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 $U$
 A: $U$ is the intersection of the subset of polynomials of degree $\leq 10$, the subset of odd  polynomials and the subset  of polynomials that have $-1$ as root. It's straightforward to check each of these subsets is a subspace, hence their intersection is a subspace.
As a polynomial $p\in U$ is odd and $p(-1)=0$, necessarily $p(1)=0$. Being odd, $0$ is also one of its roots. So a polynomial $p(x)\in U$ is divisible by $x, x-1$ and $x+1$, hence is  divisible by $x(x^2-1)$ and we can write
$$p(x)=x(x^2-1)q(x)$$
where $q(x)$ must have degree $\leq 7$ and be even; This means $q(x)=ax^6+bx^4+cx^2+d$, for some $a, b, c ,d\in \mathbf R$. 
This proves $U$ has dimension $4$. A basis is the set of polynomials:
$$\bigl\{x(x^2-1), x^3(x^2-1), x^5(x^2-1), x^7(x^2-1)\bigr\}$$
A: $f(-x)=-f(x)$ means that the subspace $U$ consists solely of odd functions. The subspace $V$ of odd functions in $P(X)$ with degree less than or equal to 10 has basis $x,x^{3},x^{5},x^{7},x^{9}$. Indeed, all these functions are linearly independent, odd and have degree less than 10. Moreover, they span $V$: If $g \in P(X)$ has degree less than or equal to $10$, then $g(x)=a_{10}x^{10}+a_{9}x^{9}+...+a_{1}x+a_{0}$ for some $a_{i}\in \mathbb{R}, 0 \leq i \leq10$. Now $g(-x)=-g(x)$ says $a_{i}=-a_{i}$ if $i$ is even, so these must be $0$. Then $g$ is an $\mathbb{R}$-linear combination of $x,x^{3},x^{5},x^{7},x^{9}$. So $dim(V)=5$.
Notice that $x^3-x, x^5-x, x^7-x, x^9-x$ all lie in $V$, are linearly independent and are all attain the value $0$ at $x=1$ (hence at $-1$ as they are all odd functions). So $dim(U)\geq 4$. But the function $x$ is odd and doesn't equal $0$ at $x=1$, so $U$ can't be all of $V$. So $dim(U)=4$, with a basis given by $x^3-x, x^5-x, x^7-x, x^9-x$. $\square$
A: 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 $(x-1)(x+1) = x^2-1$.
