Question 472568
Since there are 2 variations of signs, there are 0 or 2 positive real zeros.
If x is replaced by -x, the resulting polynomial is {{{-x^5 - 3x^4 + 2x^3 + x^2 + 5}}}, which has 1 variation of sign, hence exactly one negative zero.  Hence, either there are 
(i) 2 positive, 1 negative, and 2 complex zeros, 

OR

(ii) 0 positive, 1 negative, and 4 complex zeros.