Partial Fraction Decomposition Of (21x - 7) / ((x - 2)(x + 3)(x^2 + X + 1))

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Introduction

In the realm of algebraic manipulations, the technique of partial fraction decomposition stands out as a powerful tool. This method allows us to break down complex rational expressions into simpler fractions, making them easier to work with in various contexts, such as integration, series expansions, and solving differential equations. Our focus here is on expressing the rational function (21x - 7) / ((x - 2)(x + 3)(x^2 + x + 1)) in partial fractions. This process involves several steps, from identifying the factors in the denominator to solving for the unknown coefficients in the numerators of the partial fractions. Understanding and mastering this technique is crucial for anyone delving into advanced mathematics and engineering.

Partial fraction decomposition is not just a mathematical trick; it’s a fundamental concept rooted in the properties of polynomials and rational functions. It allows us to reverse the process of adding fractions, which is often a more straightforward task. By decomposing a complex fraction into simpler parts, we gain a clearer understanding of its behavior and can manipulate it more easily. This is particularly useful when dealing with integrals, as the integral of a sum is the sum of the integrals, and the integrals of simpler fractions are often easier to compute. Furthermore, in fields like electrical engineering and control systems, partial fraction decomposition is used extensively to analyze transfer functions and system responses. The ability to express a complex rational function in terms of its partial fractions is a cornerstone skill for engineers and mathematicians alike. Therefore, a thorough understanding of this technique is not only academically beneficial but also practically significant in various real-world applications.

To effectively decompose a rational function into partial fractions, one must first grasp the underlying principles. The denominator of the original fraction plays a crucial role in determining the form of the partial fractions. Each distinct linear factor in the denominator gives rise to a partial fraction with a constant numerator. Repeated linear factors require a series of partial fractions, each with an increasing power of the factor in the denominator. Quadratic factors, which cannot be factored further into real linear factors, result in partial fractions with linear numerators. The degree of the numerator in each partial fraction is always one less than the degree of the corresponding factor in the denominator. This rule ensures that the partial fractions are simpler than the original fraction and that the decomposition is unique. Understanding these fundamental rules is essential for setting up the partial fraction decomposition correctly and avoiding common pitfalls. The process may seem intricate at first, but with practice, it becomes a systematic procedure that can be applied to a wide range of rational functions.

Step-by-Step Decomposition

1. Factoring the Denominator

The first crucial step in partial fraction decomposition is to factor the denominator completely. In our case, the denominator is already given as (x - 2)(x + 3)(x^2 + x + 1). We observe that (x - 2) and (x + 3) are linear factors, while (x^2 + x + 1) is a quadratic factor. To confirm that the quadratic factor cannot be factored further using real numbers, we can check its discriminant, which is b^2 - 4ac = 1^2 - 4(1)(1) = -3. Since the discriminant is negative, the quadratic factor has no real roots and cannot be factored further using real numbers. This ensures that our decomposition will be accurate and complete. Factoring the denominator completely is paramount because it dictates the structure of the partial fractions. Each distinct factor in the denominator will correspond to one or more partial fractions, and the nature of these factors (linear or quadratic, repeated or distinct) determines the form of the numerators in the partial fractions. A thorough understanding of factoring techniques is therefore essential for successful partial fraction decomposition.

2. Setting Up the Partial Fractions

Given the factored denominator (x - 2)(x + 3)(x^2 + x + 1), we can now set up the partial fraction decomposition. For the linear factors (x - 2) and (x + 3), we introduce constants A and B in the numerators, respectively. For the irreducible quadratic factor (x^2 + x + 1), we introduce a linear expression Cx + D in the numerator. Thus, we express the given fraction as:

(21x - 7) / ((x - 2)(x + 3)(x^2 + x + 1)) = A / (x - 2) + B / (x + 3) + (Cx + D) / (x^2 + x + 1)

This setup is crucial because it correctly represents the original fraction as a sum of simpler fractions. The constants A, B, C, and D are the unknowns we need to determine. The form of the partial fractions is dictated by the factors in the denominator: linear factors correspond to constant numerators, and irreducible quadratic factors correspond to linear numerators. This approach ensures that we cover all possible cases and that the decomposition is both complete and accurate. Setting up the partial fractions correctly is the foundation for solving for the unknown coefficients and obtaining the final decomposition.

3. Clearing the Denominators

To solve for the constants A, B, C, and D, we need to clear the denominators in our equation.

Starting with the equation:

(21x - 7) / ((x - 2)(x + 3)(x^2 + x + 1)) = A / (x - 2) + B / (x + 3) + (Cx + D) / (x^2 + x + 1)

We multiply both sides of the equation by the common denominator (x - 2)(x + 3)(x^2 + x + 1). This gives us:

21x - 7 = A(x + 3)(x^2 + x + 1) + B(x - 2)(x^2 + x + 1) + (Cx + D)(x - 2)(x + 3)

Clearing the denominators is a critical step as it transforms the equation into a polynomial equation, which is easier to solve. By multiplying each term by the common denominator, we eliminate the fractions, making the equation more manageable. This step sets the stage for either substituting strategic values of x or expanding the equation and equating coefficients, both of which are standard techniques for solving for the unknown constants in partial fraction decomposition. The resulting polynomial equation contains all the information needed to find the values of A, B, C, and D, and is the key to completing the decomposition process.

