Problem 52 - Intermediate Algebra

Question

Simplify each function. List any restrictions on the domain. $$ f(x)=\frac{x^{2}+6 x-16}{x^{2}-4} $$

Step-by-Step Solution

Answer

The simplified function is \(f(x) = \frac{x+8}{x+2}\) with domain restrictions \(x \neq 2\) and \(x \neq -2\).

Step 1: Factor the Numerator

First, we need to factor the numerator of the function, which is a quadratic expression: \(x^{2} + 6x - 16\). To do this, we look for two numbers that multiply to \(-16\) (the constant term) and add to \(6\) (the coefficient of the middle term). These numbers are \(8\) and \(-2\). Thus, the factorization of the numerator is \((x + 8)(x - 2)\).

Step 2: Factor the Denominator

Next, we factor the denominator \(x^{2} - 4\). This is a difference of squares, which can be factored as \((x - 2)(x + 2)\).

Step 3: Simplify the Function

Now that we have factored both the numerator and the denominator, we can write the function as: \( \frac{(x+8)(x-2)}{(x-2)(x+2)} \). The \((x - 2)\) term appears in both the numerator and the denominator, allowing us to cancel it out, assuming \(x eq 2\). The simplified function is \( \frac{x+8}{x+2} \).

Step 4: Determine Domain Restrictions

To determine the domain restrictions, we set the original denominator equal to zero and solve for \(x\). \(x^2 - 4 = 0\) simplifies to \(x^2 = 4\), which gives \(x = 2\) and \(x = -2\). Thus, the domain of the function excludes \(x = 2\) and \(x = -2\).

Step 5: State the Simplified Function and Domain

The simplified function is \(f(x) = \frac{x+8}{x+2}\) with domain restrictions that \(x eq 2\) and \(x eq -2\).

Key Concepts

Factoring Quadratics

Factoring is a crucial skill when it comes to simplifying rational functions, especially those involving quadratics. In our given function, the numerator is the quadratic expression \(x^2 + 6x - 16\). To factor it, we are essentially reversing the process of expanding two binomials. The goal is to find two numbers that multiply to give the constant term (-16 in this case) and add up to the coefficient of the linear term (which is 6). After some trial and error or practice, you might recognize that 8 and -2 are the numbers that fit these requirements:

• Multiplying 8 and -2 gives \(-16\).
• Adding them gives \(6\).

Thus, the factorization of \(x^2 + 6x - 16\) is \((x + 8)(x - 2)\). Knowing how to factor efficiently can simplify the process and help uncover relationships in expressions.

Domain Restrictions

In mathematics, the domain of a function is the set of all possible input values (typically \(x\)-values) for which the function is defined. When working with rational functions, which are essentially fractions, we need to ensure the denominator never equals zero, since division by zero is undefined.

To find domain restrictions for our function, we set the denominator equal to zero and solve for \(x\), looking for values that would cause division by zero. Our denominator is \(x^2 - 4\). Setting this expression to zero gives:

• \(x^2 - 4 = 0\)
• Solving gives \(x^2 = 4\)
• This further breaks down to \(x = 2\) or \(x = -2\)

So, the domain of the original function excludes \(x = 2\) and \(x = -2\). Understanding domain restrictions ensures we only consider valid inputs for the function.

Simplification of Expressions

Once we factor both the numerator and denominator of a rational function, we can look to simplify or "reduce" the expression by cancelling out common factors. This is akin to reducing fractions to their simplest form in elementary math.

For our function, after factoring, we have the expression \( \frac{(x+8)(x-2)}{(x-2)(x+2)} \). We notice the common factor \((x-2)\) in both the numerator and the denominator. These can be cancelled out, simplifying our function to \( \frac{x+8}{x+2} \).

However, it's important to remember that cancelling factors can only be done when these elements are truly common to both the numerator and denominator. We also have to consider the domain restrictions identified earlier. Therefore, even though "2" seems allowed here, our original problem tells us that \(x eq 2\) and \(x eq -2\). Simplifying expressions helps achieve more manageable equations while maintaining mathematical legitimacy.