Problem 25 - Calculus Single Variable

Question

Use Heaviside's method to calculate the partial fraction decomposition of the given rational function. $$ \frac{7 x^{2}+9 x+5}{4 x^{3}-x} $$

Step-by-Step Solution

Answer

The partial fraction decomposition is \(\frac{-5}{x} + \frac{11.25}{2x - 1} + \frac{2.25}{2x + 1}\).

Step 1: Factor the Denominator

First, identify the denominator of the rational function. The denominator is given as \(4x^3 - x\). Factor it to make it easier to use Heaviside's method.\[4x^3 - x = x(4x^2 - 1) = x(2x - 1)(2x + 1)\]Now we can see that the denominator has three factors: \(x\), \(2x - 1\), and \(2x + 1\).

Step 2: Set Up the Partial Fraction Decomposition

Now that we have factored the denominator, set up the partial fraction decomposition using those factors. The form will be:\[\frac{7x^2 + 9x + 5}{(x)(2x - 1)(2x + 1)} = \frac{A}{x} + \frac{B}{2x - 1} + \frac{C}{2x + 1}\]where \(A\), \(B\), and \(C\) are constants that need to be determined.

Step 3: Clear the Denominator

Multiply both sides of the equation by the original denominator \((x)(2x - 1)(2x + 1)\) to clear the fractions.\[7x^2 + 9x + 5 = A(2x - 1)(2x + 1) + Bx(2x + 1) + Cx(2x - 1)\]Expand each term on the right-hand side to prepare for comparison.

Step 4: Expand and Compare Coefficients

Expand the terms on the right side:\[A(2x - 1)(2x + 1) = A(4x^2 - 1) = 4Ax^2 - A\] \[Bx(2x + 1) = Bx(2x + 1) = 2Bx^2 + Bx\]\[Cx(2x - 1) = Cx(2x - 1) = 2Cx^2 - Cx\]Combine them: \[4Ax^2 - A + 2Bx^2 + Bx + 2Cx^2 - Cx\]Compare with the original:\[7x^2 + 9x + 5 = (4A + 2B + 2C)x^2 + (B - C)x - A\]

Step 5: Solve the System of Equations

Set up equations based on equating coefficients:1. \(4A + 2B + 2C = 7\)2. \(B - C = 9\)3. \(-A = 5\)Solve equation 3: \(A = -5\)Substitute \(A = -5\) in equation 1: \[4(-5) + 2B + 2C = 7 \rightarrow -20 + 2B + 2C = 7 \rightarrow 2B + 2C = 27 \rightarrow B + C = 13.5\]Now solve the system:\(B - C = 9\)\(B + C = 13.5\)Add both equations to eliminate \(C\): \[2B = 22.5 \rightarrow B = 11.25\]Substitute \(B = 11.25\) into \(B + C = 13.5\):\[11.25 + C = 13.5 \rightarrow C = 2.25\]

Step 6: Write the Partial Fraction Decomposition

Substitute the values of \(A\), \(B\), and \(C\) back into the partial fractions:\[\frac{7x^2 + 9x + 5}{4x^3 - x} = \frac{-5}{x} + \frac{11.25}{2x - 1} + \frac{2.25}{2x + 1}\]This is the desired partial fraction decomposition.

Key Concepts

Heaviside's method

Heaviside's method offers an efficient and straightforward approach to partial fraction decomposition, especially when it involves simple linear factors. The main idea is to express a complex rational function as a sum of simpler fractions. These simpler fractions are typically easier to handle in calculus, particularly when integrating.

• Before applying Heaviside's method, ensure the denominator is fully factored into linear and irreducible quadratic factors.
• Match parts of the numerator to determine constants corresponding to each simpler fraction.
• The goal is to transform the rational function into a series of manageable algebraic expressions.

By converting a complicated function into a sum of basic fractions, Heaviside's method significantly reduces the complexity encountered in solving calculus problems.

Rational functions

Rational functions are formed when you divide one polynomial by another. These are represented as the quotient of two polynomials, such as \( \frac{P(x)}{Q(x)} \). In the given exercise, the rational function is \( \frac{7x^2 + 9x + 5}{4x^3 - x} \).

• The numerator \( P(x) = 7x^2 + 9x + 5 \) and the denominator \( Q(x) = 4x^3 - x \) are both polynomials, ensuring our function is rational.
• Understanding rational functions is crucial, as they frequently appear in algebra, calculus, and various applied fields.
• Behavior, including asymptotes and intercepts, of rational functions depends significantly on their numerator and denominator.

Rational functions are important due to their versatility and frequent appearance in mathematical problems, making them a key concept in understanding fractions and their behaviors.

Factoring polynomials

Factoring polynomials is a critical step in working with rational functions and applying methods like partial fraction decomposition. The aim is to break down a complex polynomial into simpler, easily manageable factors.

• In this exercise, the denominator \( 4x^3 - x \) was factored into \( x(2x-1)(2x+1) \), making it easier to decompose.
• Factoring involves finding roots or using techniques such as grouping or using the quadratic formula for more complex expressions.
• A fully factored polynomial unveils the hidden structure within algebraic expressions, aiding in subsequent calculations.

Mastery of polynomial factoring enables smoother progress in algebraic manipulation, paving the way for efficient application of methods such as partial fraction decomposition and simplifying expressions.

Coefficient comparison

Coefficient comparison is a technique used to solve equations deriving from partial fraction decomposition. Once polynomial expressions are simplified to linear terms, equate the coefficients of corresponding powers of \( x \) to solve for unknown constants.

• Each coefficient of\( x^2, x, \) and the constant term in your expanded expression must match corresponding terms in the original polynomial.
• For the problem at hand, equate coefficients from both sides: \(4A + 2B + 2C \) for \( x^2, \) \(B - C \) for \( x, \) and \(-A \) for constant terms.
• Solving these equations gives the values of the unknown constants \( A, B, \) and \( C \).

Using coefficient comparison simplifies solving for constants in partial fraction decomposition, thereby breaking down complex algebraic structures into manageable parts that align with given polynomial identities.