Basis of a subspace example

Question

I am having trouble understanding the reasoning in the book Linear Algebra Done Right, example 2.41. I will quote the example almost verbatim. The sentence in bold is the one I am having difficulty with.

Example

Show that $1$, $(x-5)^2$, $(x-5)^3$ is a basis of the subspace $U$ of $P_3(R)$ defined by $U = \{p \in P_3(R) : p'(5)=0\}$.

Solution. Clearly each of the polynomials is in $U$. Suppose $a + b(x-5)^2 + c(x-5)^3 = 0$ for every $x \in R$. Reasoning leading to $a=b=c$ omitted. Thus the list is linearly independent in $U$.

Thus $dim\:U \ge 3$. Because $U$ is a subspace of $P_3(R)$, we know that $dim\:U\le P_3(R)=4$. However, $dim \: U$ cannot equal 4, because otherwise when we extend a basis of $U$ to a basis of $P_3(R)$ we would get a list with length greater than $4$. Hence $dim \: U = 3$. Thus the list is a basis of $U$.

End of example

Sure, the list of linearly independent polynomials in $U$ can be extended to a basis of $P_3(R)$, but the extension can be a trivial one, with no elements being added. What am I missing?

Answer 1

The extension cannot be trivial, since $U \ne P_3(R)$, i.e., there exists some $p \in P_3(R)$ such that $p \notin U$, i.e., $p'(5) \ne 0$ (for example, let $p(x) = x+2$.) If the extension were trivial, then every $p \in P_3(R)$ could be expressed as $p(x) = a + b(x-5)^2 + c(x-5)^3$ for some constants $a,b,c$. It follows that $p'(5) = 0$, so that $P_3(R) \subset U$, which is a contradiction.