Problem 44 - Elementary and Intermediate Algebra

Question

Perform the operations. Simplify, if possible. $$ \frac{1}{c+6}+\frac{4}{c^{2}+8 c+12} $$

Step-by-Step Solution

Answer

The simplified result is \(\frac{1}{c+2}\).

Step 1: Identify the Denominators

Look at each fraction and identify their denominators. The first fraction has a denominator of \(c + 6\) and the second fraction has a denominator of \(c^2 + 8c + 12\).

Step 2: Factor the Second Denominator

To handle \(c^2 + 8c + 12\), we factor it. This can be factored as \((c + 6)(c + 2)\). The expression is a quadratic trinomial.

Step 3: Determine the Least Common Denominator (LCD)

Based on the denominators, the least common denominator is \((c + 6)(c + 2)\) because the second denominator includes both \(c + 6\) and \(c + 2\).

Step 4: Rewrite the Fractions

Rewrite the fractions with the LCD: The first fraction \(\frac{1}{c+6}\) becomes \(\frac{1 \cdot (c+2)}{(c+6)(c+2)} = \frac{c+2}{(c+6)(c+2)}\). The second fraction is already with the LCD: \(\frac{4}{(c+6)(c+2)}\).

Step 5: Combine the Fractions

Now that both fractions have the same denominator, combine them into one fraction: \(\frac{c+2 + 4}{(c+6)(c+2)} = \frac{c+6}{(c+6)(c+2)}\).

Step 6: Simplify the Fraction

Cancel out the common factor of \(c+6\) in the numerator and denominator. This leaves \(\frac{1}{c+2}\) because \(c+6\) divided by \(c+6\) is 1.

Key Concepts

Factoring Quadratics

In algebra, factoring quadratics is an essential skill that allows us to break down complex expressions into simpler binomial forms. A quadratic expression typically resembles \( ax^2 + bx + c \). Our goal is to identify values for \( a \), \( b \), and \( c \). In our example, \( c^2 + 8c + 12 \) is the quadratic component. To factor this quadratic,

• Look for two numbers that multiply to the constant term (12) and add up to the linear coefficient (8).
• These numbers are 6 and 2 because \(6 imes 2 = 12\) and \(6 + 2 = 8\).

This lets us express the quadratic as \((c+6)(c+2)\). Factoring quadratics this way helps when simplifying algebraic fractions by revealing common terms in the denominators.

Least Common Denominator

The least common denominator (LCD) is critical when working with algebraic fractions. It enables us to add or subtract fractions that initially have different denominators. The goal is to find the smallest expression that both denominators can divide into without leaving a remainder. In the given problem:

• The factors we identified were \(c+6\) and \(c+2\).
• The LCD is the product of these factors: \((c+6)(c+2)\).

Once found, the LCD becomes the new, shared denominator for both fractions in the problem, allowing us to combine them seamlessly.

Simplifying Expressions

Simplifying expressions involves reducing an algebraic expression to its simplest form, making it easier to manage and understand. After rewriting our fractions with the LCD, they share a common denominator of \((c+6)(c+2)\). This allows us to combine them into one single fraction.

• The combined fraction is \( \frac{c+6}{(c+6)(c+2)} \).
• Both the numerator and the denominator contain the factor \(c+6\), which we can cancel out.

This results in the final simplified expression \( \frac{1}{c+2} \). Simplification helps reveal the underlying simplicity in complex algebraic fractions.