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Suppose that [tex]$R(x)$[/tex] is a polynomial of degree 12 whose coefficients are real numbers. Also, suppose that [tex]$R(x)$[/tex] has the following zeros:
[tex]\[ -4, \quad -4+i, \quad 3-2i, \quad 3+2i \][/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 :

GinnyAnsweridn
Let's go through each part of the problem step by step.

### Part (a):

Since the polynomial [tex]\(R(x)\)[/tex] has real coefficients, any complex zeros must come in conjugate pairs. The zeros given are:
[tex]\[ -4, \quad -4+i, \quad 3-2i, \quad 3+2i \][/tex]

For every complex number [tex]\(a + bi\)[/tex] (where [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are real numbers), the complex conjugate [tex]\(a - bi\)[/tex] must also be a zero. Given that [tex]\(-4 + i\)[/tex] is a zero, its conjugate [tex]\(-4 - i\)[/tex] must also be a zero.

Thus, another zero of [tex]\(R(x)\)[/tex] is:
[tex]\[ -4 - i \][/tex]

### Part (b):

To determine the maximum number of real zeros, we need to list all the zeros we have so far:
[tex]\[ -4, \quad -4+i, \quad 3-2i, \quad 3+2i, \quad -4-i \][/tex]

Here, the real zeros we have are:
[tex]\[ -4 \][/tex]

So far, we have only one real zero ([tex]\(-4\)[/tex]). Since the polynomial is of degree 12, there could be additional real zeros, but we need to account for the complex zeros already given and their conjugates. Each distinct complex zero pair accounts for two of the polynomial's roots. Therefore, with the zeros given and their conjugates, the real zeros currently are:
[tex]\[ -4 \][/tex]
So, the maximum number of real zeros is:
[tex]\[ 1 \][/tex]

### Part (c):

The polynomial [tex]\(R(x)\)[/tex] is of degree 12, and the number of nonreal zeros is given by the complex pairs mentioned above. Each pair of complex zeros contributes to 2 nonreal zeros. We listed the zeros (including their conjugates) as follows:
[tex]\[ -4, \quad -4+i, \quad 3-2i, \quad 3+2i, \quad -4-i \][/tex]

The nonreal zeros are:
[tex]\[ -4+i, \quad 3-2i, \quad 3+2i, \quad -4-i \][/tex]

Counting each of these reveals that there are 4 distinct nonreal zeros.

Because each complex pair comprises 2 roots of the polynomial, the number of nonreal zeros doubles:
[tex]\[ (4 \text{ nonreal zeros}) \times 2 = 8 \][/tex]

Thus, the maximum number of nonreal zeros is:
[tex]\[ 8 \][/tex]

### Final Answer:
(a) Another zero of [tex]\(R(x)\)[/tex] is [tex]\(-4 - i\)[/tex].

(b) The maximum number of real zeros that [tex]\(R(x)\)[/tex] can have is [tex]\(1\)[/tex].

(c) The maximum number of nonreal zeros that [tex]\(R(x)\)[/tex] can have is [tex]\(8\)[/tex].
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