Partial Fraction Decomposition Of F(x) = (8 + 5x + 12x^2) / ((1-x)(2+3x)^2)

Introduction

In the realm of mathematical analysis, partial fraction decomposition stands as a pivotal technique, especially when dealing with rational functions. A rational function, defined as the ratio of two polynomials, often appears complex and unwieldy. However, the method of partial fractions allows us to break down these intricate functions into simpler components, making them far easier to manipulate and analyze. This decomposition is particularly useful in calculus for integration, in solving differential equations, and in various areas of engineering and physics. In this comprehensive guide, we will delve into the step-by-step process of expressing a given rational function, specifically $f(x) = \frac{8 + 5x + 12x^2}{(1-x)(2+3x)^2}$, in its partial fraction form. This exploration will not only elucidate the mechanics of the decomposition but also highlight the underlying principles and strategies involved. Understanding these principles is crucial for mastering the technique and applying it effectively to a wide range of problems. We will begin by understanding the fundamental concepts behind partial fractions and then methodically apply these concepts to our specific example, ensuring a clear and thorough understanding of each step.

Understanding Partial Fractions

Before diving into the specifics of our function, let's solidify our understanding of partial fractions in general. The core idea behind partial fractions is to reverse the process of adding fractions. When we add fractions, we find a common denominator and combine the numerators. Partial fraction decomposition does the opposite: it starts with a single fraction and breaks it down into a sum of simpler fractions. These simpler fractions have denominators that are factors of the original denominator. The key principle here is that any rational function where the degree of the numerator is less than the degree of the denominator can be expressed as a sum of partial fractions. This condition is crucial; if the degree of the numerator is greater than or equal to the degree of the denominator, we must first perform polynomial long division to obtain a proper rational function (where the numerator's degree is less than the denominator's). The form of the partial fractions depends on the nature of the factors in the denominator. If the denominator has linear factors (like $1-x$ or $2+3x$), each linear factor corresponds to a partial fraction of the form $\frac{A}{ax+b}$, where A is a constant. If the denominator has repeated linear factors (like $(2+3x)^2$), we need a partial fraction for each power of the repeated factor, such as $\frac{B}{2+3x}$ and $\frac{C}{(2+3x)^2}$. Quadratic factors (factors that cannot be factored further into real linear factors) require numerators of the form $Dx + E$. Understanding these rules is fundamental to setting up the correct form for the partial fraction decomposition. It's like having the right tools for a job; knowing the rules allows us to approach the problem systematically and avoid common pitfalls. The process involves not just algebraic manipulation but also a clear understanding of the underlying structure of rational functions and their factors. The goal is to find the constants in the numerators of the partial fractions, which we will do by equating coefficients or using strategic substitutions.

Setting Up the Partial Fraction Decomposition for $f(x)$

Now, let's apply these principles to our specific function, $f(x) = \frac{8 + 5x + 12x^2}{(1-x)(2+3x)^2}$. The first step is to observe the denominator, $(1-x)(2+3x)^2$. We see that it has a linear factor $(1-x)$ and a repeated linear factor $(2+3x)^2$. Based on the rules of partial fractions, this means we can express $f(x)$ as a sum of three partial fractions:

$$\frac{8 + 5x + 12x^2}{(1-x)(2+3x)^2} = \frac{A}{1-x} + \frac{B}{2+3x} + \frac{C}{(2+3x)^2}$$

Here, A, B, and C are constants that we need to determine. This setup is crucial because it defines the structure of our decomposition. Each term on the right-hand side corresponds to a factor in the original denominator. The term $\frac{A}{1-x}$ corresponds to the linear factor $(1-x)$. The terms $\frac{B}{2+3x}$ and $\frac{C}{(2+3x)^2}$ correspond to the repeated linear factor $(2+3x)^2$. Notice that we have one fraction for each power of the repeated factor. This is essential for a correct decomposition. If we had a cubic factor, we would need three fractions: one for the factor, one for the factor squared, and one for the factor cubed. Now that we have the basic structure, the next step is to find the values of the constants A, B, and C. This involves algebraic manipulation to clear the denominators and then solving a system of equations. The setup is like laying the foundation for a building; it's the critical first step that determines the success of the entire process. A clear understanding of this setup ensures that we are on the right track and can proceed confidently with the remaining steps.

Solving for the Constants A, B, and C

To find the values of the constants A, B, and C, we first need to clear the denominators in our equation:

$$\frac{8 + 5x + 12x^2}{(1-x)(2+3x)^2} = \frac{A}{1-x} + \frac{B}{2+3x} + \frac{C}{(2+3x)^2}$$

Multiply both sides of the equation by the original denominator, $(1-x)(2+3x)^2$. This gives us:

$$8 + 5x + 12x^2 = A(2+3x)^2 + B(1-x)(2+3x) + C(1-x)$$

Now, we have a polynomial equation. To solve for A, B, and C, there are two primary methods: equating coefficients and strategic substitution. Let's explore both.

