Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![**Problem 5: Polynomial with Integer Coefficients**
**Objective:**
Find a 4th degree polynomial \( f(x) \) with integer coefficients, such that \( \sqrt{2}i, -1, \) and \( 2 \) are roots of \( f(x) \) and \( f(1) = 12 \).
**Solution Steps:**
1. **Identify the Roots:**
Given the roots are \( \sqrt{2}i, -1, \) and \( 2 \). For a polynomial with integer coefficients, non-real complex roots must occur in conjugate pairs. Hence, if \( \sqrt{2}i \) is a root, its conjugate \( -\sqrt{2}i \) is also a root.
2. **Constructing the Polynomial:**
Considering the roots, the polynomial can be written as:
\[
f(x) = a(x - \sqrt{2}i)(x + \sqrt{2}i)(x + 1)(x - 2)
\]
Simplify the complex conjugate part:
\[
(x - \sqrt{2}i)(x + \sqrt{2}i) = x^2 - (\sqrt{2}i)^2 = x^2 + 2
\]
Thus,
\[
f(x) = a(x^2 + 2)(x + 1)(x - 2)
\]
3. **Finding the Coefficient \( a \):**
\( f(x) \) must satisfy \( f(1) = 12 \). Substituting \( x = 1 \) in the polynomial:
\[
f(1) = a(1^2 + 2)(1 + 1)(1 - 2)
\]
Simplify:
\[
f(1) = a(3)(2)(-1) = -6a
\]
Given \( f(1) = 12 \):
\[
-6a = 12 \implies a = -2
\]
4. **Final Polynomial:**
Substituting \( a = -2 \) back into \( f(x) \):
\[
f(x) = -2(x^2 + 2)(x + 1)(x](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3558762b-56d3-4a19-8c4b-b2d15f9a070e%2F6dcce838-eeca-47ff-8d16-a01e40359a24%2F4ahuyl3.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem 5: Polynomial with Integer Coefficients**
**Objective:**
Find a 4th degree polynomial \( f(x) \) with integer coefficients, such that \( \sqrt{2}i, -1, \) and \( 2 \) are roots of \( f(x) \) and \( f(1) = 12 \).
**Solution Steps:**
1. **Identify the Roots:**
Given the roots are \( \sqrt{2}i, -1, \) and \( 2 \). For a polynomial with integer coefficients, non-real complex roots must occur in conjugate pairs. Hence, if \( \sqrt{2}i \) is a root, its conjugate \( -\sqrt{2}i \) is also a root.
2. **Constructing the Polynomial:**
Considering the roots, the polynomial can be written as:
\[
f(x) = a(x - \sqrt{2}i)(x + \sqrt{2}i)(x + 1)(x - 2)
\]
Simplify the complex conjugate part:
\[
(x - \sqrt{2}i)(x + \sqrt{2}i) = x^2 - (\sqrt{2}i)^2 = x^2 + 2
\]
Thus,
\[
f(x) = a(x^2 + 2)(x + 1)(x - 2)
\]
3. **Finding the Coefficient \( a \):**
\( f(x) \) must satisfy \( f(1) = 12 \). Substituting \( x = 1 \) in the polynomial:
\[
f(1) = a(1^2 + 2)(1 + 1)(1 - 2)
\]
Simplify:
\[
f(1) = a(3)(2)(-1) = -6a
\]
Given \( f(1) = 12 \):
\[
-6a = 12 \implies a = -2
\]
4. **Final Polynomial:**
Substituting \( a = -2 \) back into \( f(x) \):
\[
f(x) = -2(x^2 + 2)(x + 1)(x
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