We need to search for rational zeroes. We take the factors of the coefficient of the highest and lowest degrees of x. So the highest degree is 6x4 and the lowest is 3.
The factors of 6 are 1, 2, 3 and the factors of 3 are 1, 3.
First try the easiest factor which is x+1 or x-1. If these are true factors then substituting x=-1 or 1 will give a value of zero for the polynomial. All we have to do is write down the coefficients:
6∓17+8±8-3 where ∓ and ± cover whichever we use of -1 or 1.
6-17+8+8-3=2 and 6+17+8-8-3=20; so neither yields zero.
The other rational zeroes combine the factors of the greatest and least coefficients. So we have:
1/3, 3/1, 1/2, 3/2 as possible candidates. We have to try positive and negative values of these by substitution into the polynomial. To cut to the chase, we eventually arrive at zeroes ⅓ and 3/2.
We can use synthetic division to divide by each of these:
⅓ | 6 -17 8 8 | -3
6 2 -5 1 | 3
6 -15 3 9 | 0
3/2 | 6 -15 3 9
6 9 -9 | -9
6 -6 -6 | 0 = 6(x2-x-1)
We can solve the quadratic later. Note the preponderance of the factor 6.
We have identified the factors x-⅓ and x-3/2. We had no fractional coefficients in the original polynomial, so this is where the 6 comes in handy because 6=2×3, and 2 and 3 are the numbers we need to get rid of the fractions: 3(x-⅓)=3x-1; 2(x-3/2)=2x-3. So the factorisation of the polynomial is (so far) (3x-1)(2x-3)(x2-x-1).
We can use the quadratic formula to solve x2-x-1=0: x=(1±√5)/2, so x=(1+√5)/2 or (1-√5)/2.
The complete solution is therefore: x=⅓, 3/2, (1+√5)/2 or (1-√5)/2.