Problem 20 - College Algebra

Question

Find the decomposition of the partial fraction for the repeating linear factors. \(\frac{-5 x-19}{(x+4)^{2}}\)

Step-by-Step Solution

Answer

\(\frac{-5}{x+4} + \frac{1}{(x+4)^2}\)

Step 1: Identify the Form of Decomposition

The expression has a repeating linear factor, \((x+4)^2\). Therefore, the partial fraction decomposition takes the form: \(\frac{-5x-19}{(x+4)^2} = \frac{A}{x+4} + \frac{B}{(x+4)^2}\), where \(A\) and \(B\) are constants to be determined.

Step 2: Clear the Fractions

Multiply through by \((x+4)^2\) to eliminate the fractions on both sides, resulting in: \[-5x - 19 = A(x+4) + B.\]

Step 3: Distribute and Rearrange Terms

Expand the equation:\[-5x - 19 = Ax + 4A + B.\] Reorder terms to form a single expression:\[-5x - 19 = Ax + (4A + B).\]

Step 4: Set Up the System of Equations

By comparing coefficients, we get:1. For \(x\): \(-5 = A\)2. For the constant: \(-19 = 4A + B\).

Step 5: Solve for A and B

From the first equation, \(A = -5\). Substitute \(A\) into the second equation: \(-19 = 4(-5) + B\), which simplifies to \(-19 = -20 + B\). Thus, \(B = 1\).

Step 6: State the Final Decomposition

Substitute \(A\) and \(B\) back into the partial fraction expression:\[\frac{-5x-19}{(x+4)^2} = \frac{-5}{x+4} + \frac{1}{(x+4)^2}.\]

Key Concepts

Repeating Linear Factors

When dealing with polynomials in a fraction where a linear factor is raised to a power, we refer to it as a repeating linear factor. In our example, \((x+4)^2\), the factor \(x+4\) appears more than once. Understanding repeating linear factors is crucial for partial fraction decomposition. This helps you handle polynomials with factors that are not only distinct but also repeating.A single appearance of \(x+4\) would imply a simple linear factor, but repeated instances create complexity. Here, we need two separate terms for decomposition. Consider:

• A term for the simpler single factor, \(\frac{A}{x+4}\).
• Another term for the repeating factor, \(\frac{B}{(x+4)^2}\).

Recognizing these helps set up the equation for decomposition.

System of Equations

Once you've identified the decomposition format, the next step involves creating a system of equations. This is accomplished by clearing the fractions through multiplication and rearranging terms.By expanding and equating the expressions, coefficients from each side can be compared, forming a system of equations. For example:

• From \(-5x - 19 = Ax + (4A + B)\), set up the equation \(-5 = A\) by comparing coefficients of \(x\).
• Form another equation \(-19 = 4A + B\) for constant terms.

These equations now need solving to find constants \(A\) and \(B\). Systems of equations, hence, transform the polynomial complexities into simpler arithmetic solutions.

Partial Fractions

Partial fraction decomposition is a technique that allows you to express complex rational expressions as a sum of simpler fractions. This method is invaluable for integration and solving differential equations.For any rational function, especially with repeating factors, partial fractions provide a structured method to simplify the expression.In our exercise, starting with \(\frac{-5x-19}{(x+4)^2}\), we convert it into the sum:\[\frac{-5}{x+4} + \frac{1}{(x+4)^2}\].This process enhances understanding and simplifies calculations significantly for advanced mathematical operations.

Linear Factors

Linear factors are polynomial expressions of the first degree, such as \(x+4\). When considering polynomials, they are the simplest building blocks.In partial fraction decomposition, every linear factor is central to formulating the simpler fractions in the decomposition.When these linear factors are repeated, as in \((x+4)^2\), they need special attention as separate terms are needed for each occurrence:

• One for each power or level, like \(\frac{1}{x+4}\) and \(\frac{1}{(x+4)^2}\).

Understanding linear factors and their roles helps correctly set up the necessary equations to decompose and solve complex rational expressions correctly.