Not Imaginary

Consider the given expression as $f(x)$ .

Then find its derivative $f^{\prime}(x)$ which gives all even powers of $x$ and a positive constant , implying that it is an always increasing function.

Also we know polynomials with their highest degree odd will have at least one real root.

An increasing function cuts the x-axis only once.

Hence, the given polynomial has $1~real~root$

The total number of roots is 2015

Since all the powers are odd numbers (2015, 21, 277, 99, 1), any negative value for x will yield a larger negative value, and the term (21x^2015+18x^21+6x^277+77x^99+x) will never be a positive value and it will never be equal to 88 then there is no negative real roots

for x>=1 the term (21x^2015+18x^21+6x^277+77x^99+x) will always be greater than 88 then the only real solutions available is the period 0<x<1

since complex come in pairs and the total number of roots is odd, then we must have an odd number of real roots.

since the term (21x^2015+18x^21+6x^277+77x^99+x) is always increase in the period 0<x<1 and there is an odd number of solutions in this period, then there can't be more than one solution

Descartes rule of signs? Positive roots have 1 sign change, and negative roots does not have a sign change.