Question on Spectral Mapping Theorem in Hilbert space

Question

I understand the whole proof except for the part which says:

Factor $p(x)-\mu=c(x-x_1)\dots (x-x_n) $ [...]

I understand why I can factor $p(x)$ individually but why do I obtain $p(x)-\mu$ is also factorable?

Any help would be valuable! Thanks in advance!!

Answer 1

If $p(x)$ is not constant, $p(x)-\mu$ is a polynomial of the same degree as $p(x)$, so it is also subject to the Fundamental Theorem of Algebra.