v(x)=x3-13x+12 has x=1 as a zero because v(1)=1-13+12=0. We can divide by this zero using synthetic division:
1| 1 0 -13 12
1 1 1 | -12
1 1 -12 | 0 = x2+x-12 = (x+4)(x-3).
So the zeroes of v(x) are 1, -4 and 3 (that is, v(1)=v(-4)=v(3)=0).
The question doesn't ask a specific question, but I guess this solution is probably what's expected.