-bit values) but that seems blunt!