Problem 11 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 11-14: Show that the equation has no rational root. $$ x^{3}+3 x^{2}-4 x+6=0 $$

Step-by-Step Solution

Answer

The equation has no rational roots as none of the possible rational roots satisfy the polynomial.

Step 1: Identify the Rational Root Theorem

The Rational Root Theorem states that if a polynomial has a rational root \( \frac{p}{q} \), then \( p \) is a factor of the constant term and \( q \) is a factor of the leading coefficient. In our polynomial \( x^3 + 3x^2 - 4x + 6 \), the constant term is 6 and the leading coefficient is 1. Therefore, potential rational roots are the factors of 6.

Step 2: List Possible Rational Roots

The possible rational roots, according to the Rational Root Theorem, are the factors of the constant term 6 divided by the factors of the leading coefficient 1. These factors include: \( \pm 1, \pm 2, \pm 3, \pm 6 \).

Step 3: Test Possible Roots with Synthetic Division

Use synthetic division to test each possible rational root. If division results in a remainder of zero, the number is a rational root. Start with testing \( x = 1 \):\[ 1 | 1 \quad 3 \quad -4 \quad 6 \]\[ \quad | \quad \quad 1 \quad 4 \quad 0 \]\[ \quad | 1 \quad 4 \quad 0 \quad 6 \]The remainder is 6, thus not zero, meaning \( x = 1 \) is not a root. Repeat this process for \( -1, 2, -2, 3, -3, 6, -6 \). All yield a remainder other than 0.

Step 4: Conclude No Rational Roots Exist

Since no tested factors of 6 divided by the factors of 1 resulted in a remainder of zero during synthetic division, there are no rational roots for the equation \( x^3 + 3x^2 - 4x + 6 = 0 \).

Key Concepts

Polynomial Equations

A polynomial equation is an expression set to equal zero that consists of variables and coefficients combined using addition, subtraction, multiplication, and non-negative integer exponents. These equations are prevalent in algebra and come in various degrees depending on the highest power of the variable. The general form of a polynomial equation is:
\[ a_n x^n + a_{n-1} x^{n-1} + ext{...} + a_2 x^2 + a_1 x + a_0 = 0 \]
In this form, \(a_n, a_{n-1}, ..., a_0\) are constants, and the highest exponent \(n\) determines the degree of the polynomial. For instance, our given equation \[ x^3 + 3x^2 - 4x + 6 = 0 \] is a third-degree polynomial since the highest power of \(x\) is three. Understanding polynomial equations involves recognizing their terms, constant coefficients, and how to explore their roots.

Synthetic Division

Synthetic division is a simplified method of dividing polynomials, particularly useful when testing potential roots of a polynomial equation. This technique offers a streamlined alternative to the traditional long division approach, especially when dividing by a linear factor of the form \(x - c\). To carry out synthetic division, you need:

• A polynomial, which in our case is \(x^3 + 3x^2 - 4x + 6\).
• A potential root, such as the numbers derived from the Rational Root Theorem.

Start by writing the coefficients of the polynomial in a horizontal line. If you are testing \(x = c\), list it to the left. Perform the division operation by bringing down the leading coefficient as is. Multiply it by the root and add horizontally moving across coefficients. If you end with a zero as the remainder, \(c\) is a root.
For example, testing \(x = 1\) involves using the coefficients 1, 3, -4, and 6, but yielded a remainder of 6, indicating it's not a root. Repeat similar steps with all candidates from the Rational Root Theorem to identify or rule out rational roots.

Rational Roots

Rational roots refer to the solutions of a polynomial equation that can be expressed as a fraction \(\frac{p}{q}\). These roots are specifically relevant to the Rational Root Theorem, a valuable tool in identifying possible simple, rational solutions to address and test in polynomial equations. According to the theorem:

• \(p\) is a factor of the constant term of the polynomial.
• \(q\) is a factor of the leading coefficient.

In our case, the constant term is 6 and the leading coefficient is 1. Therefore, the potential rational roots can be any factor of 6. This list includes \( \pm 1, \pm 2, \pm 3, \pm 6 \). By applying synthetic division, each of these roots was tested in our polynomial equation \(x^3 + 3x^2 - 4x + 6\). Unfortunately, none of these trials resulted in a remainder of zero, leading us to conclude there are no rational roots for this equation.