Problem 12 - A Graphical Approach to College Algebra

Question

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

Step-by-Step Solution

Answer

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

Step 1: Identify Denominator Form

The denominator of the rational expression is \((x+2)^3\). This indicates a repeated linear factor \((x+2)\) raised to the power 3.

Step 2: Set Up the Partial Fractions

Since it involves a repeated linear factor, we express the decomposition as follows:\[\frac{2x+1}{(x+2)^3} = \frac{A}{x+2} + \frac{B}{(x+2)^2} + \frac{C}{(x+2)^3}\]where \(A\), \(B\), and \(C\) are constants to be determined.

Step 3: Clear Fractions

Multiply through by \((x+2)^3\) to eliminate the denominators:\[2x + 1 = A(x+2)^2 + B(x+2) + C\]

Step 4: Expand and Simplify

Expand each term on the right-hand side:\(A(x+2)^2 = A(x^2 + 4x + 4) = Ax^2 + 4Ax + 4A\),\(B(x+2) = Bx + 2B\).Substitute these into the equation:\[2x + 1 = Ax^2 + 4Ax + 4A + Bx + 2B + C\].

Step 5: Collect Like Terms

Combine terms according to powers of \(x\):\[0x^2 + 2x + 1 = Ax^2 + (4A + B)x + (4A + 2B + C)\].

Step 6: Solve for the Coefficients

By comparing coefficients, we create the following system of equations:1. For \(x^2\): \(A = 0\),2. For \(x\): \(4A + B = 2\),3. For constant term: \(4A + 2B + C = 1\).Solve the system:- From 1, \(A = 0\).- Substitute \(A = 0\) into 2: \(B = 2\).- Substitute \(A = 0\) and \(B = 2\) into 3: \(2(2) + C = 1\), hence \(C = -3\).

Step 7: Write the Partial Fraction Decomposition

Now we have the values \(A = 0\), \(B = 2\), and \(C = -3\). Substitute them back into the partial fraction expression:\[\frac{2x+1}{(x+2)^3} = \frac{0}{x+2} + \frac{2}{(x+2)^2} + \frac{-3}{(x+2)^3}\]or simply:\[\frac{2}{(x+2)^2} - \frac{3}{(x+2)^3}\]

Key Concepts

Repeated Linear Factors

When working with partial fraction decomposition, identifying the nature of the factors in the denominator is crucial. A repeated linear factor occurs when a linear term, like \(x + a\), appears multiple times in the denominator, raised to a power \(n\) greater than 1.

In the given exercise, \( (x+2)^3 \) is a repeated linear factor because \(x+2\) is repeated three times. This indicates that, during partial fraction decomposition, we need to consider each potential power of the factor separately.

• For \((x+2)^3\), the decomposition involves fractions with denominators \(x+2\), \((x+2)^2\), and \((x+2)^3\).
• This ensures each different power of the linear factor is accounted for in the expression.

The key is to express the original rational expression as a sum of simpler fractions, each involving lower powers of the factor, which makes integration or further algebraic manipulation feasible.

Rational Expressions

Rational expressions are fractions where both the numerator and the denominator are polynomials. They play a key role in algebra because they model relationships and can be used in a wide range of applications.

The given problem involves the rational expression \(\frac{2x+1}{(x+2)^3}\). The numerator is a linear polynomial, \(2x+1\), and the denominator is a polynomial involving a repeated factor \( (x+2)^3 \). Understanding rational expressions involves:

• Factorizing polynomials to identify their roots and potential simplifications.
• Applying techniques like partial fraction decomposition to break them into simpler components.

This process is particularly useful when solving equations or integrating functions where direct methods are cumbersome. Rational expressions are ubiquitous in calculus and algebra, and mastering them unlocks a variety of mathematical tools.

System of Equations

Solving for coefficients \(A, B, \) and \(C\) in partial fraction decomposition often requires establishing a system of equations. This is done by equating the coefficients from the expanded form to those in the original polynomial.

For the exercise's solution, the following system was derived:

• Equation 1: Coefficient of \(x^2\): \A = 0\.
• Equation 2: Coefficient of \(x\): \4A + B = 2\.
• Equation 3: Constant term: \4A + 2B + C = 1\.

To solve this system:

- From Equation 1, we establish that \(A = 0\).
- Substituting \(A = 0\) into Equation 2 gives \(B = 2\).
- Using \(A = 0\) and \(B = 2\) in Equation 3, we solve for \(C\) yielding \(C = -3\).

This process of solving simultaneous equations is essential to determine all unknowns accurately in the decomposition, ensuring that the partial fractions represent exactly the original rational expression.