Intuitively, dimension is the number of degrees of freedom. The elements of $\mathcal{P}_n(\mathbb R)$ are polynomials of degree $n$ (more precisely, at most $n$), so they look like
$$ a_0 + a_1 x + a_2 x^2 + \dotsb + a_{n-1} x^{n-1} + a_n x^n $$
To specify such a polynomial, you have to specify $n+1$ numbers, the coefficients $a_0,a_1,\dotsc,a_n$. So there are $n+1$ degrees of freedom in this "space" of polynomials.
To prove that formally, you'd want to think of polynomials $a_0+\dotsb+a_nx^n$ as being linear combinations of the polynomials $1,x,x^2,\dotsc,x^n$, and show that these latter polynomials form a basis. This is done in chapter 2 of Axler.
Again intuitively, a constraint that specifies a single number reduces the number of degrees of freedom by 1. Thus imposing the constraint that we will only work with polynomials $f(x)$ satisfying $f(1)=0$ should, we expect, reduce the dimension from $n+1$ to $n$.
The formal version of this is the rank-nullity theorem (Axler's theorem 3.4), which is why everybody's giving answers involving it. I see Axler doesn't do that until chapter 3, though.
So I think the only thing you can do at this point is to produce an explicit basis for the subspace in question. Exercise 8 in chapter 2 is similar; have you tried that? (And for playing with polynomials, exercises 9 and 12 in the same chapter are good.)
(I have the 2nd edition of Axler's text; hopefully it matches yours.)
Consider the linear map $V\to \mathbb{R}^2$ given by $a+bx+cx^2\mapsto (a,b)$. It is injective because $c=-b$. Therefore it is an isomorphism and $dim V=2$.
If you want an explicit basis just pull a basis from $\mathbb{R}^2$. For instance, pulling $(1,0),(0,1)$ gives you $1,x-x^2$ as a basis for $V$.
Best Answer
There is a linear map on $\rm \mathbb{P}_n(\mathbb{R})$ given by $\rm L_a :f(x)\to f(x+a)$ for any $\rm a\in\mathbb{R}$. It is invertible because we can see $\rm L_a L_{-a}=L_{-a}L_a=Id$ as linear maps. Hence $\rm L_a$ preserves dimensions of subspaces. We may then write $\rm L_{-a} W=V=\{ p(x)\in \mathbb{P}_n(\mathbb{R}):p(0)=0\}$. It is easy to see that $\rm B=\{x,x^2,\dots,x^n\}$ is a basis for this space by showing that $\rm p(0)=0\,$ is equivalent to $\rm p_0=0$ (the constant term vanishes).
One may use this to show $\rm L_{-a}B=\{x-a,(x-a)^2,\dots,(x-a)^n\}$ is a vector space basis for $\rm W$.