Problem 25 - College Algebra with Corequisite Support

Question

For the following exercises, use the Rational Zero Theorem to find the real solution(s) to each equation. $$ x^{3}+5 x^{2}-16 x-80=0 $$

Step-by-Step Solution

Answer

The real solutions are \(x = -5, 4, -4\).

Step 1: Rational Zero Theorem

The Rational Zero Theorem states that any rational solution of a polynomial equation \(a_nx^n + a_{n-1}x^{n-1} + ... + a_0 = 0\) is a fraction \(\frac{p}{q}\), where \(p\) is a factor of the constant term \(a_0\) and \(q\) is a factor of the leading coefficient \(a_n\). For the given equation \(x^3 + 5x^2 - 16x - 80 = 0\), the constant term is \(-80\) and the leading coefficient is \(1\). Hence, the potential rational zeros are the factors of \(-80\).

Step 2: List the Possible Rational Zeros

The factors of \(-80\) are \(\pm 1, \pm 2, \pm 4, \pm 5, \pm 8, \pm 10, \pm 16, \pm 20, \pm 40, \pm 80\). Since the leading coefficient of \(x^3\) is 1, these numbers are the possible rational solutions. Thus, we have \(\pm 1, \pm 2, \pm 4, \pm 5, \pm 8, \pm 10, \pm 16, \pm 20, \pm 40, \pm 80\) as candidates for rational roots.

Step 3: Test Possible Zeros Using Synthetic Division

Test each possible rational zero by using synthetic division to determine if it is a root of the polynomial. For example, testing \(x = -5\):Set up the synthetic division with coefficients \(1, 5, -16, -80\):- \(-5\) | 1 5 -16 -80- | -5 0 80--------------------------- 1 0 -16 0Since the remainder is 0, \(x + 5 = 0\) is a factor, and \(x = -5\) is a root.

Step 4: Factor the Polynomial

After finding \(x = -5\) is a root, we can factor the polynomial using the result from synthetic division. The quotient is \(x^2 - 16\), so the polynomial is expressed as \((x+5)(x^2 - 16)\).

Step 5: Solve for Remaining Roots

Set the remaining factor \(x^2 - 16 = 0\).This can be factored further as \((x - 4)(x + 4) = 0\), giving additional roots \(x = 4\) and \(x = -4\).

Step 6: List All Real Solutions

The real solutions obtained from the polynomial equation includes \(x = -5\), \(x = 4\), and \(x = -4\).

Key Concepts

Polynomial Equations

A polynomial equation is fundamentally an equation that involves a polynomial expression. A polynomial itself is an algebraic expression composed of variables and coefficients, grouped together using addition, subtraction, and multiplication. The equation sets the polynomial equal to a specific value, often zero. For example, the equation \(x^3 + 5x^2 - 16x - 80 = 0\) represents a polynomial equation where the highest power of \(x\) indicates it is a cubic polynomial. Understanding polynomial equations:

• The highest exponent of the variable is called the 'degree' of the polynomial; here, the degree is 3.
• Polynomial equations can have multiple solutions, potentially as many as the degree of the polynomial.
• The solutions could be real or complex numbers, and finding the real solutions often involves techniques such as factoring or applying the Rational Zero Theorem.

The Rational Zero Theorem is particularly handy here, as it allows for the identification of potential rational solutions through factoring the constant term and the leading coefficient.

Synthetic Division

Synthetic division is a streamlined method for dividing a polynomial by a linear divisor of the form \(x - c\). It simplifies the division process by using only the coefficients of the polynomial. This technique is economical both in terms of calculations and space.How it Works:

• Write down the coefficients of the polynomial.
• Select a potential zero (a value from the Rational Zero Theorem).
• Use this value in the synthetic division setup to test if it yields a zero remainder.

For example, in the polynomial \(x^3 + 5x^2 - 16x - 80 = 0\), when trying \(x = -5\):

• List coefficients: 1, 5, -16, -80.
• Add results as you multiply the value by each successive new row value.
• If the remainder (last number) is 0, it indicates \(x - c=0\) is indeed a factor.

This helps break down complex polynomials into simpler, more manageable forms, making the roots easier to find.

Factoring Polynomials

Factoring polynomials involves expressing a polynomial as the product of its factors. This is crucial when solving polynomial equations because it simplifies finding roots. The primary goal is to break the equation into simpler parts, revealing solutions more directly.Steps to Factor:

• Identify a factor using methods like synthetic division or by recognizing patterns.
• Express the polynomial as a product of this factor and a remaining polynomial.
• Continue factoring the resulting polynomial until all parts are sufficiently simplified.

Using the example \((x^3 + 5x^2 - 16x - 80)\), once synthetic division confirms \(x = -5\) as a root, you factor it as \((x + 5)(x^2 - 16)\).

• The next step involves breaking down \(x^2 - 16\) further into \((x - 4)(x + 4)\) using the difference of squares.

Thus, the original polynomial can be expressed as \((x + 5)(x - 4)(x + 4)\), revealing the roots: \(-5\), \(4\), and \(-4\). Factoring in this manner simplifies the equation, making solutions readily accessible.