Given the zeros -i, -3+21, and 4+31, you can find another zero using the complex conjugate theorem. For each complex zero, its conjugate is also a zero. Therefore, the conjugate of -i is i. Hence, another zero is i.
Since -i is a complex zero, its conjugate i is also a root. Similarly, for -3+21i, its conjugate -3-21i is a root, and for 4+3i, its conjugate 4-31i is a root. This means that for every complex zero, there is a conjugate pair of roots.
So, for the given zeros, you have a total of 6 roots (including the conjugates). Since R(x) is a polynomial of degree 9, the remaining 3 roots must be real. Therefore, the maximum number of real zeros R(x) can have is 3.
For each complex zero, there is a conjugate pair. In this case, you have three complex zeros: -i, -3+21i, and 4+31i. So, there are a total of six complex roots (including their conjugates).
Since the degree of R(x) is 9, and you already have 6 complex roots, the maximum number of nonreal zeros (complex roots) that R(x) can have is 6.
