Problem 47 - Elementary and Intermediate Algebra

Question

Simplify. See Example 4. $$ \frac{x^{2}-4}{x^{2}-6 x+8} $$

Step-by-Step Solution

Answer

The simplified expression is \( \frac{x+2}{x-4} \) with restrictions \( x \neq 2 \) and \( x \neq 4 \).

Step 1: Recognize the Expression Type

The expression \( \frac{x^2 - 4}{x^2 - 6x + 8} \) is a rational expression, which means we need to factorize both the numerator and the denominator if possible.

Step 2: Factor the Numerator

The numerator \( x^2 - 4 \) is a difference of squares. It can be factored using the formula \( a^2 - b^2 = (a - b)(a + b) \). Applying this formula, we have:\[ x^2 - 4 = (x - 2)(x + 2) \]

Step 3: Factor the Denominator

The denominator \( x^2 - 6x + 8 \) is a quadratic expression. We look for two numbers whose product is 8 and sum is -6. These numbers are -4 and -2. Thus, the denominator factors as:\[ x^2 - 6x + 8 = (x - 4)(x - 2) \]

Step 4: Simplify the Expression

Now that both the numerator and denominator are factored, we rewrite the expression:\[ \frac{(x - 2)(x + 2)}{(x - 4)(x - 2)} \]Since \(x - 2\) is a common factor in both the numerator and the denominator, it can be cancelled out. Thus, we simplify the expression to:\[ \frac{x + 2}{x - 4} \]

Step 5: State any Restrictions

Before canceling common terms, note that \( x eq 2 \) and \( x eq 4 \) to avoid any undefined expressions in the original function. These restrictions arise because they make the denominator of the original expression zero.

Key Concepts

Factoring Polynomials

Factoring polynomials is a key skill in simplifying expressions, especially rational expressions. In mathematics, to factor a polynomial means to express it as a product of simpler polynomials. Think of factoring as breaking down a large number into its prime components, but instead, we're working with polynomials.
For example, consider the polynomial \(x^2 - 6x + 8\). To factor this quadratic polynomial, we look for two numbers whose product is the constant term (in this case, 8) and whose sum is the coefficient of the linear term (here, -6). For \(x^2 - 6x + 8\), those two numbers are -4 and -2. Hence, it can be factored as \((x-4)(x-2)\).
Factoring helps break down complex expressions into simpler components, making further algebraic manipulations more straightforward.

Difference of Squares

The concept of difference of squares is pivotal in factoring certain types of polynomials. It applies to any expression that can be written in the form \(a^2 - b^2\). The beauty of difference of squares is it factors into two binomials: \((a - b)(a + b)\).
In our example, the numerator \(x^2 - 4\) is a classic difference of squares. Here, \(x^2\) is \(a^2\) and 4 is \(2^2\). Applying the difference of squares formula, we factor it to \((x - 2)(x + 2)\).
This factorization is useful because it transforms the expression into a form where further simplification might be possible, helping to cancel out common factors when simplifying rational expressions.

Quadratic Expressions

Quadratic expressions are polynomials of degree two, generally in the form \(ax^2 + bx + c\). Understanding and factoring these expressions is fundamental. They are involved in numerous algebra and calculus problems.
The expression \(x^2 - 6x + 8\) is a perfect example of a quadratic expression. To factor it, we identify two numbers that multiply to give the constant term, 8, and add to give the linear coefficient, -6. These numbers are -4 and -2, allowing us to write the expression as \((x - 4)(x - 2)\).
This technique of factoring is crucial because it simplifies the equation structure, facilitating easier solutions to algebraic equations, especially when part of a rational expression.