Polynomial Subspaces Let $P_{m} := \{p(t) = a_{0} + a_{1}\cdot t + \ldots + a_{m}\cdot t^m\}$  be the set of all polynomials with degree $\leq$ $m$. 
$V := \{p(t) \in P_{3} | p(0)=p(1)=0 \}$
now I need to show that $V$ is a linear subspace of $P_{3}$ and find a basis for it. This is how I proceede but I'm not quite sure if I'm getting this right and I don't have a solution to this exercise to check. 
to prove that a vectorspace is a linear subspace we need to check the two conditions
$ \bullet x+y \in V$ 
$\bullet \alpha \cdot x \in V$
I proceeded as such:
$x(t),y(t) \in P_{3} \\$
$y_{0} + y_{1}t+y_{2}t^2+y_{3}t^3 + x_{0} + x_{1}t+x_{2}t^2+x_{3}t^3$ is obviously also in $P_{3}$. Now is this already enough or do I need to proceed by plugging in 1 and zero for t and getting something like: $\\$
$x_{0}+y_{0} = 0 $ and $x_{1}+x_{2}+x_{3} + y_{1}+y_{2}+y_{3} = 0$??. Please help me since the space $V$ kinda confuses me. I usually know how to solve this kind of problems but for some reason I can't wrap my head around this one.
Thanks a lot in advance.
edit: thank you for your help. I managed to solve this.
 A: *

*The first part is basically right. I would say show that $x(t) + y(t)$ and $\alpha \ast x(t)$ are polynomials of degree at most $3$ and vanish at $0$ and $1$, even though it seems obvious... [The sad things in life.]

*Hint: For the second part, I would consider using a result that you know about polynomials... If $p(x)$ is a polynomial that has a root at $r$, then can you express $p(x)$ in some form that reflects that fact? If $r = 0$, it is easy enough, but what about general $r$? ... To remember this result, it might be useful to think of all the easy examples of polynomials that have a root at $r$.
A: Showing it is a vector space is easy enough:
If $x,y \in V$ we know that $x(0) = x(1) = 0$ and $y(0) = y(1) = 0$ 
Now $(x+y)(t) = x(t) + y(t)$ for all $t$ by the definition of addition of polynomials. So $(x+y)(1) =x(1) + y(1) = 0 + 0 = 0$, and the same for $t=0$.
Also $(\alpha \cdot x)(t) = \alpha x(t)$, so $(\alpha \cdot x)(1)  = \alpha x(1) = \alpha 0  = 0$ etc. 
Intuitively, we lose two dimensions of the 4 of $P_3$ by imposing two linear conditions in $V$. Algebraically, if $p(0) = p(1) = 0$ for some polynomial,
$p \in V$ we know that $p(t) = t(t-1)q(t)$ for some polynomial $q(t)$ (here of degree at most $1$ or the degree is too large). So we use the basis $\{1,t\}$ for the $q$.
Show that $p_1(t) = t(t-1)$ and $p_2(t) = t^2(t-1)$ are linearly independent, and the above argument shows they span $V$, together this means they're a basis for $V$.
