Welcome to Westonci.ca, where finding answers to your questions is made simple by our community of experts. Get immediate and reliable answers to your questions from a community of experienced experts on our platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

Suppose that [tex]$R(x)$[/tex] is a polynomial of degree 7 whose coefficients are real numbers. Also, suppose that [tex]$R(x)$[/tex] has the following zeros:
[tex]$
-i, \quad -1-5i, \quad 3-2i
$[/tex]

Answer the following:

(a) Find another zero of [tex]$R(x)$[/tex].
[tex]$ \square $[/tex]

(b) What is the maximum number of real zeros that [tex]$R(x)$[/tex] can have?
[tex]$ \square $[/tex]

(c) What is the maximum number of nonreal zeros that [tex]$R(x)$[/tex] can have?
[tex]$ \square $[/tex]


Sagot :

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]
Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.