If $u = \{f \in P_{4} | f(3) = 0 \}$
$w = \{g \in P_{4} | g(2) = 0 \}$
Then $u \cap w = \{h \in P_{4} | h(3) = 0, h(2) = 0 \}$
How can we find the dimension of each of these subspaces? Dimension theorem states that the dimension of $u$ and $w$ should be (4+1), however when checking the answer, the question says the correct dimension should be 4 for both $u$ and $w$. Additionally it says $u \cap w$ should have a dimension of 5. Bit confused to why this is. Any help would be appreciated.
As I understand it, $U$ is given by polynomials of the form $(a_3x^3 + a_2x^2 + a_1x + a_0)(x-3)$, so $\{x^i(x-3)\mid i=0..4\}$ should be a basis, ie. $U$ has dimension 4. Similarly for $V$. $U \cap V$ is a subspace of both $U$ and $V$, so by the dimension axiom cannot possibly be have dimension larger than the minimum of both dimensions, i.e. has to be $\leq 4$. In fact, just using the same trick as before, elements in $U \cap V$ should have the form $(c_2x^2 + c_1x + c_0)(x-2)(x-3)$, so $U \cap V$ should have dimension 3.