Problem 20 - College Algebra

Question

For the following exercises, find the decomposition of the partial fraction for the repeating linear factors. $$\frac{-5 x-19}{(x+4)^{2}}$$

Step-by-Step Solution

Answer

\(\frac{-5}{x+4} + \frac{1}{(x+4)^2}\) is the decomposition.

Step 1: Identify the Type of Partial Fraction

The given expression is \(\frac{-5x-19}{(x+4)^2}\). The denominator \((x+4)^2\) is a repeated linear factor. This tells us we need a particular form of partial fraction decomposition.

Step 2: Set Up the Partial Fraction Decomposition

For a repeated linear factor \((x+4)^2\), the partial fraction decomposition will have the form \(\frac{A}{x+4} + \frac{B}{(x+4)^2}\), where \(A\) and \(B\) are constants we need to find.

Step 3: Create the Equation to Find Unknown Constants

Set the original expression \(\frac{-5x-19}{(x+4)^2}\) equal to the partial fraction form \(\frac{A}{x+4} + \frac{B}{(x+4)^2}\). Multiply both sides by \((x+4)^2\) to get rid of the denominators: \[-5x-19 = A(x+4) + B\].

Step 4: Expand and Simplify the Right Side

Expand the equation: \[-5x-19 = A(x+4) + B\] becomes \[-5x-19 = Ax + 4A + B\].

Step 5: Equate Coefficients

By comparing coefficients of like terms, we set up the following equations: 1. For the terms involving \(x\): \(-5 = A\).2. For the constant terms: \(-19 = 4A + B\).

Step 6: Solve for Unknown Constants

Substitute \(A = -5\) (from the first equation) into the second equation:\(-19 = 4(-5) + B\).This gives \(-19 = -20 + B\), so \(B = 1\).

Step 7: Write the Final Partial Fraction Decomposition

Substitute \(A = -5\) and \(B = 1\) back into the partial fraction decomposition:\[\frac{-5}{x+4} + \frac{1}{(x+4)^2}\].

Key Concepts

Repeated Linear Factor

When dealing with partial fraction decomposition, a repeated linear factor in the denominator plays a crucial role. Here, the expression \((x+4)^2\) signifies a repeated factor, meaning \((x+4)\) appears twice. This characteristic influences how we break down the fraction.

Instead of a single term in the decomposition, every instance of the factor will contribute. For example, for \((x+4)^2\), the decomposition must involve both \(x+4\) and \((x+4)^2\) in separate components.

• \(A/(x+4)\) - addresses the single occurrence of the factor.
• \(B/(x+4)^2\) - for its squared repetition.

This ensures we account for all power levels of the repeated factor.

Constant Terms

In a partial fraction decomposition, the constant terms are the non-variable components in the expression. These terms are essential when matching like terms during the solving process.

For this problem, when decomposing the expression \(-5x-19\), \(-19\) is the constant term. It becomes part of the equation you set up and solve for.

In the system of equations derived from equating coefficients, focusing on constant terms helps in determining unknown constants. Here, we used the equation:

• \(-19 = 4A + B\)

to solve for \(B\) once \(A\) was known.

Equivalent Expressions

The foundation of partial fraction decomposition lies in creating equivalent expressions. The goal is to express something complex as a sum of simpler fractions.

In our example, the expression \(\frac{-5x-19}{(x+4)^2}\) is transformed into \(\frac{A}{x+4} + \frac{B}{(x+4)^2}\).

• This new expression is equivalent to the original.
• Each part corresponds directly to the terms derived from the denominator's factors.

By multiplying through by the denominator, we remove it, leading to an equation: \(-5x-19 = A(x+4) + B\). This equation represents both expressions equally, enabling us to compare coefficients to find \(A\) and \(B\).

Unknown Constants

Determining the unknown constants \(A\) and \(B\) is the central task in partial fraction decomposition. These constants allow you to express a complicated rational function in terms of simpler fractions.

In our solution, these were obtained by setting up equations based on comparing coefficients:

• \(-5 = A\)
• \(-19 = 4A + B\)

These equations stem from equating the original numerator to the expanded decomposition form.
Finding these constants involves algebraic manipulation, such as substitution and solving linear equations, to identify the values that satisfy both equations.