Problem 73 - Intermediate Algebra

Question

Simplify each expression. If an expression cannot be simplified, write "Does not simplify." $$ \frac{a^{2}-4}{a^{3}-8} $$

Step-by-Step Solution

Answer

The expression simplifies to \( \frac{a+2}{a^2 + 2a + 4} \).

Step 1: Identify Common Factor

First, we need to identify and factor any common elements in the numerator and denominator. The numerator is \(a^2 - 4\) and the denominator is \(a^3 - 8\). Both are special forms that can be factored. The numerator is a difference of squares, while the denominator is a difference of cubes.

Step 2: Factor the Numerator

The numerator \(a^2 - 4\) can be factored as \((a+2)(a-2)\), using the difference of squares formula \(x^2 - y^2 = (x-y)(x+y)\).

Step 3: Factor the Denominator

The denominator \(a^3 - 8\) can be factored using the difference of cubes formula: \(x^3 - y^3 = (x-y)(x^2 + xy + y^2)\). Here, \(a^3 - 8\) becomes \((a-2)(a^2 + 2a + 4)\).

Step 4: Cancel Common Factors

Now that we have factored both parts, we rewrite the fraction as \( \frac{(a+2)(a-2)}{(a-2)(a^2 + 2a + 4)} \). We see that the common factor \((a-2)\) can be canceled from both the numerator and the denominator.

Step 5: Simplify Expression

After canceling the common factors, we are left with \( \frac{a+2}{a^2 + 2a + 4} \). This is the simplified form of the expression.

Key Concepts

Difference of Squares

The difference of squares is a special algebraic expression of the form \( x^2 - y^2 \). It represents the relationship between two square numbers that are subtracted from each other.

To factor a difference of squares, you can use the formula:

• \( x^2 - y^2 = (x-y)(x+y) \)

This form is very common and appears many times in algebraic expressions, often making it easier to simplify them.

In our example, the numerator \( a^2 - 4 \) is a classic difference of squares \( (a)^2 - (2)^2 \). This can be rewritten as \( (a-2)(a+2) \), simplifying the expression for further operations.

Difference of Cubes

The difference of cubes formula involves the subtraction of two cubes, expressed as \( x^3 - y^3 \). This type of expression also has a recognizable factoring formula. The general form is:

• \( x^3 - y^3 = (x-y)(x^2 + xy + y^2) \)

Each term is reduced into linear and quadratic factors that simplify the expression.

For instance, the denominator \( a^3 - 8 \) is a difference of cubes: \( (a)^3 - (2)^3 \). Applying the difference of cubes formula, it factors into \( (a-2)(a^2 + 2a + 4) \). Use this form to simplify and solve algebraic equations that involve cubes.

Factoring

Factoring is the process of breaking down an expression into simpler components or factors that can be multiplied to give back the original expression. This skill is crucial in algebra for simplifying expressions and solving equations.

In this exercise, recognizing common factoring patterns, such as differences of squares and cubes, was key to rewriting the numerator and the denominator.

The goal of factoring is to express complex expressions like \( a^2 - 4 \) and \( a^3 - 8 \) in a more manageable form, specifically for these types:

• \( a^2 - 4 = (a+2)(a-2) \)
• \( a^3 - 8 = (a-2)(a^2 + 2a + 4) \)

Factoring reduces the complexity of expressions and helps identify any common terms that could be further simplified.

Canceling Common Factors

Once expressions are factored, the next powerful tool is canceling common factors. This step helps reduce redundant components present in both the numerator and the denominator, simplifying the overall fraction without changing its value.

If a factor appears in both the numerator and the denominator, you can "cancel" it by essentially dividing it out of the equation. In our problem, after factoring, we find that both parts contain \( (a-2) \). By canceling \( (a-2) \), we simplify the complete expression significantly.

Canceling common factors transformed the fraction \( \frac{(a+2)(a-2)}{(a-2)(a^2 + 2a + 4)} \) into \( \frac{a+2}{a^2 + 2a + 4} \). This provides a cleaner and more concise form of the equation, reflecting the same relationship with simpler elements.