Find the dimension of a vector subspace I'm doing a problem on finding the dimension of a linear subspace, more specifically

if $\:$ {$f \in \mathcal P_n(\mathbf F): f(1)=0, f'(2)=0$} is a subspace of $P_n$, what is this dimension of this subspace? Here $\mathcal P_n(\mathbf F)$ denotes a vector space of Polynomials of degree $n$ over the real number field.

At first glance, I thought the dimension is infinity, but I think perhaps since the degree is restricted, the dimension should be finite. Yet I find it hard to specify the number of dimensions. Being a beginner of linear algebra, I would like to hear some detailed explanation on how to solve this type of problems.
Thanks in advance!

If you want to do this from scratch, it's a slog but you can actually display a basis for the given subspace: 
According to your conditions: 
$p(x)=(-a_{1}-a_{2}-\cdots -a_{n})+(-4a_{2}-12a_{3}-\cdots -n2^{n-1}a_{n})x+a_{2}x^{2}+\cdots +a_{n}x^{n}$. 
Substituting again for $a_{1}$, we have 
$(-(-4a_{2}-12a_{3}-\cdots -n2^{n-1})-a_{2}-\cdots -a_{n})+(-4a_{2}-12a_{3}-\cdots -n2^{n-1})a_{n}x+a_{2}x^{2}+\cdots +a_{n}x^{n}$. 
This is, finally, 
$(4a_{2}+12a_{3}+\cdots +n2^{n-1}a_{n}-a_{2}-\cdots -a_{n})+(-4a_{2}-12a_{3}-\cdots -n2^{n-1}a_{n})x+a_{2}x^{2}+\cdots +a_{n}x^{n}$
$=(3a_{2}+11a_{3}+\cdots +(n2^{n-1}-1)a_{n})+(-4a_{2}-12a_{3}-\cdots -n2^{n-1}a_{n})x+a_{2}x^{2}+\cdots +a_{n}x^{n}$
Factoring out the $a_{i}$, we have
$p(x)=(x^{2}-4x+3)a_{2}+(x^{3}-12x+11)a_{3}+\cdots +(x^{n}-n2^{n-1}x+(n2^{n-1}-1))a_{n}$
To finish, just observe that the set of $n-1$ vectors 
$\left \{ x^{2}-4x+3,x^{3}-12x+11,\cdots ,x^{n}-n2^{n-1}x+(n2^{n-1}-1) \right \}$ spans the subspace and is linearly independent since all the vectors have distinct degrees. Therefore the dimension of the subspace is $n-1$.
