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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]-3+i[/tex], [tex]4i[/tex].

Answer the following:

(a) Find another zero of [tex][tex]$R(x)$[/tex][/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 non-real zeros that [tex]$R(x)$[/tex] can have? [tex]\(\square\)[/tex]

Sagot :

GinnyAnswer
Let's solve the problem step-by-step.

### Given Information:
- Degree of polynomial [tex]\( R(x) \)[/tex] is 7.
- Zeros: [tex]\( -3+i \)[/tex] and [tex]\( 4i \)[/tex].
- Polynomial has real coefficients.

### Understanding the consequences:
When a polynomial has real coefficients, any nonreal complex zeros must occur in conjugate pairs. Therefore:
- If [tex]\( -3+i \)[/tex] is a zero, then its complex conjugate [tex]\( -3-i \)[/tex] must also be a zero.
- Similarly, if [tex]\( 4i \)[/tex] is a zero, then its complex conjugate [tex]\( -4i \)[/tex] must also be a zero.

### Step-by-Step Solution:

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

Given zeros: [tex]\( -3+i \)[/tex] and [tex]\( 4i \)[/tex].

By conjugate pairs:
- For [tex]\( -3+i \)[/tex], another zero is [tex]\( -3-i \)[/tex].
- For [tex]\( 4i \)[/tex], another zero is [tex]\( -4i \)[/tex].

Thus, the other zeros are:
[tex]\[ -3-i \quad \text{and} \quad -4i \][/tex]

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

Since the polynomial is of degree 7, the total number of zeros (including their multiplicities) is 7.

We have identified the following nonreal zeros (each pair counts as 2 zeros):
- [tex]\( -3+i \)[/tex] and [tex]\( -3-i \)[/tex] (2 zeros)
- [tex]\( 4i \)[/tex] and [tex]\( -4i \)[/tex] (2 zeros)

Thus, we have 4 nonreal zeros in total. This leaves:
[tex]\[ 7 - 4 = 3 \][/tex]

Therefore, the maximum number of real zeros the polynomial can have is 3.

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

Each nonreal zero must occur in conjugate pairs, so for the maximum number of nonreal zeros, we want the maximum number of such pairs.

Since each pair of nonreal zeros counts as 2 zeros, the number of nonreal zeros must be even and lie within the constraint of the polynomial's degree, which is 7.

Given that the maximum number of real zeros is 3, the remaining zeros would all be nonreal:
[tex]\[ 7 - 3 = 4 \][/tex]

Thus, the maximum number of nonreal zeros the polynomial can have is 4.

### Summary of Answers:
(a) Other zeros: [tex]\(-3-i\)[/tex] and [tex]\(-4i\)[/tex].
(b) Maximum number of real zeros: 3.
(c) Maximum number of nonreal zeros: 4.
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