Dimension of $W=\{p(x) : p(x)=p(1-x)\}$. 
Find the dimension of the subspace $W$ of $P_n(x)$ , space of all polynomials ; where $\displaystyle W=\{p(x) : p(x)=p(1-x)\}$.

I just found that the polynomials satisfies the condition $p(x)=p(1-x)$ are of the types $p(x)=x^n(1-x)^n$ for every positive integer $x$. But I don't know whether there are more than this type of polynomials or not and how I can find the dimension.
 A: $W$ consists of polynomias that are symmetric w.r.t the axis $x=\frac{1}{2}$. Thus after a simple translation, you ask for the dimension of even polynomials. The dimension is $\frac{n}{2}+1$ (Basis: $1,x^2, \dotsc, x^n$) or $\frac{n+1}{2}$ (Basis: $1,x^2, \dotsc, x^{n-1}$), depending on the parity of $n$.
A: Putting the elegant argument from
 @MooS in effect you can also use a pure linear algebra approach.
Looking at the linear map $A:p\mapsto Ap$, where $(Ap)(x) = p(1-x)$ we can prove that $A$ can be represented as a diagonal matrix. For that choose a following basis of $P_n[x]$:
$$1,\; \left(x-\frac{1}{2}\right),\;\left(x-\frac{1}{2}\right)^2,\;\left(x-\frac{1}{2}\right)^3,\;\left(x-\frac{1}{2}\right)^4,\cdots$$ 
Then $A$ has the following form:
\begin{eqnarray}
 A & = & \begin{bmatrix}1&0&0&0&\cdots\\0&-1&0&0&\cdots\\0&0&1&0&\cdots \\0&0&0&-1&\cdots\\\vdots&\vdots&\vdots&\vdots&\ddots&\end{bmatrix}
\end{eqnarray}
We know that $p(x)=p(1-x) \Leftrightarrow p = Ap \Leftrightarrow (A-id)p=0 \Leftrightarrow p\in \ker (A-id)$. Since the matrix representation of $A-id$ is given by
\begin{eqnarray}
 A-id & = & \begin{bmatrix}0&0&0&0&\cdots\\0&-2&0&0&\cdots\\0&0&0&0&\cdots \\0&0&0&-2&\cdots\\\vdots&\vdots&\vdots&\vdots&\ddots&\end{bmatrix}
\end{eqnarray}
the kernel of $A-id$ is given by $\langle 1,\left(x-\frac{1}{2}\right)^2, \left(x-\frac{1}{2}\right)^4, \cdots\rangle$. The dimension now follows directly dependent of $n$ being even or odd.
