I think you missed out the x2 term: f(x)=x5-3x4-5x3+15x2+4x-12?
f(1)=1-3-5+15+4-12=20-20=0, so x=1 is a zero.
f(-1)=-1-3+5+15-4-12=20-20=0, so x=-1 is also a zero. This means that x2-1 is a factor of f(x).
Divide by x2-1:
x3-3x2-4x+12
x2-1 ) x5-3x4-5x3+15x2+4x-12
x5 -x3
-3x4 -4x3+15x2
-3x4 +3x2
-4x3+12x2+4x
-4x3 +4x
12x2 -12
12x2 -12
0
So the quintic reduces to a cubic: x3-3x2-4x+12=x(x2-4)-3(x2-4)=(x-3)(x-2)(x+2).
The five zeroes are: -2, -1, 1, 2, 3.