The quadratic equation $x^2-mx+24 = 10$ has roots $x_1$ and $x_2$. If $x_1$ and $x_2$ are integers, how many different values of $m$ are possible?
Simplify as
x^2 -mx + 14 = 0
(x +1) (x + 14) = 0 x^2 + 15x + 14 = 0 m = 15
So proceeding likewise....
(x -14) ( x -1) = 0 m = -15
(x + 2) ( x + 7) = 0 m = 9
(x -2) (x -7) = 0 m = - 9
+346 
