Problem 82 - Algebra A Combined Function

Question

Does \(\frac{(x-3)(x+3)}{x-3}\) have the same value as \(x+3\) for all real numbers? Explain why or why not.

Step-by-Step Solution

Answer

No, they differ at \( x = 3 \); the expression is undefined there.

Step 1: Simplifying the Expression

Let's begin by simplifying the given expression. We have \( \frac{(x-3)(x+3)}{x-3} \). Assuming \( x eq 3 \), we can factor out \( x-3 \) from both the numerator and the denominator, which gives us \( (x-3) \) in both places. After canceling \( x-3 \) from the numerator and the denominator, we have \( x+3 \). Thus, for \( x eq 3 \), the expression simplifies to \( x+3 \).

Step 2: Identifying Undefined Points

Next, we need to consider what happens when \( x = 3 \). Substituting \( x = 3 \) into the original expression, the denominator becomes zero: \( x-3 = 0 \). Division by zero is undefined in mathematics, which means that the expression \( \frac{(x-3)(x+3)}{x-3} \) is undefined at \( x = 3 \).

Step 3: Concluding the Comparison

Now, we can determine if \( \frac{(x-3)(x+3)}{x-3} \) equals \( x+3 \) for all real numbers. From the simplification, we see they are equal for all \( x eq 3 \). However, they are not defined the same way for \( x = 3 \): the expression is undefined at this point, while \( x+3 \) would evaluate to \( 6 \). Thus, \( \frac{(x-3)(x+3)}{x-3} \) does not have the same value as \( x+3 \) for all real numbers.

Key Concepts

Simplifying Algebraic Expressions

Simplifying algebraic expressions is a method of rewriting expressions in a simpler form by reducing complexity. This often involves canceling terms that appear in both the numerator and denominator of a fraction. For example, consider the expression \( \frac{(x-3)(x+3)}{x-3} \). To simplify this, we look for common terms.

• The numerator \((x-3)(x+3)\) can be expanded using the distributive property to \(x^2 - 9\), but in this context, we notice that \(x-3\) is a common factor.
• As long as \(x eq 3\), we can divide both the numerator and denominator by \(x-3\), effectively canceling \(x-3\).
• What remains is the simple expression \(x+3\).

This simplified expression helps in understanding and solving algebraic problems more efficiently. By eliminating unnecessary terms, you can work with a cleaner and more manageable form of the expression.

Undefined Expressions

An undefined expression occurs when we attempt to perform an operation that is mathematically invalid. Division by zero is one such operation that leads to undefined expressions. In our example, the expression \( \frac{(x-3)(x+3)}{x-3} \) becomes undefined for \( x=3 \), because:

• When \( x=3 \), the denominator \( x-3 \) becomes zero.
• A zero denominator in a fraction results in division by zero, which is undefined in mathematical terms.

Thus, it's crucial to identify points where expressions become undefined to accurately evaluate or simplify problems. This means that while the simplified form \(x + 3\) works for most values of \(x\), it is important to point out exceptional cases like \(x = 3\). This recognition helps ensure our solutions are mathematically sound for the entire domain of possible values.

Factoring Polynomials

Factoring polynomials is an essential skill in algebra that involves expressing a polynomial as a product of its factors. This method is often the first step in simplifying expressions or solving equations. For instance, in our expression \(\frac{(x-3)(x+3)}{x-3}\):

• The polynomial \((x-3)(x+3)\) is already factored for us. This is a classic case of the difference of squares, where \( a^2 - b^2 = (a-b)(a+b) \).
• Recognizing this form allows you to factor the expression easily revealing its components \( (x-3) \) and \( (x+3) \).
• By factoring, it becomes apparent which terms can be canceled out when simplifying the expression.

Factoring is a foundational aspect of algebra that is useful not only in simplifying expressions but also in solving equations and inequalities. Mastering this skill helps to break down complex problems into manageable steps, making algebraic manipulation more intuitive.