Problem 13 - Precalculus: Functions and Graphs

Question

Find the partial fraction decomposition. \(\frac{x^{2}-6}{(x+2)^{2}(2 x-1)}\)

Step-by-Step Solution

Answer

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

Step 1: Identify the Type of Partial Fraction Decomposition

The given expression \(\frac{x^2 - 6}{(x+2)^2(2x-1)}\) has a quadratic numerator and a denominator with repeated and distinct linear factors. We decompose it into the form:\[ \frac{x^2 - 6}{(x+2)^2(2x-1)} = \frac{A}{x+2} + \frac{B}{(x+2)^2} + \frac{C}{2x-1} \] where \(A\), \(B\), and \(C\) are constants to be determined.

Step 2: Clear the Denominator

Multiply both sides of the equation by the common denominator \((x+2)^2(2x-1)\) to eliminate the denominators:\[ x^2 - 6 = A(x+2)(2x-1) + B(2x-1) + C(x+2)^2 \]

Step 3: Expand and Combine Terms

Expand each term on the right-hand side of the equation:1. \(A(x+2)(2x-1) = A(2x^2 + 3x - 2) = 2Ax^2 + 3Ax - 2A\) 2. \(B(2x-1) = 2Bx - B\)3. \(C(x+2)^2 = C(x^2 + 4x + 4) = Cx^2 + 4Cx + 4C\)Combine all terms:\[ 2Ax^2 + 3Ax - 2A + 2Bx - B + Cx^2 + 4Cx + 4C \]

Step 4: Match Coefficients

Equate the coefficients from the expanded formula to the original quadratic equation \(x^2 - 6\):- Coefficient of \(x^2\): \(2A + C = 1\)- Coefficient of \(x\): \(3A + 2B + 4C = 0\)- Constant term: \(-2A - B + 4C = -6\)

Step 5: Solve the System of Equations

Solve the system for \(A\), \(B\), and \(C\):1. From \(2A + C = 1\), rearrange to find \(C = 1 - 2A\).2. Substitute \(C = 1 - 2A\) into the other equations: - \(3A + 2B + 4(1 - 2A) = 0\) - \(-2A - B + 4(1 - 2A) = -6\)3. Simplify and solve the resulting two equations to find \(A = 2\), \(B = 3\), \(C = -3\).

Step 6: Write the Final Partial Fraction Decomposition

Substitute \(A\), \(B\), and \(C\) back into the partial fractions:\[ \frac{x^2 - 6}{(x+2)^2(2x-1)} = \frac{2}{x+2} + \frac{3}{(x+2)^2} - \frac{3}{2x-1} \]

Key Concepts

Repeated Linear Factors

In partial fraction decomposition, a **repeated linear factor** is a factor in the denominator that appears more than once. For example, in the expression \((x+2)^2(2x-1)\), \((x+2)\) is a repeated linear factor because it appears as \((x+2)^2\).
When dealing with repeated linear factors, we include separate terms for each power of the factor in the decomposition. This is because each power might contribute differently and must be accounted for individually.
The decomposition form becomes:

• \(\frac{A}{x+2}\) for the first instance of \((x+2)\)
• \(\frac{B}{(x+2)^2}\) for the second occurrence to handle the squared term

This ensures that each instance is represented and can be individually manipulated to obtain a correct decomposition.

Distinct Linear Factors

**Distinct linear factors** are factors in the polynomial's denominator that are different from one another, appearing just once. In the expression \((x+2)^2(2x-1)\), the term \((2x-1)\) is a distinct linear factor.
For each distinct linear factor, we use a separate fraction in our partial fraction decomposition to manage its influence individually.
The decomposition incorporating distinct linear factors looks like this:

• Include a fraction \(\frac{C}{2x-1}\) reflecting the direct impact of \((2x-1)\)

Including fractions for distinct linear factors allows us to isolate and solve their individual effects on the numerator.

System of Equations

A **system of equations** is a set of equations with multiple unknowns. In partial fraction decomposition, solving a system of equations allows us to find the constants, like \(A\), \(B\), and \(C\), needed in the fractions.
Once we clear the denominator and simplify, we equate the polynomial expression to form a system:

• \(2A + C = 1\) from the coefficient of \(x^2\)
• \(3A + 2B + 4C = 0\) from the coefficient of \(x\)
• \(-2A - B + 4C = -6\) from the constant term

Solving such a system yields the values of \(A\), \(B\), and \(C\), which decode the original rational expression through their respective fractions.

Coefficient Matching

**Coefficient matching** is a technique used in partial fraction decomposition to determine the values of unknown constants by comparing coefficients from both sides of an equation.
This approach works by expanding both sides of the cleared denominator equation. Then, each term from the numerator is aligned with corresponding terms on the other side. For instance:

• Compare the coefficients of like terms, such as \(x^2\), \(x\), and constant terms.
• This results in a system of equations based on the matched coefficients.

By matching coefficients, we ensure that every part of the polynomial is accounted for, allowing us to solve for unknowns easily. This method emphasizes understanding how each factor contributes to the overall polynomial.