Problem 55 - Algebra and Trigonometry

Question

\(55-58\) m Find all real solutions of the equation, rounded to two decimals. $$ x^{3}-2 x^{2}-x-1=0 $$

Step-by-Step Solution

Answer

The real solution is approximately \(x = 2.56\).

Step 1: Understand the Problem Statement

The problem is to find the real roots of the polynomial equation \( x^3 - 2x^2 - x - 1 = 0 \), which means we are looking for values of \(x\) that satisfy the equation.

Step 2: Factor the Polynomial or Use the Rational Root Theorem

To factor the polynomial or use the Rational Root Theorem, identify potential rational roots by considering the factors of the constant term (-1) and the leading coefficient (1). The potential rational roots are \( \pm 1 \).

Step 3: Test Potential Roots

Substitute the potential rational roots found in Step 2 into the polynomial:- For \(x=1\), \(1^3 - 2\times1^2 - 1 - 1 = -3\), which is not zero.- For \(x=-1\), \((-1)^3 - 2\times(-1)^2 - (-1) - 1 = -3\), which is not zero.Since neither are roots, the equation might need numerical methods.

Step 4: Use Numerical Methods

Since the polynomial does not factor easily and the Rational Root Theorem did not work, use numerical methods like the Newton-Raphson method or a graphing calculator to approximate the roots. Using a calculator, the real root is approximately \(x = 2.56\).

Step 5: Verify the Solution

Insert \(x = 2.56\) back into the original equation to verify the solution:\((2.56)^3 - 2(2.56)^2 - 2.56 - 1 \approx 0\).The result is very close to zero, confirming \(x = 2.56\) is a valid approximate solution.

Key Concepts

Polynomial Equation

A polynomial equation is an expression of mathematical terms that involves variables raised to whole number powers and coefficients. In this exercise, the polynomial equation is \( x^3 - 2x^2 - x - 1 = 0 \). Let's break this down:

• The term \( x^3 \) is the cube of \( x \), which is the highest power, making it a cubic polynomial.
• The polynomial has terms down to \( x \), the linear term, and a constant term, \(-1\).
• Coefficients are the numbers in front of each variable's power. Here, the coefficients are 1, -2, and -1.

In this cubic polynomial, we aim to find the roots, which are the solutions for \( x \) that satisfy the equation set equal to zero. These roots are where the polynomial equation equals zero when graphed on the xy-plane.
The challenge lies in finding these roots as they may not always be whole numbers or intuitive to find.

Rational Root Theorem

The Rational Root Theorem is a handy tool that suggests possible rational roots of a polynomial equation. This theorem states that any potential rational root, \( p/q \), of a polynomial must have \( p \) as a factor of the constant term and \( q \) as a factor of the leading coefficient.For our polynomial \( x^3 - 2x^2 - x - 1 \):

• The constant term is \(-1\).
• The leading coefficient is \(1\).
• The potential rational roots are therefore \( \pm 1 \), because these are the factors of both the constant term and the leading coefficient.

Testing these potential roots by substituting them back into the polynomial showed neither \( x = 1 \) nor \( x = -1 \) results in zero, indicating they are not roots.
When the Rational Root Theorem doesn't yield results, it signals that we might need more advanced methods to uncover the real roots.

Numerical Methods

Numerical methods are mathematical procedures used to find approximate solutions to complex equations that aren't easily solved by algebra alone. When methods like the Rational Root Theorem fail, numerical techniques come to the rescue.For example, the Newton-Raphson method is a popular numerical technique for finding roots. This iterative method uses calculus to hone in on a root's precise location by starting with an initial guess and refining:

• An initial estimate is made for the root.
• The derivative of the polynomial, \( f'(x) \), is calculated.
• The estimate is improved using the formula \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \).
• These steps are repeated until the value converges to a stable root.

In this problem, using such methods or a graphing calculator helped identify an approximate real root, \( x = 2.56 \).
Verifying this solution by substituting it back into the polynomial, we find that it holds true as it closely results in zero, wrapping up this journey to find the polynomial equation's real root.