d) Solve the equation for all roots x' +1 = 0
Trigonometry (11th Edition)
11th Edition
ISBN:9780134217437
Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Chapter1: Trigonometric Functions
Section: Chapter Questions
Problem 1RE:
1. Give the measures of the complement and the supplement of an angle measuring 35°.
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![### Solving Polynomial Equations
**Problem Statement:**
d) Solve the equation for all roots \( x^3 + 1 = 0 \)
**Solution Steps:**
To solve the polynomial equation \( x^3 + 1 = 0 \), follow these steps:
1. **Rewrite the Equation:**
\( x^3 + 1 = 0 \)
2. **Isolate the Polynomial:**
Subtract 1 from both sides to get:
\[
x^3 = -1
\]
3. **Factor the Left-Hand Side:**
Recognize that \( x^3 + 1 \) can be factored using the sum of cubes formula, which states:
\[
a^3 + b^3 = (a + b)(a^2 - ab + b^2)
\]
Here, \( a = x \) and \( b = 1 \). Apply the formula:
\[
x^3 + 1^3 = (x + 1)(x^2 - x \cdot 1 + 1^2)
\]
Therefore,
\[
x^3 + 1 = (x + 1)(x^2 - x + 1)
\]
4. **Set Each Factor to Zero:**
For the equation \( (x + 1)(x^2 - x + 1) = 0 \) to hold true, either \( x + 1 = 0 \) or \( x^2 - x + 1 = 0 \).
- Solve \( x + 1 = 0 \):
\[
x = -1
\]
- Solve \( x^2 - x + 1 = 0 \):
This is a quadratic equation. Use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
In this case, \( a = 1 \), \( b = -1 \), and \( c = 1 \). Substitute these values:
\[
x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}
\]
\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F39f0d4ee-d4a6-4902-9c76-b777b279390d%2F62c0029f-1673-487c-a657-5705ce78d898%2Ffzk0s8q.png&w=3840&q=75)
Transcribed Image Text:### Solving Polynomial Equations
**Problem Statement:**
d) Solve the equation for all roots \( x^3 + 1 = 0 \)
**Solution Steps:**
To solve the polynomial equation \( x^3 + 1 = 0 \), follow these steps:
1. **Rewrite the Equation:**
\( x^3 + 1 = 0 \)
2. **Isolate the Polynomial:**
Subtract 1 from both sides to get:
\[
x^3 = -1
\]
3. **Factor the Left-Hand Side:**
Recognize that \( x^3 + 1 \) can be factored using the sum of cubes formula, which states:
\[
a^3 + b^3 = (a + b)(a^2 - ab + b^2)
\]
Here, \( a = x \) and \( b = 1 \). Apply the formula:
\[
x^3 + 1^3 = (x + 1)(x^2 - x \cdot 1 + 1^2)
\]
Therefore,
\[
x^3 + 1 = (x + 1)(x^2 - x + 1)
\]
4. **Set Each Factor to Zero:**
For the equation \( (x + 1)(x^2 - x + 1) = 0 \) to hold true, either \( x + 1 = 0 \) or \( x^2 - x + 1 = 0 \).
- Solve \( x + 1 = 0 \):
\[
x = -1
\]
- Solve \( x^2 - x + 1 = 0 \):
This is a quadratic equation. Use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
In this case, \( a = 1 \), \( b = -1 \), and \( c = 1 \). Substitute these values:
\[
x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}
\]
\
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