Problem 14 - Precalculus: Functions and Graphs

Question

Find the partial fraction decomposition. \(\frac{2 x^{2}+x}{(x-1)^{2}(x+1)^{2}}\)

Step-by-Step Solution

Answer

The decomposition is \( \frac{0.5}{x-1} - \frac{0.5}{x+1} \).

Step 1: Set Up the Partial Fraction Decomposition

For the rational expression \( \frac{2x^2 + x}{(x-1)^2(x+1)^2} \), we propose the following partial fraction decomposition: \[ \frac{2x^2 + x}{(x-1)^2(x+1)^2} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+1} + \frac{D}{(x+1)^2} \]This setup is suitable for expressions with repeated linear factors.

Step 2: Clear the Denominator

Multiply both sides by \((x-1)^2(x+1)^2\) to eliminate the denominator:\[ 2x^2 + x = A(x-1)(x+1)^2 + B(x+1)^2 + C(x-1)^2(x+1) + D(x-1)^2 \]

Step 3: Expand Each Term

Expand the right-hand side:- \( A(x-1)(x+1)^2 = A(x^3 + 2x^2 + x - 1) \)- \( B(x+1)^2 = B(x^2 + 2x + 1) \)- \( C(x-1)^2(x+1) = C(x^3 - x^2 + x - 1) \)- \( D(x-1)^2 = D(x^2 - 2x + 1) \)Substitute these expansions back into the equation.

Step 4: Combine and Simplify Coefficients

Combine the terms to collect coefficients of like powers of \(x\):\[ 2x^2 + x = (A + C)x^3 + (2A + B - C + D)x^2 + (A + 2B + C - 2D)x + (-A + B - C + D) \]

Step 5: Set up the System of Equations

Equate coefficients from both sides of the equation to form a system of equations:1. \( A + C = 0 \)2. \( 2A + B - C + D = 2 \)3. \( A + 2B + C - 2D = 1 \)4. \( -A + B - C + D = 0 \)

Step 6: Solve the System of Equations

Solve the system of equations obtained in the previous step:- From 1. \( C = -A \)- Substitute \( C = -A \) into other equations.- Solve systematically: - Substitute \( C = -A \) in equation 2: \( 2A + B + A + D = 2 \) leading to \( 3A + B + D = 2 \) - Substitute \( C = -A \) in equation 3: \( A + 2B - A - 2D = 1 \) leads to \(2B - 2D = 1\) solving gives: \( B = D + 0.5 \) - Use equation 4 and previous solutions to find all values: Substitute \( B \) into \(-A + B + A + D = 0 \) gives: \( B + D = 0 \) - Solve the set to find values: \( A = 0.5, B = 0, C = -0.5, D = 0 \).

Step 7: Write the Partial Fraction Formula

With \( A = 0.5 \), \( B = 0 \), \( C = -0.5 \), and \( D = 0 \), the partial fraction decomposition is:\[ \frac{2x^2 + x}{(x-1)^2(x+1)^2} = \frac{0.5}{x-1} - \frac{0.5}{x+1} \]

Key Concepts

Rational Expressions

Rational expressions are mathematical fractions where both the numerator and the denominator are polynomials. These expressions feature prominently in algebra and calculus, primarily due to their complexity and the richness of their applications.

• The numerator may be a single term or a sum of multiple terms.
• The denominator is crucial since it cannot be zero, as division by zero is undefined.
• Rational expressions can often be simplified by factoring both the numerator and the denominator and canceling out any common factors.

In the given example, the rational expression is \( \frac{2x^2 + x}{(x-1)^2(x+1)^2} \). Here, the numerator \( 2x^2 + x \) is a polynomial of degree 2, while the denominator \((x-1)^2(x+1)^2\) involves repeated linear factors, adding complexity to the expression. The key skill in working with rational expressions is recognizing opportunities to factor and simplify the expressions effectively.

System of Equations

A system of equations is a set of multiple equations that are solved together because they share at least one common solution. These systems can vary in complexity depending on the number and type of equations involved.

• System of equations can be solved using various methods including substitution, elimination, or matrix operations.
• Consistency is important, meaning the system must have at least one solution for it to be solvable.

In the context of partial fraction decomposition, systems of equations help us find unknown coefficients. In our case, these coefficients \( A, B, C, D \) need to satisfy multiple conditions derived from equating coefficients of the same degree from both sides of the equation. Solving such a system often requires strategic substitutions and eliminations to reduce the number of unknowns step by step, like in the example where the solution was found: \( A = 0.5 \), \( B = 0 \), \( C = -0.5 \), and \( D = 0 \).

Repeated Linear Factors

Repeated linear factors occur when a polynomial denominator contains multiple powers of the same linear factor, such as \((x-1)^2\) in our given expression. This repetition affects how we set up the partial fraction decomposition.

• Each repeated factor requires its own distinct term in the decomposition.
• For a factor \( (x-a)^n \), we would use terms with denominators from \( (x-a) \) to \( (x-a)^n \).
• This ensures that the decomposition can accurately reverse back into the original expression when combined and simplified.

In our exercise, both \((x-1)^2\) and \((x+1)^2\) are repeated linear factors shown in the denominator. Each factor contributes to the structure of the decomposition, resulting in terms like \( \frac{A}{x-1}, \frac{B}{(x-1)^2}, \frac{C}{x+1}, \) and \( \frac{D}{(x+1)^2} \). Understanding repeated factors is essential since it directly influences how we break down a complex rational expression into more manageable parts.