Problem 36 - College Algebra with Corequisite Support

Question

For the following exercises, find the decomposition of the partial fraction for the irreducible nonrepeating quadratic factor. $$ \frac{4 x^{2}}{(x+5)\left(x^{2}+7 x-5\right)} $$

Step-by-Step Solution

Answer

\(\frac{4x^2}{(x+5)(x^2+7x-5)} = \frac{-20/3}{x+5} + \frac{32/3x - 20/3}{x^2+7x-5}\)

Step 1: Identify the Denominator Factors

The denominator of the given rational function is \((x + 5)(x^2 + 7x - 5)\). Notice that \(x + 5\) is linear and \(x^2 + 7x - 5\) is an irreducible quadratic factor.

Step 2: Set Up the Partial Fraction Decomposition

For the linear factor \(x + 5\), assign a constant \(A\). For the irreducible quadratic factor \(x^2 + 7x - 5\), assign a linear expression \(Bx + C\). Thus, the decomposition is: \[ \frac{4x^2}{(x + 5)(x^2 + 7x - 5)} = \frac{A}{x + 5} + \frac{Bx + C}{x^2 + 7x - 5} \]

Step 3: Clear the Denominator

Multiply both sides of the equation by the common denominator \((x + 5)(x^2 + 7x - 5)\) to eliminate the fractions. This yields: \[ 4x^2 = A(x^2 + 7x - 5) + (Bx + C)(x + 5) \]

Step 4: Distribute and Combine Like Terms

Expand the equation: - Distribute \(A\) over \(x^2 + 7x - 5\) to get \(Ax^2 + 7Ax - 5A\).- Distribute \((Bx + C)\) over \((x + 5)\) to get \(Bx^2 + 5Bx + Cx + 5C\).Combine all terms: \[ 4x^2 = (A + B)x^2 + (7A + 5B + C)x + (5C - 5A) \]

Step 5: Equate Coefficients

Compare the coefficients of corresponding powers of \(x\) on both sides of the equation:- Coefficient of \(x^2\): \( A + B = 4 \)- Coefficient of \(x\): \( 7A + 5B + C = 0 \)- Constant term: \( 5C - 5A = 0 \)

Step 6: Solve System of Equations

Solve the system of equations:1. From \(5C - 5A = 0\), deduce that \(C = A\).2. Substitute \(C = A\) into \(7A + 5B + C = 0\) to get \(7A + 5B + A = 0\), leading to \(8A + 5B = 0\).3. Use \(A + B = 4\) and \(8A + 5B = 0\) to find \(A\) and \(B\).Solving these:- From \(8A + 5B = 0\), express \(B = -\frac{8}{5}A\).- Substitute into \(A + B = 4\): \[ A - \frac{8}{5}A = 4 \]\[ \frac{-3}{5}A = 4 \]\[ A = -\frac{20}{3} \]4. Substituting \(A = -\frac{20}{3}\) into \(B = -\frac{8}{5}A \) gives \(B = \frac{32}{3}\).5. Since \(C = A\), \(C = -\frac{20}{3}\).

Step 7: Write the Final Decomposition

Substitute the values of \(A\), \(B\), and \(C\) back into the partial fraction form:\[ \frac{4x^2}{(x + 5)(x^2 + 7x - 5)} = \frac{-20/3}{x + 5} + \frac{\frac{32}{3}x - \frac{20}{3}}{x^2 + 7x - 5} \]

Key Concepts

Irreducible Quadratic Factor

When dealing with partial fraction decomposition, the term "irreducible quadratic factor" often plays a significant role. In this context, a quadratic factor refers to a polynomial of the form \(ax^2 + bx + c\) in the denominator where its discriminant \(b^2 - 4ac\) is less than zero. This means the polynomial does not have real roots and cannot be further broken down into linear factors with real coefficients.
This concept is essential because irreducible quadratic factors require special attention during decomposition. When facing such a denominator factor, instead of using simple constants like you would with linear factors, you use linear expressions like \(Bx + C\). These expressions help us represent the part of the rational function where real roots are not present, allowing us to balance the equation effectively.

System of Equations

A system of equations is a collection of multiple equations that are solved together. In partial fraction decomposition, these systems arise when you've set up an initial decomposition of the rational function. You equate coefficients from an expanded form of the rational function to solve for unknown variables like \(A\), \(B\), and \(C\).
For example, if your decomposed rational function is \( \frac{A}{x+5} + \frac{Bx + C}{x^2 + 7x - 5} \), after clearing denominators and expanding the expressions, you might end up with a system like:

• \(A + B = 4\)
• \(7A + 5B + C = 0\)
• \(5C - 5A = 0\)

Solving such systems requires using algebraic techniques like substitution or elimination to find values for the variables that satisfy all equations simultaneously.

Rational Function

A rational function is a ratio of two polynomials, typically expressed as \(\frac{P(x)}{Q(x)}\). In partial fraction decomposition, you focus on breaking this rational function into simpler fractions that are easier to integrate or differentiate.
For instance, in the exercise, the rational function is \(\frac{4x^2}{(x+5)(x^2+7x-5)}\). The goal is to rewrite this function as a sum of simpler rational expressions, each of which is easily managed. This decomposition aids in performing operations like integration, which can be challenging on more complex rational functions.
By simplifying the rational function into more digestible parts, students and mathematicians can tackle and solve problems much more effectively.

Denominator Factors

Denominator factors are the polynomials in the denominator of a rational function. Identifying these factors is the first and crucial step in partial fraction decomposition.
In the given problem, the denominator is \((x + 5)(x^2 + 7x - 5)\). Breaking this down helps us understand what types of partial fractions will be used in the decomposition. In this case, \(x + 5\) is a linear factor, and \(x^2 + 7x - 5\) is an irreducible quadratic factor.
The factorization informs the setup of the partial fraction: for linear, use constants; for irreducible quadratics, use linear expressions. Such a strategic setup is necessary for accurately solving rational functions and applying calculus operations where partial fractions come into play.