Problem 50 - Elementary and Intermediate Algebra

Question

Perform the operations and simplify, if possible. See Example 5 $$\frac{a+6}{a^{2}-16} \cdot \frac{3 a-12}{3 a+18}$$

Step-by-Step Solution

Answer

The simplified expression is \(\frac{1}{a+4}\).

Step 1: Identify Factorable Expressions

Notice that each part has expressions that may be factorable: \(a^2 - 16\) is a difference of squares, \(3a - 12\) and \(3a + 18\) can be factored by their greatest common factors.

Step 2: Factor Expressions

Factor the expressions as follows:\[a^2 - 16 = (a - 4)(a + 4)\]\[3a - 12 = 3(a - 4)\]\[3a + 18 = 3(a + 6)\]

Step 3: Rewrite Fractions with Factors

Substitute the factored expressions into the original fractions:\[\frac{a+6}{(a-4)(a+4)} imes \frac{3(a-4)}{3(a+6)}\]

Step 4: Cancel Common Factors

Identify and cancel out the common factors between the numerators and denominators in the fractions: 1. Cancel \(3\) from the second fraction.2. Cancel \(a + 6\) from the first fraction and denominator of the second fraction.3. Cancel \(a - 4\) from the second fraction:After canceling, it remains:\[\frac{1}{a+4}\]

Step 5: Write Final Simplified Expression

The simplified expression, after canceling all possible factors, is:\[\frac{1}{a+4}\]

Key Concepts

Factoring Expressions

Factoring is like breaking down a complex Lego structure into smaller pieces, which makes it easier to work with. In algebra, factoring expressions involves writing a number or an expression as a product of its factors, similar to how you might break down 12 into \(3 \times 4\). But with algebraic expressions, like \(a^2 - 16\), we use different techniques. One common method is recognizing algebraic identities, like the difference of squares:

• Difference of Squares: \(a^2 - b^2 = (a-b)(a+b)\)

So for \(a^2 - 16\), you can think of it as \(a^2 - 4^2\), which results in \((a-4)(a+4)\). This makes the expression simpler and easier to manage, crucial for further steps like simplifying rational expressions. Identifying the greatest common factor (GCF), which involves finding the largest factor that divides two or more numbers or terms, is another helpful strategy. For \(3a-12\), the GCF is 3, and factoring it out results in \(3(a-4)\). This approach is foundational for simplifying more complicated expressions efficiently.

Rational Expressions

Rational expressions are akin to fractions but with polynomials in the numerator and denominator, like \( \frac{1}{a+4} \). Understanding them is key to many algebra problems. They can look complex, but the strategies to simplify them are straightforward if you know the trick! When you multiply fractions, for example \(\frac{a+6}{(a-4)(a+4)} \cdot \frac{3(a-4)}{3(a+6)}\), you can find common factors to cancel out, aiming for a simpler form.

• Just like you simplify \(\frac{6}{8}\) to \(\frac{3}{4}\) by canceling common factors (in this case, 2), the same logic applies to rational expressions.

This cancellation is only possible when the factors are common across the numerator and denominator of either fraction being multiplied. The goal is always to simplify the rational expression to its most fundamental form, reducing complexity and making calculations or further algebraic manipulations more manageable.

Polynomial Operations

When working with polynomials, operations such as addition, subtraction, multiplication, and division become a routine yet crucial part of solving problems. These operations can transform a messy expression into something tidy and elegant. Let’s take multiplication involving complex fractions:

• Use factoring to rewrite expressions in simpler forms as discussed. For instance, turning \(a^2-16\) into \((a-4)(a+4)\).
• Multiply polynomial expressions directly if possible, as in distributing terms: for \((a-4)(a+4)\), resulting in \(a^2-16\) as a reconfirmation of identity.

Once polynomials are factored and rewritten in simpler formats, they allow for easier cancellation of terms across the numerator and denominator, especially within rational expressions. By conducting these operations, trying different algebraic chess moves becomes more intuitive, allowing an algebra student to simplify expressions seamlessly and efficiently. This procedural fluency raises confidence in tackling more advanced algebra challenges.