4. Solving for the Constants

Now that we have cleared the denominators, we have the equation:

21x - 7 = A(x + 3)(x^2 + x + 1) + B(x - 2)(x^2 + x + 1) + (Cx + D)(x - 2)(x + 3)

There are two common methods to solve for the constants A, B, C, and D: the substitution method and the equating coefficients method. Let's explore both.

a. Substitution Method

The substitution method involves choosing specific values of x that will eliminate some of the terms, making it easier to solve for the remaining constants. We can choose values that make the linear factors zero. For example:

  • Let x = 2: 21(2) - 7 = A(2 + 3)(2^2 + 2 + 1) + 0 + 0 35 = A(5)(7) 35 = 35A A = 1

  • Let x = -3: 21(-3) - 7 = 0 + B(-3 - 2)((-3)^2 + (-3) + 1) + 0 -70 = B(-5)(7) -70 = -35B B = 2

Now that we have A and B, we can substitute other values of x to solve for C and D. Let's use x = 0 and x = 1:

  • Let x = 0: 21(0) - 7 = 1(3)(1) + 2(-2)(1) + (C(0) + D)(-2)(3) -7 = 3 - 4 - 6D -7 = -1 - 6D -6 = -6D D = 1

  • Let x = 1: 21(1) - 7 = 1(4)(3) + 2(-1)(3) + (C + 1)(-1)(4) 14 = 12 - 6 - 4(C + 1) 14 = 6 - 4C - 4 14 = 2 - 4C 12 = -4C C = -3

Using the substitution method, we strategically chose values of x to eliminate terms and solve for the constants one by one. This method is particularly effective when dealing with linear factors in the denominator.

b. Equating Coefficients Method

The equating coefficients method involves expanding the right side of the equation and then equating the coefficients of like powers of x on both sides. Let's expand the right side:

21x - 7 = A(x + 3)(x^2 + x + 1) + B(x - 2)(x^2 + x + 1) + (Cx + D)(x - 2)(x + 3)

21x - 7 = A(x^3 + x^2 + x + 3x^2 + 3x + 3) + B(x^3 + x^2 + x - 2x^2 - 2x - 2) + (Cx + D)(x^2 + x - 6)

21x - 7 = A(x^3 + 4x^2 + 4x + 3) + B(x^3 - x^2 - x - 2) + (Cx + D)(x^2 + x - 6)

21x - 7 = Ax^3 + 4Ax^2 + 4Ax + 3A + Bx^3 - Bx^2 - Bx - 2B + Cx^3 + Cx^2 - 6Cx + Dx^2 + Dx - 6D

Now, we group the terms by powers of x:

21x - 7 = (A + B + C)x^3 + (4A - B + C + D)x^2 + (4A - B - 6C + D)x + (3A - 2B - 6D)

Next, we equate the coefficients of like powers of x:

  • x^3: A + B + C = 0
  • x^2: 4A - B + C + D = 0
  • x: 4A - B - 6C + D = 21
  • Constant: 3A - 2B - 6D = -7

We now have a system of four linear equations with four unknowns. We already found A = 1 and B = 2 using the substitution method, so we can substitute these values into the system:

  • 1 + 2 + C = 0 => C = -3
  • 4(1) - 2 + (-3) + D = 0 => D = 1

The equating coefficients method involves expanding the polynomial and creating a system of linear equations. This method is more systematic but can be more computationally intensive, especially for higher-degree polynomials. Both methods, however, lead to the same solution for the constants. The choice between the two often depends on the specific problem and personal preference.

5. Writing the Partial Fraction Decomposition

Having found the constants A = 1, B = 2, C = -3, and D = 1, we can now write the partial fraction decomposition:

(21x - 7) / ((x - 2)(x + 3)(x^2 + x + 1)) = 1 / (x - 2) + 2 / (x + 3) + (-3x + 1) / (x^2 + x + 1)

This final expression represents the original rational function as a sum of simpler fractions. Each term is now in a form that is easier to analyze and manipulate. The partial fraction decomposition allows us to break down a complex rational function into its fundamental components, making it more accessible for further mathematical operations.

Conclusion

In conclusion, we have successfully expressed the rational function (21x - 7) / ((x - 2)(x + 3)(x^2 + x + 1)) in partial fractions as 1 / (x - 2) + 2 / (x + 3) + (-3x + 1) / (x^2 + x + 1). This process involved several key steps, including factoring the denominator, setting up the partial fractions, clearing the denominators, and solving for the constants using both the substitution method and the equating coefficients method. Partial fraction decomposition is a powerful technique with numerous applications in mathematics, engineering, and other fields. It allows us to simplify complex rational expressions, making them easier to integrate, differentiate, and analyze. Mastering this technique is an invaluable skill for anyone working with rational functions and their applications. The ability to break down a complex problem into simpler parts is a hallmark of mathematical proficiency, and partial fraction decomposition exemplifies this principle. By understanding and applying this method, we can tackle a wide range of problems that would otherwise be intractable. The decomposition not only provides a more manageable form of the expression but also offers insights into the behavior and properties of the original function. Therefore, a thorough grasp of partial fraction decomposition is essential for anyone seeking to excel in mathematical analysis and its applications.