A function f has a simple zero (or zero of multiplicity 1) at x0 if f is differentiable in a neighborhood of x0 and f(x0) = 0 while f(x0) ≠ 0.
Prove that f has a simple zero at x0 iff f(x) = g(x)(x - x0), where g is continuous at x0 and differentiable in a deleted neighborhood of x0, and g(x0) ≠ 0.
The Attempt at a Solution
You need to find a function g(x) such that it fufills the properties in the problem. I thought of using the function g(x) = f(x)/(x-x0), but that is not continuous at x0. Am I close in trying to find this function g or should the proof go in a different direction?