Why is $1, (x-5)^2, (x-5)^3$ a basis of $U=\{p \in \mathcal P_3(\mathbb R) \mid p'(5)=0\}$? $\mathcal P_3(\mathbb R)$ is the set of polynomials with degree at most $3$ with coefficients in $\mathbb R$.
In the last paragraph is says $U$ cannot be extended to a basis of $\mathcal P_3(\mathbb R)$. I do not understand why not. Why would we get a list with length greater than $4$? Why can't we add $(x-5)$ to the list?

From Linear Algebra Done Right
 A: As we have already determined that $U$ is not equal to $\mathcal{P}_3(\mathbf{R})$, and that $U$ is a subspace of $\mathcal{P}_3(\mathbf{R})$, we know that $U$ cannot have dimension 4 (the same dimension as $\mathcal{P}_3(\mathbf{R})$).
The only way that $U$ would have the same dimension as $\mathcal{P}_3(\mathbf{R})$ would be if the two were equal (i.e. the basis vectors for $U$ would also span $\mathcal{P}_3(\mathbf{R})$), as we can always add vectors to the basis of a strict subspace to obtain a basis of the vector space.
Edit: In Linear Algebra Done Right 3e, theorem 2.33 states "Every linearly independent list of vectors in a finite-dimensional vector space can be extended to a basis of the vector space", with the proof following. The basis for $U$ is indeed a linearly independent list of vectors.
A: We know, If $\dim U = 4 $ and $U \subseteq \mathcal{P}_3$ then $U = \mathcal{P}_3$.
But, is easy to show a polynomial in $\mathcal{P}_3$. such that $p'(5)\neq 0 $ (for example $p(x)= x-3$).
that's mean $U$ isn't $\mathcal{P}_3$ finally $\dim U < 4$.
