Problem 89 - Elementary Algebra

Question

For the following problems, convert the given rational expressions to rational expressions having the same denominators. $$ \frac{2}{a^{2}+a}, \frac{a+3}{a^{2}-1} $$

Step-by-Step Solution

Answer

Given fractions: $$ \frac{2}{a^2 + a}, \frac{a+3}{a^2 - 1} $$ Equivalent rational expressions with the same denominator: $$ \frac{2(a-1)}{a(a-1)(a+1)}, \frac{a(a+3)}{a(a-1)(a+1)} $$

Step 1: Factor the denominators

Factor the given denominators: $$ a^2 + a = a(a + 1) $$ $$ a^2 - 1 = (a - 1)(a + 1) $$

Step 2: Calculate the least common multiple (LCM)

Now, find the LCM of the denominators by considering the unique factors and their highest power. The LCM of \(a(a+1)\) and \((a-1)(a+1)\) is \(a(a-1)(a+1)\).

Step 3: Convert both rational expressions to have the same denominator

To convert the given rational expressions to have the same denominator, multiply the numerator and the denominator of each fraction by the missing factor(s) to reach the LCM. For the first fraction, the missing factor is \((a-1)\), so we have: $$ \frac{2}{a(a+1)} \cdot \frac{(a-1)}{(a-1)} = \frac{2(a-1)}{a(a-1)(a+1)} $$ For the second fraction, the missing factor is \(a\), so we have: $$ \frac{a+3}{(a-1)(a+1)} \cdot \frac{a}{a} = \frac{a(a+3)}{a(a-1)(a+1)} $$

Step 4: Write the final answer

Now that both fractions have the same denominator, we can write the final expressions: $$ \frac{2(a-1)}{a(a-1)(a+1)}, \frac{a(a+3)}{a(a-1)(a+1)} $$

Key Concepts

Factoring Polynomials

Factoring polynomials forms the foundation for simplifying and manipulating expressions in algebra. A polynomial is an expression consisting of variables and coefficients, involving operations like addition, subtraction, and multiplication. To factor a polynomial is to express it as a product of its factors, which can simplify expressions or help solve equations.

• Let's start with the first denominator from the exercise: \(a^2 + a\). To factor this, look for common factors in each term: here, \(a\) is common. Hence, it can be factored as \(a(a + 1)\).
• The second denominator is \(a^2 - 1\). This is a difference of squares, which can be factored as \((a - 1)(a + 1)\).

Factoring in this way helps us see the simplest building blocks—or factors—of each polynomial, allowing us to continue solving or manipulating the expression.

Least Common Multiple (LCM)

Finding the least common multiple (LCM) of polynomials is a critical step for operations that require a common denominator, such as addition and subtraction of rational expressions. The LCM is the smallest polynomial that is a multiple of each polynomial in a given set.
To find the LCM of the denominators \(a(a + 1)\) and \((a-1)(a + 1)\), follow these steps:

• Identify the unique factors: \(a\), \(a + 1\), and \(a - 1\).
• For each factor, take the highest power that appears in any of the factorizations. Here each appears in their first power.
• The LCM, therefore, is \(a(a - 1)(a + 1)\).

This new denominator allows both rational expressions to be rewritten and compared or combined.

Common Denominators

Creating a common denominator for rational expressions involves rewriting each expression so they're all over the same denominator. This is crucial for operations like addition or subtraction.
Start by taking the original rational expressions:

• The first rational expression is \(\frac{2}{a(a+1)}\).
• The second rational expression is \(\frac{a+3}{(a-1)(a+1)}\).

For the first fraction, we multiply both the numerator and the denominator by the missing factor \((a-1)\) to get:

• \(\frac{2(a-1)}{a(a-1)(a+1)}\)

For the second fraction, multiply numerator and denominator by the missing factor \(a\):

• \(\frac{a(a+3)}{a(a-1)(a+1)}\)

With the same denominator, these expressions can now be easily combined or compared, facilitating further algebraic manipulation or the solving of equations.