Does $B = \{x-2, x(x-2), x^2(x-2)\}$ span $\{p(x)\in P_3(\mathbb{R})|p(2) = 0\}$?

Question

Let $P_3(\mathbb{R}) = \operatorname{Span} \{1, x, x^2, x^3\}$.

$W$ is a subspace of $P_3(\mathbb{R})$, $W = \{p(x)\in P_3(\mathbb{R})|p(2) = 0\}$.

Find a basis and the dimension of $W$.

I chose the set $B = \{x-2, x(x-2), x^2(x-2)\}$ as a basis because $(x-2)$ is a factor of $p(x)$. So the dimension would be $3$.

How can I prove $W \subseteq \operatorname{Span} (B)$ and $\operatorname{Span} (B) \subseteq W$?

Any feedback is appreciated.

Answer 1

You can use a general argument :

First prove that span(B) is indeed of dimension 3 (that your vectors are lineary independant)

Next, span(B) $\subset W$, because every vector of B is in W, so every linear combinaison of vectors from B is in W, hence the result

Then, because $1$ is not in W, the dimension of W < 4, but W contain a subspace of dimension 3, so it's of dimension 3 and B is a base of W : span(B) = W