# Basis of a subspace example

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?

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.