Problem 30 - Algebra A Combined Function

Question

Simplify each expression. $$ \frac{x+5}{x^{2}-25} $$

Step-by-Step Solution

Answer

The expression simplifies to \( \frac{1}{x-5} \), where \( x \neq 5 \) or \( x \neq -5 \).

Step 1: Identify Factoring Possibilities

Recognize that the denominator \( x^2 - 25 \) is a difference of squares. The difference of squares formula is \( a^2 - b^2 = (a-b)(a+b) \). Here, \( a = x \) and \( b = 5 \).

Step 2: Factor the Denominator

Apply the difference of squares formula: \( x^2 - 25 = (x-5)(x+5) \). Now, the expression looks like \( \frac{x+5}{(x-5)(x+5)} \).

Step 3: Simplify the Expression

Since \( x+5 \) is a common factor in both the numerator and the denominator, it can be cancelled out. The expression simplifies to \( \frac{1}{x-5} \).

Step 4: State any Restrictions

Note that for the original expression, \( x+5 = 0 \) and \( x-5 = 0 \) would make the denominator zero, so \( x eq 5 \) and \( x eq -5 \). Correspondingly, \( x eq 5 \) for the simplified expression.

Key Concepts

Difference of Squares

The difference of squares is a special algebraic method used for simplifying certain polynomial expressions. It's applicable when your expression can be identified in the form \( a^2 - b^2 \). In this equation, \( a \) and \( b \) represent terms that are squared and then subtracted.

For example:

• \( a^2 - b^2 = (a-b)(a+b) \)

The purpose of recognizing a difference of squares is to facilitate simplification by transforming the expression into a product of two binomials.

In the problem \( x^2 - 25 \), you can see that \( (x)^2 - (5)^2 \) perfectly fits the difference of squares pattern. Applying the formula gives us \( (x - 5)(x + 5) \), which simplifies the entire process when dealing with fractions or other algebraic manipulations.

Factoring

Factoring is an essential skill in algebra. It involves breaking down complex expressions into simpler multipliers.

When you "factor" something, you find values (i.e., factors or roots) that, when multiplied, give your original number or expression.

• Why is factoring useful? It simplifies expressions, making it easier to tackle algebraic problems like simplifying fractions or solving equations.
• In the given example, factoring the denominator as a difference of squares helped in identifying common terms that could be cancelled out.

Using the earlier example \( x^2 - 25 \), identifying and applying the technique of difference of squares led directly to factoring the expression into \( (x - 5)(x + 5) \).

The benefit this provides is it makes the simplification straightforward, as any common factors in the numerator and denominator can then be cancelled, leading us to a much simpler form.

Algebraic Fractions

Algebraic fractions are like regular fractions, but they contain variables. These expressions can often be simplified if you know the right techniques.

Just like simplifying numerical fractions by finding the greatest common factor, algebraic fractions require identifying common terms in the numerator and the denominator.

• Consider the expression \( \frac{x + 5}{(x-5)(x+5)} \).
• After factoring, it becomes clear that \( x + 5 \) is a common factor and can be cancelled out.

Simplifying involves canceling out these common factors, resulting in \( \frac{1}{x-5} \). Keep in mind, when simplifying algebraic fractions, it's important to identify any restrictions on the variables, such as values that make the denominator zero.

In this problem, \( x eq 5 \) and \( x eq -5 \) were the restrictions. This is because the denominator cannot be zero without making the expression undefined.