Problem 28 - College Algebra

Question

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

\(\frac{1}{5x} - \frac{1}{3(3x+5)} + \frac{20}{3(3x+5)^2}\)

Step 1: Identify the denominators in the partial fraction form

The denominator of the given fraction is \(5x(3x+5)^2\). Recognize that it consists of a linear factor \(5x\) and a repeating linear factor \((3x+5)^2\).

Step 2: Write the general form of the partial fraction decomposition

For a repeating linear factor, we use separate terms for each power, up to the power of the factor. This gives the decomposition:\[\frac{A}{5x} + \frac{B}{3x+5} + \frac{C}{(3x+5)^2}\]

Step 3: Combine the partial fractions into a single fraction

Multiply each term by the common denominator \(5x(3x+5)^2\) to combine them:\[\frac{A(3x+5)^2 + B(5x)(3x+5) + C(5x)}{5x(3x+5)^2}\]

Step 4: Equate the numerators

Set the numerator of the original fraction \(4x^2 + 55x + 25\) equal to the expanded form of the combined numerator:\[A(3x+5)^2 + B(5x)(3x+5) + C(5x) = 4x^2 + 55x + 25\]

Step 5: Expand and simplify

First, expand \(A(3x+5)^2 = A(9x^2 + 30x + 25)\),\(B(5x)(3x+5) = B(15x^2 + 25x)\),and\(C(5x) = 5Cx\).Now simplify the equation to get:\(9Ax^2 + 30Ax + 25A + 15Bx^2 + 25Bx + 5Cx = 4x^2 + 55x + 25\).

Step 6: Collect like terms and equate coefficients

Combine like terms:\((9A + 15B)x^2 + (30A + 25B + 5C)x + 25A = 4x^2 + 55x + 25\).Now, equate the coefficients:1. \(9A + 15B = 4\)2. \(30A + 25B + 5C = 55\)3. \(25A = 25\)

Step 7: Solve the system of equations

From equation 3, solve for \(A\):\(25A = 25 \Rightarrow A = 1\).Substitute \(A = 1\) in equations 1 and 2:1. \(9(1) + 15B = 4 \Rightarrow 15B = -5 \Rightarrow B = -\frac{1}{3}\)2. \(30(1) + 25(-\frac{1}{3}) + 5C = 55\)Simplify and solve for \(C\):\(30 -\frac{25}{3} + 5C = 55\Rightarrow 5C = 55 - 30 + \frac{25}{3} \Rightarrow 5C = 25 + \frac{25}{3} \Rightarrow 5C = \frac{75}{3} + \frac{25}{3}\Rightarrow 5C = \frac{100}{3}\Rightarrow C = \frac{20}{3}\).

Step 8: Write the final partial fraction decomposition

Substitute \(A = 1\), \(B = -\frac{1}{3}\), and \(C = \frac{20}{3}\) back into the partial fraction decomposition:\[\frac{1}{5x} - \frac{1}{3(3x+5)} + \frac{20}{3(3x+5)^2}\]

Key Concepts

Repeating Linear Factors

Repeating linear factors appear in partial fraction decomposition when a factor is raised to a power greater than one. In the given exercise, we encounter the repeating linear factor \((3x + 5)^2\).
To tackle this, we have to take into account every power of the repeating factor, starting from the first up to the highest power. This means that our decomposition includes separate terms for \((3x + 5)\) and \((3x + 5)^2\).
The general decomposed form for a repeating factor \((ax + b)^n\) up to power \(n\) uses each decreasing power of the factor:

• \( \frac{B}{ax + b} \, \text{for power 1} \)
• \( \frac{C}{(ax + b)^2} \, \text{for power 2} \)

This pattern continues if the power is higher. In our example, since it's squared, we only consider these two terms.

Linear Factors

A linear factor is a polynomial of the form \(ax + b\), where \(a\) and \(b\) are constants, and \(x\) is a variable. Linear factors are simple yet critical components in partial fraction decomposition.
In our problem, we identify a simple linear factor \(5x\) in the denominator. Linear factors are much easier to handle than repeating ones as they require only one term for decomposition.
When dealing with linear factors, the partial fraction corresponding to \(5x\) is of the form:

• \( \frac{A}{5x} \)

It's crucial to isolate and write out each linear factor to accurately decompose the initial fraction.

Equating Coefficients

Equating coefficients is a powerful technique used in partial fraction decomposition to solve for unknown constants. Once you have established the general partial fraction form, you aim to match it to the original fraction's numerator.
For our exercise, after combining and expanding the partial fractions, we obtained:\[9Ax^2 + 15Bx^2 + 30Ax + 25Bx + 5Cx + 25A = 4x^2 + 55x + 25\]The next step is to combine like terms on one side, leading to a new equation based on powers of \(x\). Equate the coefficients of similar terms to those on the right-hand side:

• The coefficient of \(x^2\) gives us: \(9A + 15B = 4\)
• The coefficient of \(x\) yields: \(30A + 25B + 5C = 55\)
• Finally, the constant term provides: \(25A = 25\)

Through solving these equations, we determine the values of the unknown constants \(A\), \(B\), and \(C\), effectively completing the decomposition.