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,...
Related questions
Question
Solve for X
![The equation provided is a cubic polynomial equation:
\[ x^3 - 6x^2 + 5x = 0 \]
This is a standard form of a cubic equation where:
- The highest degree term is \( x^3 \).
- The equation is set equal to zero, indicating solutions for \( x \).
### Solving the Equation
To solve for \( x \), we can factor the equation:
1. Factor out the common term \( x \):
\[ x(x^2 - 6x + 5) = 0 \]
2. Solve for \( x \) by setting each factor equal to zero:
- \( x = 0 \)
- \( x^2 - 6x + 5 = 0 \)
3. Solve the quadratic equation \( x^2 - 6x + 5 = 0 \) using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -6 \), and \( c = 5 \).
4. Calculate the discriminant:
\[ b^2 - 4ac = (-6)^2 - 4 \cdot 1 \cdot 5 = 36 - 20 = 16 \]
5. Find the roots:
\[ x = \frac{6 \pm \sqrt{16}}{2} \]
\[ x = \frac{6 \pm 4}{2} \]
- \( x = \frac{6 + 4}{2} = 5 \)
- \( x = \frac{6 - 4}{2} = 1 \)
### Solutions
The solutions to the equation are:
- \( x = 0 \)
- \( x = 1 \)
- \( x = 5 \)
The roots of the equation indicate the points where the curve \( y = x^3 - 6x^2 + 5x \) intersects the \( x \)-axis.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5ae18fd2-7331-48d4-90a1-bc335103e09b%2F594393ce-87d9-4807-84f8-fa37c27d8390%2F88jzsv8_processed.png&w=3840&q=75)
Transcribed Image Text:The equation provided is a cubic polynomial equation:
\[ x^3 - 6x^2 + 5x = 0 \]
This is a standard form of a cubic equation where:
- The highest degree term is \( x^3 \).
- The equation is set equal to zero, indicating solutions for \( x \).
### Solving the Equation
To solve for \( x \), we can factor the equation:
1. Factor out the common term \( x \):
\[ x(x^2 - 6x + 5) = 0 \]
2. Solve for \( x \) by setting each factor equal to zero:
- \( x = 0 \)
- \( x^2 - 6x + 5 = 0 \)
3. Solve the quadratic equation \( x^2 - 6x + 5 = 0 \) using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -6 \), and \( c = 5 \).
4. Calculate the discriminant:
\[ b^2 - 4ac = (-6)^2 - 4 \cdot 1 \cdot 5 = 36 - 20 = 16 \]
5. Find the roots:
\[ x = \frac{6 \pm \sqrt{16}}{2} \]
\[ x = \frac{6 \pm 4}{2} \]
- \( x = \frac{6 + 4}{2} = 5 \)
- \( x = \frac{6 - 4}{2} = 1 \)
### Solutions
The solutions to the equation are:
- \( x = 0 \)
- \( x = 1 \)
- \( x = 5 \)
The roots of the equation indicate the points where the curve \( y = x^3 - 6x^2 + 5x \) intersects the \( x \)-axis.
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 3 steps with 3 images
Recommended textbooks for you
Algebra and Trigonometry (6th Edition)
Algebra
ISBN:
9780134463216
Author:
Robert F. Blitzer
Publisher:
PEARSON
Contemporary Abstract Algebra
Algebra
ISBN:
9781305657960
Author:
Joseph Gallian
Publisher:
Cengage Learning
Linear Algebra: A Modern Introduction
Algebra
ISBN:
9781285463247
Author:
David Poole
Publisher:
Cengage Learning
Algebra and Trigonometry (6th Edition)
Algebra
ISBN:
9780134463216
Author:
Robert F. Blitzer
Publisher:
PEARSON
Contemporary Abstract Algebra
Algebra
ISBN:
9781305657960
Author:
Joseph Gallian
Publisher:
Cengage Learning
Linear Algebra: A Modern Introduction
Algebra
ISBN:
9781285463247
Author:
David Poole
Publisher:
Cengage Learning
Algebra And Trigonometry (11th Edition)
Algebra
ISBN:
9780135163078
Author:
Michael Sullivan
Publisher:
PEARSON
Introduction to Linear Algebra, Fifth Edition
Algebra
ISBN:
9780980232776
Author:
Gilbert Strang
Publisher:
Wellesley-Cambridge Press
College Algebra (Collegiate Math)
Algebra
ISBN:
9780077836344
Author:
Julie Miller, Donna Gerken
Publisher:
McGraw-Hill Education