Given the polynomial [tex]\( R(x) \)[/tex] of degree 7 with real coefficients and the zeros provided:
[tex]\[
-i, \quad -1-5i, \quad 3-2i
\][/tex]
Let's address each part of the question step-by-step.
### (a) Find another zero of [tex]\( R(x) \)[/tex].
Since the coefficients of [tex]\( R(x) \)[/tex] are real, the non-real zeros must occur in conjugate pairs. This means that for each non-real zero [tex]\( a + bi \)[/tex], [tex]\( a - bi \)[/tex] must also be a zero.
The given zeros are:
- [tex]\( -i \)[/tex]
- [tex]\( -1 - 5i \)[/tex]
- [tex]\( 3 - 2i \)[/tex]
Their conjugates are:
- [tex]\( i \)[/tex] (conjugate of [tex]\( -i \)[/tex])
- [tex]\( -1 + 5i \)[/tex] (conjugate of [tex]\( -1 - 5i \)[/tex])
- [tex]\( 3 + 2i \)[/tex] (conjugate of [tex]\( 3 - 2i \)[/tex])
Therefore, the additional zeros must be:
[tex]\[
i, \quad -1 + 5i, \quad 3 + 2i
\][/tex]
So, the additional zeros are:
[tex]\[
i, \quad -1 + 5i, \quad 3 + 2i
\][/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, meaning it can have a total of 7 zeros (real or complex). We already identified 6 non-real zeros (3 given and their 3 conjugates).
The number of non-real zeros:
[tex]\[
6
\][/tex]
Since the polynomial is of degree 7, the remaining [tex]\( 7 - 6 = 1 \)[/tex] zero must be real.
Maximum number of real zeros:
[tex]\[
1
\][/tex]
### (c) What is the maximum number of nonreal zeros that [tex]\( R(x) \)[/tex] can have?
From part (b), we have already determined that out of the 7 zeros, 6 are non-real because they appear in conjugate pairs. This is the maximum number of non-real zeros for this polynomial.
Maximum number of nonreal zeros:
[tex]\[
6
\][/tex]
To summarize:
- (a) Find another zero of [tex]\( R(x) \)[/tex].
[tex]\[
i, -1 + 5i, 3 + 2i
\][/tex]
- (b) What is the maximum number of real zeros that [tex]\( R(x) \)[/tex] can have?
[tex]\[
1
\][/tex]
- (c) What is the maximum number of nonreal zeros that [tex]\( R(x) \)[/tex] can have?
[tex]\[
6
\][/tex]
