Problem 33 - Algebra Form and Function

Question

Without solving the equation, decide how many solutions it has. $$ \left(x^{2}-4\right)(x+5)=0 $$

Step-by-Step Solution

Answer

Based on the given equation \(\left(x^{2}-4\right)(x+5)=0\), we analyzed and factored the equation and found that it has three distinct solutions, corresponding to \(x = 2, x = -2\), and \(x = -5\).

Step 1: Analyze the equation

Observe the given equation: $$ \left(x^{2}-4\right)(x+5)=0 $$ Since the equation is already factored, we can identify the factors as \((x^2-4)\) and \((x+5)\). Our goal is to find the number of solutions, which will correspond to the number of times the equation equals zero.

Step 2: Identify the factors resulting in zero

Since we have two factors, let's examine each one separately: 1. \((x^2-4)\) 2. \((x+5)\) The equation equals zero if at least one of the factors equals zero.

Step 3: Analyze the factors

Let's analyze each factor: 1. \((x^2-4)\): This is a difference of squares; it can be factored as \((x-2)(x+2)\). Considering this factorization, we can see that \((x^2-4)\) equals zero when \(x=2\) or \(x=-2\). 2. \((x+5)\): This factor equals zero when \(x=-5\).

Step 4: Count the solutions

Given the analysis of each factor, we find three distinct values of \(x\) that make the equation equals zero: \(x=2, x=-2, x=-5\). Therefore, the equation has 3 solutions.

Key Concepts

Factoring

Factoring is a mathematical process where we express a polynomial as a product of its factors. It is an essential technique in solving quadratic equations, especially when working with polynomial expressions like the one in our exercise: \( (x^2-4)(x+5) = 0 \).
Starting with the expression \( x^2 - 4 \), we can recognize it as a difference of squares. Factoring it gives us two linear expressions: \( (x-2)(x+2) \). This reveals that \( x^2-4 \) can be broken down into simpler factors, making it easier to solve the equation.

• Identify terms that can be grouped — here, \( x^2 \) and \(-4\) are naturally grouped due to the nature of the difference of squares.
• Write the expression as a product of binomials — \( (x-2)(x+2) \).
• Apply similar steps to other factors when applicable — the factor \( x+5 \) is already simplified.

Factoring simplifies complex equations and is often the first step in finding solutions to quadratics. It's like unboxing the equation to see what you truly have to work with.

Zero Product Property

The Zero Product Property is a fundamental principle in algebra used for solving equations. It states that if the product of several factors is zero, then at least one of the factors must be zero. This concept is the key to finding solutions in our exercise.
For the expression \( (x^2-4)(x+5) = 0 \), we set each factor to zero, using this property to determine the values of \( x \) that satisfy the equation:

• \( x^2-4 = 0 \) leads to separate equations \( (x-2) = 0 \) and \( (x+2) = 0 \), yielding solutions \( x = 2 \) and \( x = -2 \).
• \( x+5 = 0 \) provides another solution \( x = -5 \).

By using this simple but powerful principle, we determine three distinct solutions to the equation. The Zero Product Property ensures that once we've factored the equation fully, finding solutions is straightforward by checking each factor independently.

Difference of Squares

The "difference of squares" is a particular form of polynomial, which appears frequently in algebraic expressions. This form is characterized by two squares separated by a subtraction sign, such as \( x^2 - 4 \). Recognizing the difference of squares is crucial because it has a standard factorization pattern: \[ a^2 - b^2 = (a-b)(a+b) \] Here, \( x^2 - 4 \) fits this pattern where \( a = x \) and \( b = 2 \). Therefore, it factors to \( (x-2)(x+2) \).

• Identify square terms in the equation — \( x^2 \) and \( 4 \) (since \( 4 = 2^2 \)).
• Apply the formula and break the expression into its factors.
• Recognize how this makes solving for \( x \) manageable, reducing a quadratic into simpler linear equations.

The difference of squares is a powerful shortcut that streamlines solving equations, allowing us to identify solutions quickly and efficiently. Understanding this concept helps in easily factoring and finding solutions without unnecessary complexity.