Method 1: Equating Coefficients

First, expand the right side of the equation:

$$8 + 5x + 12x^2 = A(4 + 12x + 9x^2) + B(2 + x - 3x^2) + C(1-x)$$

$$8 + 5x + 12x^2 = 4A + 12Ax + 9Ax^2 + 2B + Bx - 3Bx^2 + C - Cx$$

Now, group the terms by powers of x:

$$8 + 5x + 12x^2 = (9A - 3B)x^2 + (12A + B - C)x + (4A + 2B + C)$$

For this equation to hold for all values of x, the coefficients of the corresponding powers of x must be equal. This gives us a system of three linear equations:

1. $9A - 3B = 12$ (coefficients of $x^2$)
2. $12A + B - C = 5$ (coefficients of $x$)
3. $4A + 2B + C = 8$ (constant terms)

We can solve this system of equations using various methods, such as substitution or elimination.

Method 2: Strategic Substitution

This method involves choosing specific values of x that simplify the equation and allow us to solve for the constants more easily. Look back at the equation before expanding:

$$8 + 5x + 12x^2 = A(2+3x)^2 + B(1-x)(2+3x) + C(1-x)$$

Notice that if we choose $x = 1$, the terms with B and C will become zero, leaving us with an equation involving only A. Similarly, if we choose $x = -\frac{2}{3}$, the terms with A and B will become zero, leaving us with an equation involving only C.

• Let $x = 1$:

  $$8 + 5(1) + 12(1)^2 = A(2+3(1))^2$$

  $$25 = 25A$$

  $$A = 1$$

• Let $x = -\frac{2}{3}$:

  $$8 + 5(-\frac{2}{3}) + 12(-\frac{2}{3})^2 = C(1-(-\frac{2}{3}))$$

  $$8 - \frac{10}{3} + \frac{16}{3} = C(\frac{5}{3})$$

  $$\frac{24 - 10 + 16}{3} = \frac{5}{3}C$$

  $$\frac{30}{3} = \frac{5}{3}C$$

  $$10 = \frac{5}{3}C$$

  $$C = 6$$

Now that we have A and C, we can substitute them back into one of the equations from Method 1 to solve for B. Let's use equation (3):

$$4A + 2B + C = 8$$

$$4(1) + 2B + 6 = 8$$

$$10 + 2B = 8$$

$$2B = -2$$

$$B = -1$$

Thus, we have found the values of the constants: A = 1, B = -1, and C = 6. This is the heart of the partial fraction decomposition process. Once we have these constants, we can express the original function as a sum of simpler fractions.

Expressing $f(x)$ in Partial Fractions

Now that we have found the values of A, B, and C, we can express $f(x)$ in its partial fraction form. Recall our setup:

$$\frac{8 + 5x + 12x^2}{(1-x)(2+3x)^2} = \frac{A}{1-x} + \frac{B}{2+3x} + \frac{C}{(2+3x)^2}$$

Substitute the values A = 1, B = -1, and C = 6 into the equation:

$$\frac{8 + 5x + 12x^2}{(1-x)(2+3x)^2} = \frac{1}{1-x} + \frac{-1}{2+3x} + \frac{6}{(2+3x)^2}$$

Therefore, the partial fraction decomposition of $f(x)$ is:

$$\frac{8 + 5x + 12x^2}{(1-x)(2+3x)^2} = \frac{1}{1-x} - \frac{1}{2+3x} + \frac{6}{(2+3x)^2}$$

This is the final result. We have successfully decomposed the original rational function into a sum of simpler fractions. Each term is now easier to handle individually, which is particularly useful in calculus when integrating or differentiating. This decomposition also provides insights into the behavior of the function. For instance, we can easily identify the vertical asymptotes of the function from the denominators of the partial fractions. The term $\frac{1}{1-x}$ has a vertical asymptote at $x=1$, and the terms involving $(2+3x)$ have a vertical asymptote at $x=-\frac{2}{3}$. The partial fraction decomposition is not just a mathematical trick; it's a powerful tool for understanding and manipulating rational functions. It allows us to break down complex expressions into manageable parts, making them accessible to further analysis and application. The ability to perform this decomposition is a valuable skill in many areas of mathematics and its applications.

Conclusion

In conclusion, we have successfully decomposed the rational function $f(x) = \frac{8 + 5x + 12x^2}{(1-x)(2+3x)^2}$ into its partial fractions form:

$$\frac{8 + 5x + 12x^2}{(1-x)(2+3x)^2} = \frac{1}{1-x} - \frac{1}{2+3x} + \frac{6}{(2+3x)^2}$$

We began by understanding the fundamental principles of partial fraction decomposition, emphasizing the importance of the degree of the numerator and denominator and the role of linear and repeated factors. We then methodically applied these principles to our specific function, setting up the correct form for the decomposition and identifying the constants A, B, and C. We explored two methods for solving for these constants: equating coefficients and strategic substitution. Both methods provide a pathway to the solution, and choosing the most efficient method often depends on the specific problem. Finally, we substituted the values of A, B, and C back into our setup to obtain the partial fraction decomposition. This process not only simplifies the function but also reveals important information about its behavior, such as its vertical asymptotes. The technique of partial fraction decomposition is a cornerstone in various mathematical disciplines, including calculus, differential equations, and complex analysis. Its ability to transform complex rational functions into simpler components makes it an indispensable tool for mathematicians, engineers, and scientists alike. Mastering this technique opens doors to solving a wide range of problems and provides a deeper understanding of the properties of rational functions. The journey through this example has highlighted the power and elegance of partial fraction decomposition, solidifying its place as a vital concept in mathematical analysis. With a firm grasp of these principles, one can confidently tackle more complex rational functions and apply this technique to diverse mathematical challenges.