Find a basis for the space of polynomials of degree $3$ such that $p(1)=p(-1)=0$ What I have is V=$\Re_3[x]$ W is a subspace of V.
$$ W= \{p \in V | p(-1)=p(1)=0\} $$
So I need to find a number of vectors that are linearly independent and span $p$.
I have only been able to come up with one vector that could be in the basis for $p$:
$b_1=\{x^2-1\}$
What's the strategy for trying to find a basis for a for a tricky subspace (in particular polynomials.)
Also, what exactly is the dimension of W? I know it has to be $ \le 4$ but is there any way to find out what it is?
 A: Write $$p(X) =aX^3+bX^2+cX+d.$$ $p(-1)=p(1)=0$ i.e $a+b+c+d=0$ and $-a+b-c+d=0$ thus $b+d=-a-c=a+c$. This is equivalent to $b+d=a+c=0$, $p(x) =aX^3+bX^2-aX-b= a(X^3-X)+b(X^2-1)$.
The base is $X^3-X, X^2-1$
A: Let 
$$
p(t)=a_0+a_1t+a_2t^2+a_3t^3
$$
be an element of $\Bbb R_3[t]$. Note that $p\in W$ if and only if
\begin{array}{rcrcrcrc}
a_0 &-&a_1&+& a_2&-& a_3 &=& 0 \\
a_0&+&a_1&+&a_2&+&a_3&=& 0
\end{array}
Solving this system can be done with row reductions
$$
\DeclareMathOperator{rref}{rref}
\rref
\left[\begin{array}{rrrr|r}
1 & -1 & 1 & -1 & 0 \\
1 & 1 & 1 & 1 & 0
\end{array}\right]=
\left[\begin{array}{rrrr|r}
1 & 0 & 1 & 0 & 0 \\
0 & 1 & 0 & 1 & 0
\end{array}\right]
$$
This proves that $p\in W$ if and only if
$$
\begin{bmatrix}a_0\\a_1\\a_2\\a_3\end{bmatrix}=
\begin{bmatrix}-a_2\\-a_3\\a_2\\a_3\end{bmatrix}=
\begin{bmatrix}-1\\0\\1\\0\end{bmatrix}a_2+
\begin{bmatrix}0\\-1\\0\\1\end{bmatrix}a_3
$$
That is, $p\in W$ if and only if
$$
p(t)=a_2(-1+t^2)+a_3(-t+t^3)
$$
This proves that 
$$
W=\DeclareMathOperator{Span}{Span}\Span\{t^2-1,t^3-t\}
$$
Can you prove that $t^2-1$ and $t^3-t$ are linearly independent?
