Answer
False.
Work Step by Step
A first example is $f(x)=(x-2)(x+2)$. The solutions can be found by setting $f(x)=0$ and getting $$(x-2)(x+2)=0$$ $$x=\{-2,2\}$$ Next, use a completely different quadratic function, $g(x)=2(x-2)(x+2)$. Find the solutions of this function the same way $$2(x-2)(x+2)=0$$ $$x=\{-2,2\}$$ These different functions both have zeroes at $x=-2$ and $x=2$. Therefore, there exists more than one function with those zeroes, making the statement false.
