List all the possible rational zeros of the function f(x) = 18x4 + 12x3 - 15x2 + 2x - 7.1. 1, -1, 2, -2, 7, -7, 9, -9, 18, -18.2. 1, -1, 2, -2, 9, -9, 18, -18.3. 1, -1, ½, -1/2, 1/9, -1/9, 1/18, -1/18, 7, -7, 7/2, -7/2, 7/9, -7/9, 7/18, -7/184. 1, -1, 7, -7

f(x) = 18x^4 + 12x^3 - 15x^2 + 2x - 7.

Independent term: -7
Divisors of the independent term (a): 1, -1, 7, -7

Leading coefficient: 18
Divisors of the leading coefficient (b): 1, -1, 2, -2, 3, -3, 6, -6, 9, -9, 18, -18

The possible rational zeros of the function: a/b
1/1=1
1/(-1)=-1
1/2
1/(-2)=-1/2
1/3
1/(-3)=-1/3
1/6
1/(-6)=-1/6
1/9
1/(-9)=-1/9
1/18
1/(-18)=-1/18 

-1/1=-1
-1/(-1)=1
-1/2
-1/(-2)=1/2
-1/3
-1/(-3)=1/3
-1/6
-1/(-6)=1/6
-1/9
-1/(-9)=1/9
-1/18
-1/(-18)=1/18

7/1=7
7/(-1)=-7
7/2
7/(-2)=-7/2
7/3
7/(-3)=-7/3
7/6
7/(-6)=-7/6
7/9
7/(-9)=-7/9
7/18
7/(-18)=-7/18

-7/1=-7
-7/(-1)=7
-7/2
-7/(-2)=7/2
-7/3
-7/(-3)=7/3
-7/6
-7/(-6)=7/6
-7/9
-7/(-9)=7/9
-7/18
-7/(-18)=-1/18

Then, the possible rational zeros of the function are:
1,-1,1/2,-1/2,1/3,-1/3,1/6,-1/6,1/9,-1/9,1/18,-1/18
7,-7,7/2,-7/2,7/3,-7/3,7/6,-7/6,7/9,-7/9,7/18,-7/18

Answer: Option 3. 
1, -1, ½, -1/2, 1/9, -1/9, 1/18, -1/18, 7, -7, 7/2, -7/2, 7/9, -7/9, 7/18, -7/18