There is at least one root since $f$ has odd degree and in order to be unique, it suffices for Velly to pick $f(0) = a_{2015} > 0$ and then to aim for $f'(x) = 2015x^{2014} + 2014a_1x^{2013} + \cdots + a_{2014}$ to be positive. (Note that when all but one of the coefficients are chosen, the resulting function has a minimum since it is a polynomial of even degree with positive leading coefficient.) This can be done exactly as in this problem.