Problem 27 - College Algebra with Corequisite Support

Question

For the following exercises, 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 final partial fraction decomposition is \( \frac{4}{x} - \frac{3}{2(x+1)} + \frac{7}{2(x+1)^2} \).

Step 1: Understand the Problem

We need to decompose the given rational expression into partial fractions. The expression is \( \frac{5x^2 + 20x + 8}{2x(x+1)^2} \). The denominator consists of a linear factor \( 2x \) and a repeated linear factor \( (x+1)^2 \).

Step 2: Setup the Partial Fractions

For the decomposition, we write the expression as: \( \frac{A}{2x} + \frac{B}{x+1} + \frac{C}{(x+1)^2} \). Here, A, B and C are constants to be determined.

Step 3: Clear the Denominator

Multiply through by the denominator \( 2x(x+1)^2 \) to clear the fractions:\[ 5x^2 + 20x + 8 = A(x+1)^2 + B(2x)(x+1) + C(2x) \]

Step 4: Expand and Collect Terms

Expand each term on the right:- \( A(x+1)^2 = A(x^2 + 2x + 1) \)- \( B(2x)(x+1) = B(2x^2 + 2x) \)- \( C(2x) = 2Cx \)Combine these to form: \( Ax^2 + 2Ax + A + 2Bx^2 + 2Bx + 2Cx \)Which simplifies to: \( (A + 2B)x^2 + (2A + 2B + 2C)x + A \)

Step 5: Compare Coefficients

Compare the coefficients of each power of \( x \) from both sides:- For \( x^2 \): \( A + 2B = 5 \)- For \( x \): \( 2A + 2B + 2C = 20 \)- Constant term: \( A = 8 \)

Step 6: Solve the System of Equations

From \( A = 8 \), substitute this into the other equations:- \( 8 + 2B = 5 \) gives \( B = -\frac{3}{2} \)- Substitute \( A = 8 \) and \( B = -\frac{3}{2} \) into \( 2A + 2B + 2C = 20 \): \( 16 - 3 + 2C = 20 \) \( 2C = 7 \) results in \( C = \frac{7}{2} \)

Step 7: Write the Final Decomposition

Substitute \( A = 8 \), \( B = -\frac{3}{2} \), and \( C = \frac{7}{2} \) into the partial fractions:\[ \frac{5x^2 + 20x + 8}{2x(x+1)^2} = \frac{8}{2x} + \frac{-\frac{3}{2}}{x+1} + \frac{\frac{7}{2}}{(x+1)^2} \]Simplified, this becomes:\[ \frac{4}{x} - \frac{3}{2(x+1)} + \frac{7}{2(x+1)^2} \]

Key Concepts

Rational Expressions

A rational expression is similar to a fraction, but instead of numbers, it contains polynomials in the numerator and the denominator. Just like numerical fractions, rational expressions have a top part and a bottom part, known respectively as the numerator and the denominator.
Understanding rational expressions is essential because they allow us to simplify complex equations, solve polynomial equations, and even help in calculus. A rational expression looks like \( \frac{P(x)}{Q(x)} \), where \( P(x) \) and \( Q(x) \) are polynomials and \( Q(x) eq 0 \). When we talk about partial fraction decomposition, we are essentially trying to express the rational expression as a sum or difference of simpler fractions—what we call partial fractions.
Partial fraction decomposition is specifically helpful when integrating rational expressions. It breaks down complex fractions, so they are easier to work with. In the example provided, we decomposed the expression \( \frac{5x^2 + 20x + 8}{2x(x+1)^2} \) to make it much simpler.

Linear Factors

Linear factors are building blocks for polynomials. A linear factor is a polynomial of degree one, meaning it can be written in the form \( ax + b \), where both \( a \) and \( b \) are constants. When dealing with rational expressions, particularly in partial fraction decomposition, identifying linear factors in the denominator is a crucial first step.In the problem you provided, the denominator \( 2x(x+1)^2 \) has linear factors: \( 2x \) and \( x+1 \). The factor \( (x+1)^2 \) indicates that the linear factor \( x+1 \) is repeated. Recognizing repeated and distinct linear factors is key, as it influences how you set up the partial fractions.
Repetitive linear factors, like \( (x+1)^2 \), require a special approach in decomposition, where you write separate terms for each power in the expression. This is why the example uses \( \frac{A}{2x} + \frac{B}{x+1} + \frac{C}{(x+1)^2} \), allowing us to properly handle both the individual and repeated linear factors.

System of Equations

A system of equations consists of two or more equations with the same set of unknowns. Solving a system of equations means finding the values for the unknowns that make all equations true simultaneously. In partial fraction decomposition, once you've set up your partial fractions, finding the constants requires solving a system of equations.Here's how it works in our problem:

• First, we multiply through to get rid of the denominator, leaving us with an equation where each side is a polynomial.
• Next, we expand and combine the terms to create a new polynomial that matches the original numerator after equalizing both sides.

By comparing the coefficients of corresponding powers of \( x \), we create separate equations. From our example, we received:

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

Each equation corresponds to a different term's coefficient. Solving these equations provides the values for constants \( A \), \( B \), and \( C \), allowing us to construct the partial fraction decomposition accurately. This analytical step converts a theoretical problem into a practical solution by breaking down complex mathematical expressions into easy-to-solve parts.