Problem 27 - College Algebra

Question

Find the decomposition of the partial fraction for the repeating linear factors. \(\frac{5 x^{2}+20 x+8}{2 x(x+1)^{2}}\)

Step-by-Step Solution

Answer

The partial fraction decomposition is \(\frac{8}{2x} - \frac{3}{2(x+1)} + \frac{7}{2(x+1)^2}\).

Step 1: Identify the Denominator

The denominator of the given fraction is \(2x(x+1)^2\). This indicates that we have repeating linear factors. Specifically, \(x\) is a simple linear factor, and \((x+1)^2\) is a repeating linear factor of degree 2.

Step 2: Set Up the Partial Fraction Form

For the partial fraction decomposition, express the fraction as a sum of simpler fractions with unknown coefficients. Use: \[ \frac{A}{2x} + \frac{B}{x+1} + \frac{C}{(x+1)^2} \] where \(A\), \(B\), and \(C\) are constants to be determined.

Step 3: Combine Fractions on the Right Side

To find \(A\), \(B\), and \(C\), combine the fractions into a single fraction on the right: \[ \frac{A(x+1)^2 + B(2x)(x+1) + C(2x)}{2x(x+1)^2} \] This should equal the original fraction \(\frac{5x^2 + 20x + 8}{2x(x+1)^2}\).

Step 4: Set Numerators Equal

Set the numerators equal since the denominators of the entire equation are already equal: \[ A(x+1)^2 + B(2x)(x+1) + C(2x) = 5x^2 + 20x + 8 \]

Step 5: Expand the Equation

Expand the left-hand side to simplify the equation:- \( A(x^2 + 2x + 1) = Ax^2 + 2Ax + A \)- \( B(2x^2 + 2x) = 2Bx^2 + 2Bx \)- \( C(2x) = 2Cx \)Add these together: \[ (A + 2B)x^2 + (2A + 2B + 2C)x + A \]

Step 6: Equate Coefficients

Equate the coefficients of the same powers of \(x\) from each side of the equation:1. For \(x^2\) term: \( A + 2B = 5 \)2. For \(x\) term: \( 2A + 2B + 2C = 20 \)3. For the constant term: \( A = 8 \)

Step 7: Solve the System of Equations

From the equations:- \( A = 8 \)- Substitute \( A = 8 \) into the first equation: \( 8 + 2B = 5 \), giving \( 2B = -3 \), hence \( B = -\frac{3}{2} \)- Substitute \( A = 8 \) and \( B = -\frac{3}{2} \) into the second equation: \( 2(8) + 2(-\frac{3}{2}) + 2C = 20 \)- Simplify: \( 16 - 3 + 2C = 20 \)- Solve for \( C \): \( 2C = 7 \) so \( C = \frac{7}{2} \)

Step 8: Write the Final Decomposition

With the solved values, write the decomposition:\[ \frac{8}{2x} - \frac{3}{2(x+1)} + \frac{7}{2(x+1)^2} \]

Key Concepts

Repeating Linear Factors

In partial fraction decomposition, repeating linear factors are polynomial expressions that appear more than once in the denominator. In this exercise, we have the denominator \(2x(x+1)^2\).
The factor \((x+1)^2\) is considered repeating because it is squared, indicating it repeats twice.
This requires us to approach the decomposition differently than we would for non-repeating factors.

When we deal with repeating linear factors, each factor gets its own distinct fraction in the partial decomposition. For example, a factor like \((x+1)^2\) would result in two separate terms:

• \(\frac{B}{x+1}\)
• \(\frac{C}{(x+1)^2}\)

This ensures that each potential part of the repeating factor is accounted for correctly in the decomposition.
It allows for more precise solving by considering all possible versions of the factor's influence.

System of Equations

An essential step in solving partial fraction decomposition is setting up a system of equations to find unknown coefficients, such as \(A\), \(B\), and \(C\) in this exercise.
After expressing the original polynomial in terms of simpler fractions, we obtain a structure whose sum should reproduce the original formula when combined.

We equate the numerators of both fractions:

• The numerator from the decomposed expression must match the original polynomial's numerator expression.
• This yields a polynomial equation with various terms that should match coefficients on both sides of the equality.

We derive separate equations for each coefficient of the term, like \(x^2\), \(x\), and the constant term, leading to a system of equations:

• \(A + 2B = 5\)
• \(2A + 2B + 2C = 20\)
• \(A = 8\)

Solving these equations simultaneously allows us to find the values of \(A\), \(B\), and \(C\).

Polynomial Division

Polynomial division is sometimes required in partial fraction decomposition when the degree of the numerator is equal to or greater than the degree of the denominator.
It ensures that the expression is properly simplified before starting the decomposition process.
For fractions where this isn't necessary, as is in this problem, it is important to understand why this step is included in other problems.

Let’s look into it briefly:

• If the degree of the numerator is higher or equal, division helps split it into a sum of a polynomial and a fraction where the numerator is lesser than the denominator.
• This makes it easier to work with and apply further partial fraction methods.

Avoid frequent confusion by always checking the degrees. If division is needed, practicing with polynomial long division or synthetic division methods can be very useful in college algebra.

College Algebra

In college algebra, partial fraction decomposition is a crucial technique that students often encounter when working with complex rational expressions.
It's especially useful in calculus and higher-level mathematics, such as when integrating rational functions.

Mastering this method assists in simplifying complex expressions and solving integral problems that involve fractions.

• Understanding linear and repeating factors enhance this learning process.
• Skill in solving systems of equations is a vital part for determining coefficients in fractions.

In addition, practicing different scenarios, like finding decompositions with one or more repeating factors, reinforces foundational algebra concepts.
It is a stepping stone to handle more advanced topics in mathematics and helps improve overall problem-solving skills in algebra.