Problem 10 - Algebra A Combined Function

Question

Perform each indicated operation. Simplify if possible. \(\frac{5}{x-4}+\frac{4 x}{x^{2}-16}\)

Step-by-Step Solution

Answer

\(\frac{9x + 20}{(x-4)(x+4)}\).

Step 1: Identify the Operation and Denominators

We are given two fractions: \(\frac{5}{x-4}\) and \(\frac{4x}{x^2-16}\). The task is to add these two fractions. When adding fractions, we need a common denominator. The denominators here are \(x-4\) and \(x^2-16\).

Step 2: Factor the Denominator

Notice that \(x^2-16\) is a difference of squares. We can factor it as \((x-4)(x+4)\). This means the new, factored denominator for \(\frac{4x}{x^2-16}\) is \((x-4)(x+4)\).

Step 3: Determine the Common Denominator

The common denominator for the two fractions needs to include \(x-4\) and \(x+4\). Therefore, the common denominator is \((x-4)(x+4)\).

Step 4: Rewrite Each Fraction with the Common Denominator

For the first fraction, \(\frac{5}{x-4}\), multiply both the numerator and the denominator by \(x+4\) to get \(\frac{5(x+4)}{(x-4)(x+4)}\). The second fraction is already expressed with the common denominator: \(\frac{4x}{(x-4)(x+4)}\).

Step 5: Add the Fractions

Now that both fractions have the common denominator, add them: \[\frac{5(x+4)}{(x-4)(x+4)} + \frac{4x}{(x-4)(x+4)} = \frac{5(x+4) + 4x}{(x-4)(x+4)}\].

Step 6: Simplify the Numerator

Expand and combine the terms in the numerator:- Expanding: \(5(x+4) = 5x + 20\)- Combine with \(4x\): \(5x + 20 + 4x = 9x + 20\)So the expression becomes \(\frac{9x + 20}{(x-4)(x+4)}\).

Step 7: Check for Further Simplification

The expression in the numerator \(9x + 20\) cannot be factored further, and it shares no common factors with the denominator \((x-4)(x+4)\). Thus, the expression is fully simplified.

Key Concepts

Common Denominator

To add fractions, no matter how complex, the first thing you need is a common denominator. Think of denominators as the shared foundation for your fractions. This common base ensures you can perform operations between fractions accurately and smoothly.

When the fractions have different denominators, it disrupts the balance. In algebraic fractions, finding a common denominator involves factoring. Knowing what the denominator is made of helps in aligning the fractions back on the same page.

In our specific problem, we see denominations such as \(x - 4\) and \(x^2 - 16\). To find a common denominator, it's crucial to spot differences of squares or other factors that make combining easier. Factoring reveals an expression's nature and provides a path to this common base.

Sum of Fractions

Adding fractions involves creating a unified whole from different parts. For a successful addition, both fractions need to stand on a common platform, hence the term 'common denominator'.

Once the denominators match, simply add the numerators. Imagine combining like terms in an equation; the same logic applies here. The real trick is to rewrite each fraction so they share the same bottom part (denominator). For instance, rewriting:

• Modify \(\frac{5}{x-4}\) into \(\frac{5(x+4)}{(x-4)(x+4)}\)
• The \(\frac{4x}{x^2-16}\) fraction already fits the new scenario.

Now both fractions look similar in the denominator, making addition straightforward. Combine the numerators over the shared denominator. For example, once rewritten, you add: \[\frac{5(x+4) + 4x}{(x-4)(x+4)}\].

By simplifying the numerator, find your solution!

Difference of Squares

The difference of squares is a special mathematical pattern where two terms are squared

• It follows the form \(a^2 - b^2\)
• This can be factored into \((a-b)(a+b)\)

This pattern saves time in algebra, especially with polynomials. It's like a shortcut, revealing hidden factors in an expression quickly.

In the given exercise, recognize \(x^2 - 16\) as a difference of squares. It factors to \((x-4)(x+4)\). This discovery makes finding a common denominator possible and simplifies the algebraic expression. Such spotting saves steps, and knowing these patterns accelerates problem-solving!

Factoring Polynomials

Factoring polynomials is unlocking the doors to simplification. It involves breaking expressions down into their component multiplicative parts or factors. Factoring transforms complex polynomials into easily manageable pieces.

When you look at \(x^2 - 16\), recognize it as a classic example. You convert it from a whole path to stepping stones of \((x-4)\) and \((x+4)\). These factors tell you a lot about the expression's behavior and interactions with others separated by denominators.

Understanding factoring gives you the tools to control the equation, and not let it control you. Dive into the polynomial world confidently, knowing it can be simplified to understandable elements.