Chapter 5: Problem 65
Solve each equation in the real number system. $$ x^{3}-\frac{2}{3} x^{2}+\frac{8}{3} x+1=0 $$
Short Answer
Expert verified
No rational root; use numerical methods for real roots.
Step by step solution
01
Recognize the polynomial
The given equation is a cubic polynomial of the form \[ x^{3}-\frac{2}{3} x^{2}+\frac{8}{3} x+1=0 \]. First, identify the standard form of the polynomial.
02
Use Rational Root Theorem
To find possible rational roots, use the Rational Root Theorem. The theorem states that any rational solution, \( p/q \), of \( a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0 = 0 \) is a factor of the constant term \( a_0 \) divided by a factor of the leading coefficient \( a_n \).Here, the constant term \( a_0 = 1 \) and the leading coefficient \( a_n = 1 \). Therefore, the possible rational roots are \( \pm 1 \).
03
Test possible roots
Substitute the possible roots into the polynomial to find the actual root:\[ f(1) = 1^3 - \frac{2}{3} \cdot 1^2 + \frac{8}{3} \cdot 1 + 1 = 1 - \frac{2}{3} + \frac{8}{3} + 1 = \frac{10}{3} eq 0 \]\[ f(-1) = (-1)^3 - \frac{2}{3} \cdot (-1)^2 + \frac{8}{3} \cdot (-1) + 1 = -1 - \frac{2}{3} - \frac{8}{3} + 1 = -\frac{12}{3} = -4 eq 0 \]Neither of these give us zero, so there are no rational roots.
04
Use Any Other Method
Since the Rational Root Theorem does not give us a rational root, proceed using numerical methods or factorization techniques such as synthetic division or the cubic formula, or transform it if possible (though it can be complex).
05
Confirmation
To completely solve using the applicable method and confirm real roots, ensure all steps for the chosen method are accurately followed through calculations. Given example may need numerical approximation like Newton's method if algebraic factoring is too complex.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Polynomial Equations
A polynomial equation is an expression consisting of variables and coefficients, involving terms where the variables are raised to whole number exponents. The general form of a polynomial equation is: $$a_nx^n + a_{n-1}x^{n-1} + \textellipsis + a_1x + a_0 = 0$$ In this exercise, we are given a cubic polynomial equation: $$x^3 - \frac{2}{3} x^2 + \frac{8}{3} x + 1 = 0$$ A cubic polynomial, being of degree 3, means the highest power of the variable is 3. To solve this, we often look for rational roots using some theorems and methods.
Rational Root Theorem
The Rational Root Theorem is a useful tool for finding possible rational solutions to a polynomial equation. It tells us that any rational solution of a polynomial equation $$a_nx^n + a_{n-1}x^{n-1} + \textellipsis + a_1x + a_0 = 0$$ can be expressed as $$\frac{p}{q}$$ where:
- \(p\) is a factor of the constant term \(a_0\)
- \(q\) is a factor of the leading coefficient \(a_n\)
Numerical Methods
When rational roots are not found, we proceed with numerical methods to find approximate solutions to polynomial equations. Numerical methods involve iterative algorithms that can yield highly accurate solutions. Common methods include:
- Newton's Method
- Bisection Method
- Secant Method
Synthetic Division
Synthetic division is a streamlined method for dividing a polynomial by a binomial of the form \(x-c\). This method simplifies the process of polynomial division and is useful for finding roots or factors. Here's the procedure in steps:
- Write down the coefficients of the polynomial.
- Use a potential root according to Rational Root Theorem.
- Perform synthetic division to test this root.
- If the remainder is zero, the root is valid.
Newton's Method
Newton's Method, also known as the Newton-Raphson Method, is an iterative technique to approximate the roots of a real-valued function. Here’s a step-by-step guide:
- Start with an initial guess \(x_0\) for the root.
- Compute the function value and its derivative at \(x_0\).
- Update the guess using the formula: $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$
- Repeat this process until \(x_n\) converges to a stable value.