f(x) = 2x^{3} - 17x^{2} + 61x + 18

These are the possible rational roots: x = ±1, ±2, ±3, ±6, ±9, ±18, ±1/2, ±3/2, and ±9/2.

By process elimination, we know 2 and 9 are two of the possible solutions since they are the factors of 18. Now let's look at the rational numbers (fractions). 2 is in the denominator since that represents the leading coefficient in the cubic term (x^{3}). 3/2 and -9/2 are the other possible solutions that can work.

So, 5/2 is not one of the rational zeros because 5 is not one of the factors of 18.