Problem 15 - Elementary Algebra

Question

For the following problems, find the domain of each of the rational expressions. $$ \frac{x-1}{x^{2}-4} $$

Step-by-Step Solution

Answer

$$ Answer: The domain of the given rational expression is all real numbers except for 2 and -2: $$ \text{Domain} = \{x \in \mathbb{R} \mid x \neq 2, -2\} $$

Step 1: Identify the denominator

First, we need to identify the denominator of the given rational expression. In this case, the denominator is: $$ x^2 - 4 $$

Step 2: Find values of x that make the denominator equal to zero

Next, we need to find the values of x that make the denominator equal to zero. To do that, we set the denominator equal to zero and solve for x: $$ x^2 - 4 = 0 $$

Step 3: Factor the quadratic equation

As the quadratic equation is a difference of squares, we can factor it easily: $$ (x - 2)(x + 2) = 0 $$

Step 4: Solve for x

Now, we solve for x in the factored equation: $$ x - 2 = 0 \Rightarrow x = 2 \\ x + 2 = 0 \Rightarrow x = -2 $$

Step 5: Determine the domain

Finally, we determine the domain by excluding the values of x that make the denominator equal to zero. Since x = 2 and x = -2 make the denominator zero, the domain of the rational expression is all real numbers except for those values: $$ \text{Domain} = \{x \in \mathbb{R} \mid x \neq 2, -2\} $$

Key Concepts

Factoring Polynomials

Factoring polynomials is a fundamental concept in algebra that involves breaking down a polynomial into simpler components called factors. This process can simplify expressions and helps in solving equations, especially when dealing with quadratic polynomials. Consider the polynomial in our example, \( x^2 - 4 \).
This expression is a quadratic polynomial since the highest power of \( x \) is 2. Polynomials like these can often be factored into the product of linear terms if they follow specific patterns.

• First, identify the structure of the polynomial: For \( x^2 - 4 \), notice it resembles the identity \( a^2 - b^2 \).
• Rewrite it as \( (x - 2)(x + 2) \), which is simpler and shows that the expression is a product of factors.

Once factored, these expressions are more manageable, especially when they emerge in larger algebraic computations.

Solving Quadratic Equations

A quadratic equation is a second-degree polynomial equation in a single variable, usually written in the form \( ax^2 + bx + c = 0 \). Solving such equations can be achieved using a variety of methods, one of which involves factoring, especially when the quadratic is factorable like in our example.
For the equation \( x^2 - 4 = 0 \), after factoring it, we get \( (x - 2)(x + 2) = 0 \). The solution involves finding values of \( x \) that make the equation true. To solve, apply the zero-product property: if the product of two factors is zero, at least one of the factors must be zero.

• Set each factor equal to zero: \( x - 2 = 0 \) and \( x + 2 = 0 \).
• Solve these linear equations to find \( x = 2 \) and \( x = -2 \).

This process determines the roots or solutions of the quadratic equation, which are critical in defining the expression's behavior.

Difference of Squares

The difference of squares is a special factoring pattern for polynomials that simplifies expressions into two binomials. It applies to expressions in the form \( a^2 - b^2 \). Recognizing and utilizing this pattern can significantly simplify solving and factoring processes.
In our context, \( x^2 - 4 \) fits the pattern \( a^2 - b^2 \) because \( x^2 \) is \( x \) squared, and \( 4 \) is \( 2^2 \). You can factor this as \( (x - 2)(x + 2) \).

• Set the expression as the difference of squares where \( a=x \) and \( b=2 \).
• Apply the identity: \( a^2 - b^2 = (a - b)(a + b) \).
• You now have the factors \( (x-2) \) and \( (x+2) \).

Using this technique allows us to quickly factor polynomials into simpler expressions, aiding in finding solutions and determining domains.