Bases for space of polynomials I'm facing an exercise to determine basis for some spaces of polynomials. Here they are

Consider the space of polynomials of degree equal or less than 3
$U =${$p(t) \in \mathbb{R_3}[t]$ | $p(0)=0$}
$U =${$p(t) \in \mathbb{R_3}[t]$ | $p(1)=0$}
$U =${$p(t) \in \mathbb{R_3}[t]$ | $p(0)=p(1)$}

So my answer was $(x^3,x^2,x)$ for the first one $(x^3-1,x^2-1,x-1$) for the second one and $(x^3-x, x^2-x, 1)$ for the last one.
When I checked the key for this question I found out they erased the vectors with $x^3$...
Can someone explain me why?
I tried to write my vectors in vectores of coordinates through the canonical basis and then apply Gauss elimination but I didn't reach to linearly dependent vectors so I think we can't erase one, can we?
Thanks!
 A: The maps $f,g,h\colon \mathbb{R}_3[x]\to\mathbb{R}$ defined by
\begin{align}
f(p)&=p(0)\\
g(p)&=p(1)\\
h(p)&=p(1)-p(0)
\end{align}
are easily seen to be linear and surjective. So their kernels (null spaces) have dimension $3$. The kernels are precisely the subspaces you have to find bases of, in the same order.
Since clearly all three subspaces contain polynomials of degree $3$ (and you found them), a bases for each of them must contain a polynomial of degree $3$.

Your solution is, as far as I can see, correct, but let's check it.
The first set is clearly contained in $\ker f$ and also linearly independent. The second set is contained in $\ker g$ and the matrix of coordinates with respect to the basis $\{x^3,x^2,x,1\}$ is
$$
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
-1 & -1 & -1 \\
\end{bmatrix}
$$
that's easily seen to have rank $3$, so the set is linearly independent.
The third set is contained in $\ker h$ and the matrix is
$$
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
-1 & -1 & 0 \\
0 & 0 & 1
\end{bmatrix}
$$
that has rank $3$.
