Partial Fraction Decomposition With Quadratic Factors

Partial fraction decomposition is a powerful technique used in calculus and algebra to simplify rational functions, especially when dealing with integration or solving differential equations. When the denominator of a rational function contains quadratic factors, the process becomes slightly more complex but remains a fundamental tool for simplifying expressions. This article provides a comprehensive guide to partial fraction decomposition with quadratic factors, covering the underlying principles, step-by-step methods, and illustrative examples.

Understanding Partial Fraction Decomposition

Partial fraction decomposition is the process of breaking down a complex rational function into simpler fractions. A rational function is a function that can be expressed as the ratio of two polynomials, P(x) and Q(x), where Q(x) is not equal to zero:

f(x) = P(x) / Q(x)

The goal is to express f(x) as a sum of simpler fractions, each with a denominator that is a factor of Q(x). This is particularly useful when Q(x) can be factored into linear and/or quadratic factors.

Why Use Partial Fraction Decomposition?

• Integration: Simplifying rational functions makes integration easier. Integrals involving complex rational functions can be broken down into simpler integrals that are straightforward to solve.
• Solving Differential Equations: Many techniques for solving differential equations rely on decomposing rational functions into simpler forms.
• Simplifying Algebraic Expressions: Partial fraction decomposition can make complex algebraic expressions more manageable.

Basic Principles

Before diving into quadratic factors, it's crucial to understand the basic principles of partial fraction decomposition.

1. Proper Fraction: The degree of P(x) must be less than the degree of Q(x). If not, perform long division first to obtain a proper fraction plus a polynomial.

2. Factor the Denominator: Factor Q(x) completely into linear and irreducible quadratic factors. An irreducible quadratic factor is a quadratic expression that cannot be factored into linear factors with real coefficients (e.g., x² + 1).

3. Set Up the Decomposition: For each factor in Q(x), set up the corresponding terms in the decomposition.

   • For a linear factor (ax + b), the term is A / (ax + b), where A is a constant.
   • For an irreducible quadratic factor (ax² + bx + c), the term is (Ax + B) / (ax² + bx + c), where A and B are constants.

4. Solve for the Constants: Multiply both sides of the equation by Q(x) to clear the denominators. Then, solve for the unknown constants (A, B, C, etc.) by either:

   • Substituting Values: Plug in values of x that make the factors equal to zero (if possible).
   • Equating Coefficients: Equate the coefficients of like terms on both sides of the equation.

5. Write the Decomposition: Substitute the values of the constants back into the decomposition.

Partial Fraction Decomposition with Quadratic Factors: A Detailed Approach

When Q(x) contains irreducible quadratic factors, the decomposition involves terms of the form (Ax + B) / (ax² + bx + c). Here's a detailed step-by-step approach:

Step 1: Check if the Fraction is Proper

Ensure that the degree of the numerator P(x) is less than the degree of the denominator Q(x). If not, perform polynomial long division to express the rational function as:

f(x) = Quotient(x) + Remainder(x) / Q(x)

Then, focus on decomposing the Remainder(x) / Q(x) part.

Step 2: Factor the Denominator

Factor Q(x) completely into linear and irreducible quadratic factors. For example:

Q(x) = (x - 1)(x² + x + 1)

Here, (x - 1) is a linear factor, and (x² + x + 1) is an irreducible quadratic factor because its discriminant (b² - 4ac = 1 - 4 = -3) is negative.

Step 3: Set Up the Partial Fraction Decomposition

For each linear factor (ax + b), include a term of the form A / (ax + b). For each irreducible quadratic factor (ax² + bx + c), include a term of the form (Ax + B) / (ax² + bx + c).

For example, if:

Q(x) = (x - 1)(x² + x + 1)

The decomposition would be:

P(x) / Q(x) = A / (x - 1) + (Bx + C) / (x² + x + 1)

Step 4: Clear the Denominators

Multiply both sides of the equation by Q(x) to eliminate the denominators:

P(x) = A(x² + x + 1) + (Bx + C)(x - 1)

Step 5: Solve for the Constants

Expand the right side of the equation:

P(x) = A(x² + x + 1) + Bx(x - 1) + C(x - 1)
P(x) = Ax² + Ax + A + Bx² - Bx + Cx - C
P(x) = (A + B)x² + (A - B + C)x + (A - C)

Now, equate the coefficients of like terms on both sides of the equation. Suppose P(x) = 2x² + x + 3:

2x² + x + 3 = (A + B)x² + (A - B + C)x + (A - C)

This gives us the following system of equations:

• A + B = 2
• A - B + C = 1
• A - C = 3

Solve this system of equations for A, B, and C.

From the first equation, B = 2 - A. From the third equation, C = A - 3.

Substitute these into the second equation:

A - (2 - A) + (A - 3) = 1
A - 2 + A + A - 3 = 1
3A - 5 = 1
3A = 6
A = 2

Now find B and C:

B = 2 - A = 2 - 2 = 0
C = A - 3 = 2 - 3 = -1

So, A = 2, B = 0, and C = -1.

