If we add all the coefficients we get 2-17+59-83+39=100-100=0, so x=1 is a zero.
Divide by this zero using synthetic division:
1 | 2 -17 59 -83 39
2 2 -15 44 | -39
2 -15 44 -39 | 0 = 2x3-15x2+44x-39 = g(x)
Add the coefficients of g(x): 2-15+44-39=46-54=-8, so x=1 is not another zero, neither is -1, because all the terms would be negative. The factors of 39 other than 1 and 39 are 3 and 13, and the factors of the x3 term are only 2 and 1. So we could have zeroes ½, 1½, 6½, 3, 13, 19½. Another clue is to plug in x=2 and see if there's a change of sign:
16-60+88-39=104-99=5. So g(1)=-8 and g(2)=5, therefore there is a zero between x=1 and x=2. It's possible that x=1½=3/2, so let's try it:
2(27/8)-15(9/4)+44(3/2)-39=(27-135+264-156)/4=(291-291)/4=0, therefore x=1½ is a zero. Divide by this zero using synthetic division:
1½ | 2 -15 44 -39
2 3 -18 | 39
2 -12 26 | 0 = 2x2-12x+26 = h(x) which has no real zeroes.
f(x)=(x-3/2)(x-1)(2x2-12x+26)=(2x-3)(x-1)(x2-6x+13).
Therefore the real zeroes are 1 and 1½ (1.5).
If we solve the quadratic x2-6x+13=0 we get the complex zeroes:
x2-6x+9-9+13=0, (x-3)2=-4, x-3=±i√4, giving conjugate zeroes 3+2i and 3-2i.