Problem 10 - A Graphical Approach to Precalculus with Limits

Question

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

Step-by-Step Solution

Answer

The decomposition is \( \frac{-1}{2(x+1)} + \frac{3}{2(x-1)} \).

Step 1: Identify the Denominator Factors

The given rational expression is \( \frac{x+2}{(x+1)(x-1)} \). Here, the denominator \((x+1)(x-1)\) is already factored into linear terms.

Step 2: Write the Partial Fraction Decomposition Form

Since the denominator is the product of linear factors \((x+1)\) and \((x-1)\), we can write the partial fraction decomposition as: \[ \frac{A}{x+1} + \frac{B}{x-1} \] where \(A\) and \(B\) are constants to be determined.

Step 3: Set Up an Equation to Solve for A and B

We equate the original fraction to the partial fraction form: \[ \frac{x+2}{(x+1)(x-1)} = \frac{A}{x+1} + \frac{B}{x-1} \] Multiply both sides by \((x+1)(x-1)\) to clear the fractions: \[ x+2 = A(x-1) + B(x+1) \]

Step 4: Expand and Collect Like Terms

Expand the right-hand side: \( A(x-1) + B(x+1) = Ax - A + Bx + B \) Combine like terms: \( (A + B)x + (B - A) \)

Step 5: Equate Coefficients

Match the coefficients from both sides of the equation: 1. Coefficient of \(x\): \( A + B = 1 \) 2. Constant term: \( B - A = 2 \)

Step 6: Solve for A and B

Solve the system of equations: From \( A + B = 1 \) and \( B - A = 2 \), we add both equations: \[ (A + B) + (B - A) = 1 + 2 \] which simplifies to: \[ 2B = 3 \] Thus, \( B = \frac{3}{2} \). Substitute \(B\) into \(A + B = 1\): \[ A + \frac{3}{2} = 1 \] implies \( A = 1 - \frac{3}{2} = -\frac{1}{2} \).

Step 7: Write the Final Partial Fraction Decomposition

Substitute \(A\) and \(B\) back into the partial fraction form: \[ \frac{x+2}{(x+1)(x-1)} = \frac{-\frac{1}{2}}{x+1} + \frac{\frac{3}{2}}{x-1} \] Which simplifies to: \[ \frac{x+2}{(x+1)(x-1)} = \frac{-1}{2(x+1)} + \frac{3}{2(x-1)} \]

Key Concepts

Rational Expressions

Rational expressions are fractions where both the numerator and the denominator are polynomials. In the expression \( \frac{x+2}{(x+1)(x-1)} \), you have a polynomial \( x+2 \) as the numerator and another polynomial \((x+1)(x-1)\) as the denominator. Understanding rational expressions is like understanding how fractions work in arithmetic, but instead with polynomials.

Key things to remember about rational expressions include:

• The denominator cannot be zero. This is because division by zero is undefined.
• If the numerator is zero and the denominator is not zero, the entire expression equals zero.
• Simplifying rational expressions may involve factoring polynomials and canceling common factors.

Rational expressions are foundational not just in algebra, but in calculus and beyond, where they model real-world scenarios and provide solutions in engineering and physics.

Linear Factors

Linear factors are polynomial expressions of the first degree. In the context of partial fraction decomposition, knowing how to identify linear factors in the denominator is crucial. For example, in the expression \((x+1)(x-1)\), both \(x+1\) and \(x-1\) are linear factors.

Here's what makes a factor linear:

• It's an algebraic expression of the form \(ax+b\), where \(a\) and \(b\) are constants and \(a eq 0\).
• Graphically, it represents a straight line when plotted.

When dealing with linear factors in rational expressions, they can be easily dealt with through methods like partial fraction decomposition. Each linear factor in the denominator can be split into separate fractions, simplifying the terms for integration or further algebraic manipulation. This makes solving or simplifying complex problems more manageable.

System of Equations

A system of equations involves finding values for variables that satisfy all equations simultaneously. In partial fraction decomposition, once the form is set up with unknown constants (like \(A\) and \(B\)), a system of equations helps find their specific values.

For example, from the equation \( (A + B)x + (B - A) = x + 2 \), matching coefficients gives:

• For the \(x\) terms: \( A + B = 1 \)
• For the constant terms: \( B - A = 2 \)

These equations together form a simple linear system.Solving a system of equations requires:

• Substituting the values into one of the equations to find one variable.
• Using the found value to calculate the other variable(s).

In the example here, finding \(B\) as \(\frac{3}{2}\) first allowed us to back substitute and find \(A\) as \(-\frac{1}{2}\). The system of equations approach is direct and effective, helping bridge the gap from setup to solution in expressions like these.