Basis of a vector space of polynomials that evaluate to zero. $$V = \{ f \in P_n(F) \: : \: f(a) = 0 \} .$$
I'm trying to find a basis for the vector space V but I have no idea how to approach it.
At first I thought it might be $$(0,x-x,x^2-x^2,...,x^n-x^n)$$ But this subset isn't linearly independent, since $$x-x = x^2 - x^2$$ Come to think of it, the basis can't possibly be more than one element: the zero polynomial, since all the polynomials evaluate to zero. I'm probably just misunderstanding this somehow.

Essentially, you're looking for a basis for the vector space of all $n^{\text{th}}$ order polynomials over $\mathbb{F}$ which have a zero at $a$. Consider a general $g\in P_n(\mathbb{F})$ written
$$
g(x)=b_0x^n+b_1x^{n-1}+b_2x^{n-2}+\cdots+b_{n-2}x^2+b_{n-1}x+b_n.
$$
If $g\in V$, then
$$
g(a)=b_0a^n+b_1a^{n-1}+\cdots+b_{n-1}a+b_n=0.
$$
Notice that
$$
b_n=-b_0a^n-b_1a^{n-1}-\cdots-b_{n-1}a.
$$
You're getting some information on the basis for this space now. We know that $b_n$ will just be a linear combination of the other coefficients, so $V$ does not require a constant in its basis. $V$ has at most $n$ elements in its basis then, instead of the $n+1$ basis elements that general vector spaces of $n^{\text{th}}$ order polynomials have. Now think about factoring $(x-a)$ from $g(x)$. This will hopefully lead you to writing down the basis for $V$.
