Let $f: \mathbb R \to \mathbb R$ be a polynomial of degree $\geq 2$. Fix arbitrary $a \in \mathbb R$, and let $$c(x) = f(x) - [f(a) + f'(a)(x-a)].$$ Prove there exists a polynomial $p$ such that $c(x) = p(x)(x-a)^2$ with $c(x)$ having a Double root at $a$. (Source: Based on Spivak.)
My proof is below.
I request verification and suggestions on the proof and its exposition. Note: Other proofs exists. This question is to verify or critique this proof.
By the Fundamental Theorem of Algebra, since $a$ is a root of the polynomial $f(x) - f(a)$, there exists a polynomial $p$ such that $$f(x) - f(a) = p(x)(x-a).$$
Let $q(x) = p(x) - f'(a)$. Since $c(x) = q(x)(x-a)$, it suffices to show $q(a) = 0$.
For $x \neq a$, $q(x) = \frac {c(x)}{x-a}$, and since $q$ is continuous, $$\begin{align*} q(a) &= \lim_{x \to a} \frac {c(x)}{x-a} \\ &= \lim_{x \to a} \frac {f(x) - f(a) -f'(a)(x-a)}{x-a} \\ &= f'(a) - f'(a) \\ = 0. \end{align*}$$ completing the proof.