The part of the proof I am asked to show from my book states:
Let $P(z)$ be a non constant polynomial in $z$. Assume that $P(z) \neq 0$ for all $z \in \mathbb{C}$.
Show that $P(z) = P(0) +zQ(z)$ Where $Q(z)$ is also a polynomial.
Here is my attempt, I would just like to know if this is the right approach or if I have no clue what I'm doing:
As $P(z)$ is a non constant polynomial it has the form of $P(z)=a_nz^n + a_{n-1}z^{n-1} …. a_1z +a_0$. It can be seen that $P(0)=a_0$ which is constant, so we can let $Q(z)= a_nz^{n-1} + a_{n-1}z^{n-2} …. + a_1$ and when we multiply this by $z$ we obtain $zQ(z)= a_nz^{n} + a_{n-1}z^{n} …. a_1z$.
Thus we can rewrite $P(z) = P(0) +zQ(z)$ as required.
I understand that this process might not be correct, but maybe someone can nudge me in the right direction to approach.
Best Answer
Your approach is correct. Here is a more algebraic proof.
Consider $P \in \mathbb{K}[X]$, where $\mathbb{K}$ is a field. Then $P(X) - P(0)$ has $0$ as a root. Thus, $X$ divides $P-P(0)$, that is, there exists $Q \in \mathbb{K}[X]$ such that $P(X) -P(0) = XQ(X)$. Hence $P(X) = P(0) + XQ(X)$.