The statement $f(x)$ is a power series with only coefficients one digit nonnegative integers means that $f\left(\frac{1}{10}\right)$ is equal to $.a_1a_2a_3a_4\ldots$ and the fact that this is rational means that for some $k$ and large enough $N$, $a_{k+n}=a_n$ for all $n>N$. We can separate the power series into the two polynomials $f(x)=g(x)+h(x)$ where $g(x)$ has degree $N$ and $h(x)$ is divisible by $x^{N+1}$. Then we know that $h(x)=a_{N+1}(x^{N+1}+x^{N+k+1}+x^{N+2k+1}+\ldots)+a_{N+2}(x^{N+2}+x^{N+k+2}+x^{N+2k+2}+\ldots)+a_{N+3}(x^{N+3}+x^{N+k+3}+x^{N+2k+3}+\ldots)+\ldots+a_{N+k}(x^{N+k}+x^{N+2k}+x^{N+3k}+\ldots)$ $h(x)\cdot(1-x^k)=a_{N+1}x^{N+1}+a_{N+2}x^{N+2}+a_{N+3}x^{N+3}+\ldots+a_{N+k}x^{N+k}:=r(x)$ Now we let $q(x)=1-x^k$ and $p(x)=g(x)q(x)+r(x)$. We know that $f(x)=g(x)+h(x)=g(x)+\frac{r(x)}{q(x)}=\frac{p(x)}{q(x)}$. We know that $r(x)$ has finite degree and so does $q(x)$, and thus so does $p(x)$. Thus these are the polynomials we're looking for.