First check out the rational zeroes by looking at the factors of the coefficients of the highest and lowest degrees of the polynomial, that is, factors of 2 (coefficient of x4) and factors of 39 (constant).
39 has 1, 39, 3 and 13 as factors; 2 has only 1 and 2.
Rational zeroes are:
1, 3, 13, 39, 1/2, 3/2, 13/2, 39/2 (positive and negative forms).
So there are 16 rational zeroes. We set x to each of these values and evaluate the polynomial. If we find any values that evaluate to zero, these are true zeroes (roots). To cut a long story short we can identify 1/2 and 3 as true zeroes. Using synthetic division we can reduce the polynomial to a quadratic.
3 | 2 -19 71 -109 39
2 6 -39 96 | -39
½ |2 -13 32 -13 | 0
2 1 -6 | 13
2 -12 26 | 0 = 2(x2-6x+13) which has no real zeroes.
The common factor 2 in this quadratic can be used to convert factor x-½ into 2x-1.
So complete factorisation is (2x-1)(x-3)(x2-6x+13).
The quadratic has a complex solution:
x2-6x=-13,
x2-6x+9=-13+9=-4,
(x-3)2=-4, x-3=±2i, so x=3+2i or 3-2i.
The zeroes are: ½, 3, 3+2i, 3-2i.