Deconvolution and Polynomial division
In addition to the answer from Matt L., I want to emphasize the fact why the comment mentions the polynomial division. Consider to finite-length sequences $h[n]$ and $x[n]$, with length $N$ (one sequence can be zero-padded to length $N$). 

Now, look at the convolution of both:

$$y[n]=h[n]*x[n]=\sum_{n'=0}^{N-1}h[n']x[n-n']$$

As an example, consider the two sequences $h[n]=[1, 2, 3]$ and $x[n]=[4, 6, 5]$. Their convolution is (in Matlab)

&gt;&gt; conv([1 2 3], [4 6 5])

ans =

     4    14    29    28    15


Now, consider the product of the two polynomials $h(x)=1+2x+3x^2$ and $x(x)=4+6x+5x^2$ (in Mathematica):

Expand[(1 + 2 x + 3 x^2) (4 + 6 x + 5 x^2)]
&gt;&gt;&gt; 4+14 x+29 x^2+28 x^3+15 x^4


As you can see, the coefficients of the polynomial of the product are equal to the values of the convolution of both sequences. 

So, now to get $x[n]$ from $y[n]$, you need to perform the polynomial division of $y(x)/h(x)$:

PolynomialQuotient[4 + 14 x + 29 x^2 + 28 x^3 + 15 x^4, (1 + 2*x + 3*x^2), x]
&gt;&gt;&gt; 4+6 x+5 x^2


Hence, the coefficients of the polynomial yield the original sequence.

This technique can be extended to infinite sequences. 

To get the connection to the Z-Transform here: The Z-Transform of the finite sequences $x[n]$ and $h[n]$ are given by

$$H(z)=1+2z^{-1}+3z^{-2}\\ X(z)=4+6z^{-1}+5z^{-2}$$

And the Z-Transform of the output $y[n]$ is given by

$$Y(z)=H(z)X(z).$$

Here, you again have the polynomial multiplication. Also, here it becomes even easier to spot, why it also works with infinite sequences: Then both $H(z)$ and $X(z)$ become fractions of polynomials, but the principle of the polynomial multipication remains.