Factoring the roots of function in general

Question

When finding derivative of $x^2$ or $x^3$. It always show that the infinitesimal small quantity h is canceled. Further more, special function like $\sin x$ and $\cos x$ can be computed and derivative of $\vert x-a \vert$ or $x \sin \frac {1}{x-a}$ at $a$ do not exist only because of $\lim_{x\to a+} \neq \lim_{x \to a-}$ and discontinuous at $a$. So the derivative of f(x) at a must cancel the infinitesimal small quantity h otherwise $\lim_{h \to 0} \frac {f(a+h)-f(a)}{h}$ do not make sense. Then in general if $G(\alpha)=0$ does it implies that $G(x)=(x-\alpha)h(x)$ for some $h(x)$?