Let's solve each part of the question step-by-step.
### Given:
- [tex]\( R(x) \)[/tex] is a polynomial of degree 8 with real coefficients.
- [tex]\( R(x) \)[/tex] has the zeros: [tex]\( 3 \)[/tex], [tex]\( -5 \)[/tex], [tex]\( 9 \)[/tex], and [tex]\( -2 - 4i \)[/tex].
### (a) Find another zero of [tex]\( R(x) \)[/tex]:
Since [tex]\( R(x) \)[/tex] has real coefficients, any non-real (complex) zeros must come in conjugate pairs. This ensures that the polynomial remains with real coefficients. Given that [tex]\( -2 - 4i \)[/tex] is a zero, its conjugate [tex]\( -2 + 4i \)[/tex] must also be a zero.
Thus, another zero of [tex]\( R(x) \)[/tex] is:
[tex]\[ -2 + 4i \][/tex]
### (b) What is the maximum number of real zeros that [tex]\( R(x) \)[/tex] can have?
- We know that the polynomial [tex]\( R(x) \)[/tex] is of degree 8, meaning it can have at most 8 zeros in total.
- Complex roots occur in conjugate pairs.
Given the zero [tex]\( -2 - 4i \)[/tex] and its conjugate [tex]\( -2 + 4i \)[/tex], we already have one pair of complex zeros, which accounts for 2 of the 8 possible zeros.
The remaining zeros must be real since the degree of the polynomial is 8.
Thus, the maximum number of real zeros is:
[tex]\[ 8 (total degree) - 2 (complex zeros) = 6 \][/tex]
### (c) What is the maximum number of nonreal zeros that [tex]\( R(x) \)[/tex] can have?
- As mentioned, [tex]\( R(x) \)[/tex] is of degree 8.
- We already identified that there can be at most 6 real zeros.
The total number of zeros must be 8, so the remaining can be nonreal.
Thus, the maximum number of nonreal zeros is:
[tex]\[ 8 (total degree) - 6 (max number of real zeros) = 2 \][/tex]
### Summary:
(a) Another zero of [tex]\( R(x) \)[/tex] is:
[tex]\[ -2 + 4i \][/tex]
(b) The maximum number of real zeros that [tex]\( R(x) \)[/tex] can have is:
[tex]\[ 6 \][/tex]
(c) The maximum number of nonreal zeros that [tex]\( R(x) \)[/tex] can have is:
[tex]\[ 2 \][/tex]