Step 6: Write the Partial Fraction Decomposition

Substitute the values of A, B, and C back into the original decomposition:

(2x² + x + 3) / ((x - 1)(x² + x + 1)) = 2 / (x - 1) + (0x - 1) / (x² + x + 1)
= 2 / (x - 1) - 1 / (x² + x + 1)

Examples

Let's work through a few examples to illustrate the process.

Example 1

Decompose the following rational function:

f(x) = (5x² + 3x - 2) / (x³ + 2x)

1. Check if Proper: The degree of the numerator (2) is less than the degree of the denominator (3), so it's a proper fraction.

2. Factor the Denominator:

   x³ + 2x = x(x² + 2)
   

   Here, x is a linear factor, and (x² + 2) is an irreducible quadratic factor.

3. Set Up the Decomposition:

   (5x² + 3x - 2) / (x(x² + 2)) = A / x + (Bx + C) / (x² + 2)
   

4. Clear the Denominators:

   5x² + 3x - 2 = A(x² + 2) + (Bx + C)x
   5x² + 3x - 2 = Ax² + 2A + Bx² + Cx
   5x² + 3x - 2 = (A + B)x² + Cx + 2A
   

5. Solve for the Constants:

   Equate the coefficients:

   • A + B = 5
   • C = 3
   • 2A = -2

   From 2A = -2, A = -1. From A + B = 5, B = 5 - A = 5 - (-1) = 6. C = 3.

   So, A = -1, B = 6, and C = 3.

6. Write the Partial Fraction Decomposition:

   (5x² + 3x - 2) / (x³ + 2x) = -1 / x + (6x + 3) / (x² + 2)
   

Example 2

Decompose the following rational function:

f(x) = (x² + 2x + 3) / ((x - 1)(x² + 1))

1. Check if Proper: The degree of the numerator (2) is equal to the degree of the denominator (3), so we actually don't need to do long division.

2. Factor the Denominator:

   The denominator is already factored: (x - 1)(x² + 1), where (x - 1) is a linear factor and (x² + 1) is an irreducible quadratic factor.

3. Set Up the Decomposition:

   (x² + 2x + 3) / ((x - 1)(x² + 1)) = A / (x - 1) + (Bx + C) / (x² + 1)
   

4. Clear the Denominators:

   x² + 2x + 3 = A(x² + 1) + (Bx + C)(x - 1)
   x² + 2x + 3 = Ax² + A + Bx² - Bx + Cx - C
   x² + 2x + 3 = (A + B)x² + (-B + C)x + (A - C)
   

5. Solve for the Constants:

   Equate the coefficients:

   • A + B = 1
   • -B + C = 2
   • A - C = 3

   From the first equation, B = 1 - A. From the third equation, C = A - 3.

   Substitute these into the second equation:

   -(1 - A) + (A - 3) = 2
   -1 + A + A - 3 = 2
   2A - 4 = 2
   2A = 6
   A = 3
   

   Now find B and C:

   B = 1 - A = 1 - 3 = -2
   C = A - 3 = 3 - 3 = 0
   

   So, A = 3, B = -2, and C = 0.

6. Write the Partial Fraction Decomposition:

   (x² + 2x + 3) / ((x - 1)(x² + 1)) = 3 / (x - 1) + (-2x + 0) / (x² + 1)
   = 3 / (x - 1) - (2x) / (x² + 1)
   

Dealing with Repeated Quadratic Factors

If the denominator has repeated quadratic factors, such as (ax² + bx + c)², the decomposition must include terms for each power of the factor up to the highest power present. For example, if:

Q(x) = (x² + 1)²

The decomposition would be:

P(x) / Q(x) = (Ax + B) / (x² + 1) + (Cx + D) / (x² + 1)²

The process for solving the constants remains the same, but the algebraic manipulation becomes more involved.

Practical Applications

Partial fraction decomposition with quadratic factors is invaluable in various fields:

• Calculus: As previously mentioned, it simplifies integration. For example, integrating a rational function with an irreducible quadratic factor in the denominator often requires completing the square and using trigonometric substitution after partial fraction decomposition.
• Engineering: It's used in circuit analysis, control systems, and signal processing to simplify transfer functions and analyze system behavior.
• Physics: It arises in problems involving electromagnetic fields and wave propagation.

Common Mistakes to Avoid

• Forgetting to Check if the Fraction is Proper: Always ensure that the degree of the numerator is less than the degree of the denominator.
• Incorrectly Factoring the Denominator: Make sure that the denominator is factored completely and that all quadratic factors are irreducible.
• Setting Up the Wrong Decomposition: Ensure that each factor in the denominator has the correct corresponding term in the decomposition.
• Algebraic Errors: Be careful with algebraic manipulations when clearing denominators and solving for the constants.
• Incorrectly Solving the System of Equations: Double-check your work when solving for the constants to avoid errors.

Conclusion

Partial fraction decomposition with quadratic factors is a crucial technique for simplifying rational functions, especially when dealing with integration or solving differential equations. By understanding the underlying principles and following the step-by-step methods outlined in this article, you can confidently decompose complex rational functions into simpler forms, making them easier to work with in various mathematical and engineering applications. The key is to practice and pay close attention to detail in each step of the process.