K Find a polynomial function of degree 3 with the given numbers as zeros. Assume that the leading coefficient is 1. -3, 4i, -4i The polynomial function is f(x)= (Simplify your answer. Use integers or fractions for any numbers in the expression.)
K Find a polynomial function of degree 3 with the given numbers as zeros. Assume that the leading coefficient is 1. -3, 4i, -4i The polynomial function is f(x)= (Simplify your answer. Use integers or fractions for any numbers in the expression.)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Question
![**Task:**
Find a polynomial function of degree 3 with the given numbers as zeros. Assume that the leading coefficient is 1.
**Zeros:**
\( -3, 4i, -4i \)
**Instructions:**
The polynomial function is \( f(x) = \, \_\_ \)
(Simplify your answer. Use integers or fractions for any numbers in the expression.)
---
**Explanation:**
To construct the polynomial, use the fact that if \( a \) is a zero, then \( (x-a) \) is a factor of the polynomial. Since complex zeros occur in conjugate pairs, both \( 4i \) and \( -4i \) must be used.
### Factors:
- \( (x + 3) \) for the zero \( -3 \)
- \( (x - 4i) \) for the zero \( 4i \)
- \( (x + 4i) \) for the zero \( -4i \)
### Polynomial Expansion:
1. Multiply the complex factors:
\[
(x - 4i)(x + 4i) = x^2 - (4i)^2 = x^2 + 16
\]
(since \( (i)^2 = -1 \) and thus \( (4i)^2 = -16 \))
2. Multiply with the real factor:
\[
f(x) = (x + 3)(x^2 + 16)
\]
Expand further:
\[
f(x) = x(x^2 + 16) + 3(x^2 + 16) \\
= x^3 + 16x + 3x^2 + 48
\]
3. Simplify:
\[
f(x) = x^3 + 3x^2 + 16x + 48
\]
Thus, the polynomial function is:
\[ f(x) = x^3 + 3x^2 + 16x + 48 \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdf429993-eecf-4773-959c-7bbde76a8695%2Fc26adbac-c207-4cf4-91d1-71e3b0a043f2%2Fy6lzsd9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Task:**
Find a polynomial function of degree 3 with the given numbers as zeros. Assume that the leading coefficient is 1.
**Zeros:**
\( -3, 4i, -4i \)
**Instructions:**
The polynomial function is \( f(x) = \, \_\_ \)
(Simplify your answer. Use integers or fractions for any numbers in the expression.)
---
**Explanation:**
To construct the polynomial, use the fact that if \( a \) is a zero, then \( (x-a) \) is a factor of the polynomial. Since complex zeros occur in conjugate pairs, both \( 4i \) and \( -4i \) must be used.
### Factors:
- \( (x + 3) \) for the zero \( -3 \)
- \( (x - 4i) \) for the zero \( 4i \)
- \( (x + 4i) \) for the zero \( -4i \)
### Polynomial Expansion:
1. Multiply the complex factors:
\[
(x - 4i)(x + 4i) = x^2 - (4i)^2 = x^2 + 16
\]
(since \( (i)^2 = -1 \) and thus \( (4i)^2 = -16 \))
2. Multiply with the real factor:
\[
f(x) = (x + 3)(x^2 + 16)
\]
Expand further:
\[
f(x) = x(x^2 + 16) + 3(x^2 + 16) \\
= x^3 + 16x + 3x^2 + 48
\]
3. Simplify:
\[
f(x) = x^3 + 3x^2 + 16x + 48
\]
Thus, the polynomial function is:
\[ f(x) = x^3 + 3x^2 + 16x + 48 \]
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