since the sign changed from negative to positive and positive to negative twice, then there are 2 or zero positive real roots.
notice that if x is replaced with -x, the function becomes
and notice also that there are two changes of sign, which implies that there 2 or no negative real roots..
the complex roots depends on the exixtence of positive and negative real roots. if there is no positive real root but 2 negatve real roots, then there must be 2 complex roots. also if there is no negative real root but 2 positive real roots, then there must be 2 complex roots. however, if there are no negative and positive real roots, then there must be 4 complex roots.
for your last question, there is no complex roots (assuming you got the right real roots) Ü note that if a function is of degree n (n is positive integer), then the function have exactly n roots in the system of Complex Numbers (of the form of a + bi). (remark: for real roots, the b there is 0). (Fundamental Theorem of Algebra)