Problem 12 - Precalculus

Question

Show that each equation has no rational roots. $$x^{3}+8 x^{2}-1=0$$

Step-by-Step Solution

Answer

The equation has no rational roots.

Step 1: Identify Possible Rational Roots

The Rational Root Theorem states that any rational solution of a polynomial equation is a fraction \( \frac{p}{q} \), where \( p \) is a factor of the constant term, and \( q \) is a factor of the leading coefficient. For the equation \( x^3 + 8x^2 - 1 = 0 \), the constant term is \(-1\) and the leading coefficient is \(1\). Therefore, the possible rational roots are \( \pm 1 \).

Step 2: Test Possible Rational Roots

Now we test the possible rational roots identified in the previous step. Substitute \( x = 1 \) into the equation: \[ 1^3 + 8 \cdot 1^2 - 1 = 1 + 8 - 1 = 8 eq 0 \] Now, substitute \( x = -1 \) into the equation: \[ (-1)^3 + 8 \cdot (-1)^2 - 1 = -1 + 8 - 1 = 6 eq 0 \] Neither \( x = 1 \) nor \( x = -1 \) is a root of the equation.

Step 3: Conclude on Rational Roots

Since neither of the possible rational roots is an actual root of the equation, we conclude that \( x^3 + 8x^2 - 1 = 0 \) has no rational roots.

Key Concepts

Polynomials

Polynomials are mathematical expressions involving a sum of powers in one or more variables multiplied by coefficients. The basic form of a polynomial in one variable, say \( x \), is given as: \[ a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 \] Here, \( a_n, a_{n-1}, \ldots, a_0 \) are coefficients, and \( n \) is a non-negative integer representing the degree of the polynomial. The degree is determined by the highest power of \( x \) present in the polynomial.
In the equation \( x^3 + 8x^2 - 1 = 0 \), we observe it is a cubic polynomial due to the highest power being 3. Each term is a component of this polynomial, such as \( x^3 \) and \( 8x^2 \), with their respective coefficients. They are main elements to identify when analyzing the behavior and roots of the polynomial.

Rational Roots

Rational roots are solutions to a polynomial equation that can be expressed as a fraction, where both the numerator and denominator are integers. The Rational Root Theorem helps us determine potential rational roots based on the polynomial's coefficients. According to this theorem, any rational root of a polynomial \( \frac{p}{q} \) is such that \( p \) is a factor of the constant term and \( q \) is a factor of the leading coefficient.
In the polynomial \( x^3 + 8x^2 - 1 = 0 \), the constant term is \(-1\), and the leading coefficient is \(1\). Therefore, the possible rational roots are limited to factors of \(-1\) over factors of \(1\), giving us potential roots such as \( \pm 1 \). These potential roots are the candidates we then test to determine if any are actual solutions of the polynomial.

Factor Theorem

The Factor Theorem is pivotal in understanding polynomials and their roots. It provides a connection between the roots of a polynomial and its factors. According to this theorem, a polynomial \( f(x) \) has a factor \((x - c)\) if and only if \( c \) is a root of \( f(x) = 0 \). This means when you substitute \( x = c \) into the polynomial, the result should be zero.
In the context of our equation \( x^3 + 8x^2 - 1 = 0 \), we tested the potential rational roots \( x = 1 \) and \( x = -1 \). Upon substitution, both resulted in non-zero values, indicating they are not roots. Thus, \( (x - 1) \) and \((x + 1) \) are not factors of the polynomial. The Factor Theorem confirms our findings, helping cement that there are no rational roots in this polynomial equation.