Sure, let's go through the problem step by step.
### (a) Find another zero of [tex]\( R(x) \)[/tex].
Since the polynomial [tex]\( R(x) \)[/tex] has real coefficients, any complex roots must occur in conjugate pairs. The given zeros are:
[tex]\[ -4 + i, \quad -5i, \quad 3 + 2i \][/tex]
The conjugates of these zeros would be:
[tex]\[ -4 - i, \quad 5i, \quad 3 - 2i \][/tex]
Thus, another zero of [tex]\( R(x) \)[/tex] is one of these conjugate pairs. For example, one possible answer is:
[tex]\[ -4 - i \][/tex]
So, the answer is:
[tex]\[ \boxed{-4 - i} \][/tex]
### (b) What is the maximum number of real zeros that [tex]\( R(x) \)[/tex] can have?
The polynomial [tex]\( R(x) \)[/tex] is of degree 7, which means it has 7 zeros (counting multiplicity). As observed in part (a), there are already 3 pairs of complex zeros (which account for 6 zeros in total). Since these must occur in pairs, there can only be one remaining zero, which must be real.
So, the maximum number of real zeros that [tex]\( R(x) \)[/tex] can have is:
[tex]\[ \boxed{1} \][/tex]
### (c) What is the maximum number of nonreal zeros that [tex]\( R(x) \)[/tex] can have?
Given that the polynomial [tex]\( R(x) \)[/tex] has a total degree of 7 and we know that complex zeros occur in pairs, the maximum number of non-real zeros would be 6 (3 pairs of complex zeros).
So, the maximum number of non-real zeros that [tex]\( R(x) \)[/tex] can have is:
[tex]\[ \boxed{6} \][/tex]
In summary:
(a) Another zero of [tex]\( R(x) \)[/tex] is: [tex]\( \boxed{-4 - i} \)[/tex]
(b) The maximum number of real zeros that [tex]\( R(x) \)[/tex] can have is: [tex]\( \boxed{1} \)[/tex]
(c) The maximum number of nonreal zeros that [tex]\( R(x) \)[/tex] can have is: [tex]\( \boxed{6} \)[/tex]
