We look at the factors of 6 (constant term) and the factors of 3 (highest power of x).
6 has factors 1, 2, 3, 6 and 3 has factors 1 and 3.
(a) This means that possible zeroes are 1, ⅓, 2, ⅔, 3, 6, ⅙. We can put plus or minus in front of each of these.
(b) Try x=1: 3-8-6+17+6=12, so x-1 is not a factor.
Try x=-1: 3+8-6-17+6=-6, so x+1 is not a factor.
Try x=2: 48-64-24+34+6=0, so x-2 is a factor. We divide by x-2 using synthetic division.
3x³-2x²-10x-3, so now we have rational zeroes ±⅓, ±3. x=±3 doesn’t work, but x=-⅓ does work, so 3x+1 is a factor. Using synthetic division we get x²-x-3 as the remaining factor, which has no rational zeroes.
Therefore p(x)=(x-2)(3x+1)(x²-x-3). We can check this by multiplying the factors.
[We can find the irrational zeroes by completing the square or using the quadratic formula: ½(1±√13).]