Problem 12 - Algebra for College Students

Question

Find the partial fraction decomposition for each rational expression. See answers below. \(\frac{-3 x+1}{(x+1)^{2}}\)

Step-by-Step Solution

Answer

\(\frac{-3}{x+1} + \frac{4}{(x+1)^2}\)

Step 1: Identify the Form

The rational expression is given as \( \frac{-3x + 1}{(x+1)^2} \). Since the denominator is \( (x+1)^2 \), there is a repeated linear factor. We need to express this in the form \( \frac{A}{x+1} + \frac{B}{(x+1)^2} \).

Step 2: Setup the Partial Fractions

Write the expression as a sum of two fractions: \( \frac{-3x + 1}{(x+1)^2} = \frac{A}{x+1} + \frac{B}{(x+1)^2} \).

Step 3: Clear the Denominator

Multiply every term by \( (x+1)^2 \) to eliminate the denominators: \(-3x + 1 = A(x+1) + B\).

Step 4: Expand and Equate Coefficients

Expand the right-hand side: \(A(x + 1) + B = Ax + A + B\).Equate coefficients with the left-hand side expression, \(-3x + 1 = Ax + A + B\).

Step 5: Solve for A and B

Equate the coefficients of like terms:1. From the \(x\)-terms: \(-3 = A\)2. From the constant terms: \(1 = A + B\)Substituting \(A = -3\) into the second equation gives:\(1 = -3 + B\)Solving for \(B\), we find \(B = 4\).

Step 6: Write the Final Decomposition

Substitute \(A\) and \(B\) back into the partial fraction form:\(\frac{-3x + 1}{(x+1)^2} = \frac{-3}{x+1} + \frac{4}{(x+1)^2}\).

Key Concepts

Rational Expressions

Rational expressions are fractions where both the numerator and the denominator are polynomials. Understanding these expressions is key to many algebraic processes, including partial fraction decomposition. When you work with rational expressions, the goal is often to simplify them or perform operations like addition, subtraction, and decomposition.

In the given rational expression, \( \frac{-3x + 1}{(x+1)^2} \), we have a degree 1 polynomial, \(-3x + 1\), in the numerator. The denominator \((x+1)^2\) is a degree 2 polynomial because it's a square of a linear factor. Such expressions are common in calculus and other advanced mathematics courses, which emphasize studying their behavior and simplifying them for further computation.

• Numerator: The polynomial on the top, here \(-3x + 1\).
• Denominator: The polynomial on the bottom, here \((x+1)^2\).

Rational expressions can often be broken down into simpler fractions, especially when the denominator includes factors that can be repeated or expanded. This breakdown is crucial for understanding the function better in mathematical analysis.

Repeated Linear Factors

A repeated linear factor occurs when a linear polynomial (like \(x+1\)) appears more than once in the factorization of the denominator. In our case, \((x+1)^2\) shows repetition because the linear factor \(x+1\) is squared.

When dealing with partial fraction decomposition, repeated linear factors require special attention because each "duplicate" of the factor corresponds to a separate term in the decomposition. For \((x+1)^2\), our expression is thus decomposed to include:\[ \frac{A}{x+1} + \frac{B}{(x+1)^2} \]
This setup ensures that each instance of \(x+1\) in the denominator is adequately represented, expanding the expression's flexibility in terms of integration or simplification.

Take note of these aspects:

• A single linear factor \(x+a\) is represented as \(\frac{A}{x+a}\).
• A repeated linear factor \((x+a)\) appears with increasing powers: \(\frac{A}{x+a} + \frac{B}{(x+a)^2}\) and so on.

This helps account for all possible polynomial reductions that can be made from the factors available in the denominator.

Equating Coefficients

Equating coefficients is a method used to solve the values of constants in partial fraction decomposition. After expressing the initial rational expression in terms of partial fractions, you clear the denominators by multiplying through by the original denominator. This leaves you with a polynomial equation, where equating the coefficients of corresponding powers of \(x\) from both sides of the equation allows you to solve for unknown constants.

In our problem, after clearing the denominators, we reach the equation:\(-3x + 1 = A(x+1) + B\). Expanding this on the right gives \(Ax + A + B\).
From here:

• coefficients of \(x\) are equated: \(-3 = A\).
• the constant terms are equated: \(1 = A + B\).

This process is systematic: you handle each different power of \(x\) independently, leading to a set of simpler equations. Solving these gives the values of the constants used in the partial fractions, completing the decomposition. In this example, solving these equations yields \(A = -3\) and \(B = 4\). Through equating coefficients, the final partial fraction decomposition makes the initial expression easier to understand and manipulate in future calculations.