"Modernization" of an old AIME problem

Let us consider $f(x) = a(x-p)(x-q)$ . hence $f(0) = apq = 2014$ . $apq = 2*19*53$ .

First we consider the cases where $a$ , $p$ and $q$ are all positive. The number of such possibilities is $3^{3} = 27$ .BUt we must also consider the cases where the values of $p$ and $q$ are simply interchanged as they lead to the same equation.There is only one value where $p = q = 1$ All others possibilities are double counted.So the number of remaining possibilities $\frac{27-1}{2}= 13$ .

Now we must consider the signs,it is to be noted that whatever we choose the signs of $p$ and $q$ ,the sign of $a$ is automatically determined by the choice to make the the RHS positive.So considering signs There are $2*2 = 4$ possibilities for each value.But the case in which $|p| = |q|=1$ ,there are 3 possibilities as $(p,q) = (1,-1)$ and $(p,q) = (-1,1)$ are indistinguishable.

So the total number of possibilities = $13*4+3 = \boxed{55}$ .