Almost certainly, your textbook is refering to the set of all polynomials of degree at most $n$, rather than the polynomials of degree exactly $n$. The latter is indeed not a vector space under the usual addition of polynomials: it doesn't contain the $0$ polynomial (which either has no degree, or has some degree smaller than $0$), and it's not closed under sums: both $x^n+x^{n-1}$ and $-x^n$ are polynomials of degree $n$, but their sum is not.
So I suspect your textbook is asking for the vector space that consists of all polynomials of degree at most $3$, which further satisfy $f(1)=0$ and $\int_{-1}^1 f(t)\,dt = 0$. That is, if we let $\mathbf{P}_3$ denote the vector space of polynomials of degree at most $3$, then it wants you to find a basis for $\mathbf{W}$, where
$$\mathbf{W} = \left\{ f(t)\in\mathbf{P}_3\;\left|\; f(1)=0\text{ and }\int_{-1}^1f(t)\,dt = 0\right.\right\}.$$
You say you want to solve this via a matrix; well: the map $\mathbf{P}_3\to\mathbb{R}$ given by "evaluation at $1$" ($p(t)\longmapsto p(1)$) is a linear transformation. And the map $\mathbf{P}_3\to\mathbb{R}$ given by "integrate over $[-1,1]$ ($p(t)\longmapsto \int_{-1}^1p(t)\,dt$) is also a linear transformation. Call the first linear transformation $E$ and the second $T$. The polynomials that are in the nullspace/kernel of the first map are precisely the ones that satisfy $f(1) = 0$. The ones that are in the kernel of the second one are precisely the ones that satisfy $\int_{-1}^1f(t)\,dt = 0$. You want the ones that are in the kernel of both.
The standard way of achieving this is to map to the product: take the map $L\colon\mathbf{P_3}\to \mathbb{R}\times\mathbb{R}=\mathbb{R}^2$, given by:
$$L(p(t)) = \Bigl( E(p),\; T(p)\Bigr) = \left(p(1),\quad\int_{-1}^1p(t)\,dt\right).$$
Then the subspace you want is precisely the kernel of $L$.
To find the kernel of $L$, one possibility is to pick a basis for $\mathbf{P}_3$ (say, the standard basis $[1,t,t^2,t^3]$), a basis for $\mathbb{R}^2$ (say, the standard basis $[(1,0), (0,1)]$), construct the matrix representation for $L$, and find the nullspace of that matrix. Interpreting this nullspace in terms of the basis $[1,t,t^2,t^3]$ will give you the kernel you want, and finding a basis for that nullspace will yield a basis for the subspace of $\mathbf{P}_3$ that you are looking at.
Best Answer
To offer a different perspective and to be a bit nit-picky:
The zero vector would be the zero polynomial, which is just $0$. The reason why I am pointing this out is because you don't specify the field $F$ over which you are working. In general, polynomials are not the same as polynomial functions, meaning that you "cannot plug in values into polynomials and say that the polynomial $f$ is the same as the collection $(f(x))_{x \in F}$."
For an example, assume that $F$ is the finite field with two elements. Then $f = x^2 + x$ is a polynomial different from $0$. However, if you plug in both elements $0$ and $1$, you get $f(0) = f(1) = 0$.
If you are working over an infinite field, e.g. the real numbers, then what you do is okay and not at all problematic. In this situation, we can identify polynomials with polynomial functions. (One can show that one can identify polynomials and polynomial functions if and only if the field one is working with is infinite.)