Examine the factors of 30 (the constant) and 2 (the coefficient of the highest power).
30=1×2×3×5; 2=1×2. Let a be the set {1,2,3,5} and b be the set {1,2}. Then we divide each element of a by each of b, and these form the set r of rational zeroes (positive and negative forms).
If it factorises into rational factors, the above may qualify as actual zeroes.
Take the smallest and work out f(½) and f(-½):
f(½)=-2/8+3/4+59/2-30=0, so ½ is an actual zero. We need not work out any more, because we can use this zero to reduce the cubic to a quadratic, using synthetic division:
½ | -2 3 59 -30
-2 -1 1 | 30
-2 2 60 | 0 = -2x2+2x+60=-2(x2-x-30)=-2(x-6)(x+5).
Therefore the other zeroes are 6 and -5.
f(x)=-2(x-½)(x-6)(x+5)=-(2x-1)(x-6)(x+5).