Problem 1 - College Algebra and Calculus: An Applied Approach

Question

Write the partial fraction decomposition for the expression. $$ \frac{2(x+20)}{x^{2}-25} $$

Step-by-Step Solution

Answer

The partial fraction decomposition for the given expression is \( \frac{2(x+20)}{(x-5)(x+5)} = \frac{5}{x-5} + \frac{4}{x+5} \)

Step 1: Factorize the Denominator

The denominator here is \( x^{2} - 25 \). Use the difference of squares factoring formula, \( a^{2} - b^{2} = (a-b)(a+b) \), that is, \( x^{2} - 25 = (x-5)(x+5) \).

Step 2: Initialize Your Partial Fraction Decomposition

The general form of a partial fraction for a quadratic when the denominator contains real and distinct roots, \( p \) and \( q \) is \( \frac{A}{(x-p)} + \frac{B}{(x-q)} \). So in this case the expression becomes \( \frac{2(x+20)}{(x-5)(x+5)} = \frac{A}{x-5} + \frac{B}{x+5} \).

Step 3: Determine the Coefficients A and B

Multiply through by the common denominator, \( (x-5)(x+5) \), and simplify to get the equation: \( 2(x+20) = A(x+5) + B(x-5) \). Now choose suitable values for \( x \) to make the calculations simpler and to solve for \( A \) and \( B \). If we choose \( x=5 \), the second term on the right side will become zero leaving us with \( 50=10A \). Solving this, we get \( A=5 \). Similarly, we can choose \( x=-5 \), then the first term on the right becomes zero leaving us with \( -40= -10B \). Solving this gives \( B=4 \).

Step 4: Write the Final Answer

Plug the values for \( A \) and \( B \) back into the equation \( \frac{2(x+20)}{(x-5)(x+5)} = \frac{A}{x-5} + \frac{B}{x+5} \) to get the solution: \( \frac{2(x+20)}{(x-5)(x+5)} = \frac{5}{x-5} + \frac{4}{x+5} \)

Key Concepts

Difference of Squares

Understanding the difference of squares is crucial when factoring expressions in algebra. It is a pattern for factoring expressions where two terms are squared and subtracted from each other. The formula is given by \( a^2 - b^2 = (a - b)(a + b) \). In practice, when you encounter a quadratic expression such as \( x^2 - 25 \), recognize that it represents the difference of squares. Here, \( a = x \) and \( b = 5 \) (since \( 25 = 5^2 \) ). Thus, it can be factored into \( (x - 5)(x + 5) \).

Using this approach simplifies complex algebraic fractions into something more manageable, as demonstrated in the partial fraction decomposition exercise. Recognizing and applying the difference of squares pattern can also greatly assist in solving equations, finding roots, and simplifying algebraic expressions in a variety of contexts.

Factoring Polynomials

The term factoring polynomials refers to breaking down a polynomial into a product of its simplest parts, which are usually easier to work with. In the context of the given exercise, after identifying the difference of squares, the next step is to decompose the algebraic fraction. Polynomials can often be factored using several techniques, including finding the greatest common factor, grouping, and using special patterns like difference of squares or perfect square trinomials.

Effective factoring is like solving a puzzle, where identifying particular patterns can simplify potentially complex problems. For example, once the denominator is factored and the roots are revealed, we can move on to the process of setting up partial fractions, which greatly simplifies integration and helps in solving differential equations.

Algebraic Fractions

Algebraic fractions are expressions that contain a numerator and a denominator with polynomial terms. Simplifying these fractions often requires techniques from various areas of algebra such as factoring polynomials and solving linear equations. Partial fraction decomposition is a method for breaking down complex algebraic fractions into a sum or difference of simpler fractions, where the denominators are polynomials that are factorable.

In the original exercise, the algebraic fraction \(\frac{2(x+20)}{x^{2}-25}\) is decomposed into simpler fractions. This decomposition is essential in calculus, particularly when integrating rational functions. Understanding algebraic fractions and how to deconstruct them is therefore not just a skill for algebra classes, but a foundational tool for higher mathematics.

Linear Algebra

Although the term linear algebra often brings to mind the study of vectors, matrices, and systems of linear equations, its techniques and concepts also play a role in simplifying algebraic expressions. While the exercise does not directly involve matrices or vector spaces, the process of finding coefficients A and B in partial fraction decomposition resembles solving linear equations, which is a fundamental aspect of linear algebra.

In the exercise, we seek values of A and B that make the equation \(2(x+20) = A(x+5) + B(x-5)\) true for all values of x. This is akin to solving a system of linear equations where A and B are the unknowns. By strategically choosing values for x that simplify the equation, we can easily solve for A and B. This problem-solving strategy—the ability to simplify and solve equations—is at the heart of linear algebra and reflects its power in a variety of mathematical tasks.