I'm not sure if I'm interpreting the question correctly but I am reading it as follows.
Suppose that $D:\mathbb{R}\rightarrow\mathbb{R}$ is a polynomial with real coefficients. Does there exist a polynomial $H:\mathbb{R}\rightarrow\mathbb{R}$ such that $D(x)=H(x)+H(x-1)$.
Let $D(x)=\sum_{i=0}^na_ix^i$ ($a_n\neq0$). It's pretty clear that we need $H(x)$ to be of degree $n$ also. Let $H(x)=\sum_{i=0}^nb_ix^i$. Now
$H(x)+H(x-1)=\sum_{i=0}^n(b_ix^i+b_i(x-1)^i)$ $=\sum_{i=0}^n\left(b_ix^i+b_i\left(\sum_{j=0}^i{i\choose j}x^{i-j}(-1)^j\right)\right)$
Now we can ask the question what is the coefficient of $x^m$? Now the indices $i$, $j$ satisfy $0\leq j\leq i\leq n$. Hence for the $x^{i-j}$, we need $i-j=m$. This can only happen for $(i,j)\in{(m,0),(m+1,1),\dots,(m+(n-m),n-m)}$. Introduce the index $l=0,1,\dots,n-m$. Using this we can write the coefficient of $x^m$ and hence we solve the system of equations
$a_m=b_m+\sum_{l=0}^{m-n}b_{m+l}{m+l\choose l}(-1)^{m+1}$ for $m=0,1,\dots,n$.
These are equations of the form
$a_m=\sum_{k=m}^n\alpha^m_{k}b_k$. $(*)_m$
Hence the equation $a_n=\alpha^m_nb_n$ determines $b_n$ and we can solve sequentially for $b_{n-1},b_{n-2},\dots,b_0$ so that when we solve the $(*)_m$ the $b_k$ for $k>m$ have already been found. This implies that $H$ is in fact unique.
I hope this is what you were looking for and sorry it's a bit messy --- it's 2 a.m. in Ireland!