Problem 28 - College Algebra with Corequisite Support

Question

For the following exercises, find the decomposition of the partial fraction for the repeating linear factors. $$ \frac{4 x^{2}+55 x+25}{5 x(3 x+5)^{2}} $$

Step-by-Step Solution

Answer

\(\frac{1}{5x} - \frac{1}{3(3x+5)} + \frac{14}{3(3x+5)^2}\)

Step 1: Identify the Type of Fraction

The given fraction is \(\frac{4x^2+55x+25}{5x(3x+5)^2}\). The denominator has a repeating linear factor \((3x+5)^2\) and a separate linear factor \(5x\). Thus, it fits the partial fraction decomposition for repeating linear factors.

Step 2: Set Up the Partial Fraction Form

Express the partial fraction as a sum of terms that correspond to the factors in the denominator. For \((3x+5)^2\), we use components for both \((3x+5)^2\) and \((3x+5)\), leading to:\[ \frac{A}{5x} + \frac{B}{3x+5} + \frac{C}{(3x+5)^2} \]

Step 3: Clear the Denominator

Multiply both sides by the denominator \(5x(3x+5)^2\) to eliminate the fractions:\[ 4x^2 + 55x + 25 = A(3x+5)^2 + B(5x)(3x+5) + C(5x) \]

Step 4: Expand and Simplify Terms

Expand each term on the right-hand side:- For \(A(3x+5)^2\): \(A(9x^2 + 30x + 25)\).- For \(B(5x)(3x+5)\): \(B(15x^2 + 25x)\).- For \(C(5x)\): \(5Cx\).Resulting equation:\[ A(9x^2 + 30x + 25) + B(15x^2 + 25x) + 5Cx = 4x^2 + 55x + 25 \]

Step 5: Combine Like Terms

Combine like terms on the right-hand side to match coefficients with the left-hand side:\[ (9A + 15B)x^2 + (30A + 25B + 5C)x + 25A = 4x^2 + 55x + 25 \]

Step 6: Set Up System of Equations

By comparing coefficients for terms \(x^2\), \(x\), and the constant terms, create a system of equations:1. \(9A + 15B = 4\)2. \(30A + 25B + 5C = 55\)3. \(25A = 25\)

Step 7: Solve the System of Equations

Solve these equations:- From Equation 3: \(25A = 25\) results in \(A = 1\).Substitute \(A = 1\) in the other equations:- In Equation 1: \(9(1) + 15B = 4\) simplifies to \(15B = -5\), so \(B = -\frac{1}{3}\).- Substitute \(A = 1\) and \(B = -\frac{1}{3}\) in Equation 2: \[ 30(1) + 25(-\frac{1}{3}) + 5C = 55 \] \[ 30 - \frac{25}{3} + 5C = 55 \] Solving gives \(5C = 55 - 30 + \frac{25}{3}\) and \(C = \frac{14}{3}\).

Step 8: Write the Partial Fraction Decomposition

Substitute the values of \(A\), \(B\), and \(C\) into the partial fraction form:\[ \frac{1}{5x} - \frac{1}{3(3x+5)} + \frac{14}{3(3x+5)^2} \]

Key Concepts

Repeating Linear Factors

When dealing with partial fraction decomposition, one of the intriguing scenarios is managing repeating linear factors in the denominator of a fraction. These repeating factors require special attention to ensure the fraction is accurately decomposed.
In our example, the denominator consists of

• a single linear factor: \(5x\), and
• a repeating linear factor: \((3x+5)^2\).

The presence of the \((3x+5)^2\) term implies that we have a root repeated twice, meaning the factor

• \(3x+5\)

exists two times in the polynomial. This repetition directs us to incorporate separate terms for each power of the factor in the decomposition formula:

• \(\frac{B}{3x+5}\)
• \(\frac{C}{(3x+5)^2}\)

These separate terms allow for capturing the effect of each repeated factor, enabling a complete and consistent decomposition of the original fraction.

System of Equations

Once the appropriate forms of terms have been established for the partial fraction decomposition, the next step involves forming a system of equations based on matching coefficients. This step harnesses the power of algebraic equations to solve for unknown constants.
As seen in the solution, by comparing equivalent coefficients between both sides of the equation, we can construct a system that comprises:

• \(9A + 15B = 4\) for \(x^2\) terms,
• \(30A + 25B + 5C = 55\) for \(x\) terms, and
• \(25A = 25\) for constant terms.

These equations arise from equating the coefficients of like terms. Solving this system allows us to find the exact values for \(A\), \(B\), and \(C\), which are needed for reconstructing the decomposed fractions.

Algebraic Fractions

Algebraic fractions like the one in this exercise are expressions involving variables in the numerator and the denominator. Understanding how to manipulate algebraic fractions is critical for solving advanced algebra problems and performing operations such as addition, subtraction, or decomposition.
The example fraction \(\frac{4x^2 + 55x + 25}{5x(3x+5)^2}\) exemplifies an algebraic fraction where the numerator and denominator are both polynomials.
Handling algebraic fractions effectively requires determining common divisors, simplifying expressions, and in this case, decomposing the fraction into simpler parts that can be managed individually or applied to solve integrals.

Coefficient Matching

Coefficient matching is the essential technique used in partial fraction decomposition to find the unknowns involved in the expression.
Once the equation is cleared of fractions by multiplying through by the denominator, you are left with a polynomial equation. Here, the method of coefficient matching comes in:

• Expand the polynomial terms.
• Collect like terms on each side of the equation.
• Set coefficients of corresponding terms equal to form equations.

By matching the coefficients of corresponding powers of \(x\) across both sides, as shown here:

• \( (9A + 15B)x^2 = 4x^2 \)
• \((30A + 25B + 5C)x = 55x \)
• \( 25A = 25 \)

You are led to a system of linear equations. Solving these equations provides the necessary coefficients \(A, B,\) and \(C\), enabling you to write the equivalent decomposed, simpler fraction expressions.