Prove that odd polynomials $f(x)$ of degree $\leq 10$ with $f(-1) = 0$ form a vector space. 
Let $P(X)$ be the usual vector space of polynomials in $x$ with real coefficients.
Let $U$ denote the subset of $P(X)$ consisting of those elements $f(x)$ which have degree less than or equal to $10$, satisfy $f(-x)=-f(x)$ for all $x$ and also satisfy $f(-1)=0$.
Prove $U$ is a subspace of $P(x)$.
Find a basis for $U$ and the dimension of $U$.

I have proved that $U$ is a subspace of $P(X)$ but am not sure how to find a basis and dimension of $U$

Polynomials in $U$ must also have a root at $1$ by symmetry so a superspace is $\langle x^3, x^5, x^7, x^9\rangle$. In fact a basis is given by
$$U = \langle x(x-1)(x+1), x^3(x-1)(x+1), x^5(x-1)(x+1), x^7(x-1)(x+1)\rangle$$
Since the given superspace is a subspace of $P[X]$, so is $U$. The above basis of $U$ can also be written as
$$U = (x-1)(x+1) \cdot \langle x,x^3,x^5,x^7 \rangle$$
That is each element of $U$ is an odd polynomial of degree $\le 7$ multiplied by $(x-1)(x+1) = x^2-1$.
