Problem 19 - Precalculus: Mathematical for Calculus

Question

Find the partial fraction decomposition of the rational function. $$\frac{x+14}{x^{2}-2 x-8}$$

Step-by-Step Solution

Answer

The partial fraction decomposition is \(\frac{3}{x-4} - \frac{2}{x+2}.\)

Step 1: Factor the Denominator

Start by factoring the quadratic denominator, \(x^2 - 2x - 8\). We need two numbers that multiply to \(-8\) and add to \(-2\). These numbers are \(2\) and \(-4\). Thus, the factorization is \((x - 4)(x + 2)\).

Step 2: Set Up Partial Fractions

Since the denominator \((x - 4)(x + 2)\) consists of two linear factors, the partial fraction decomposition takes the form: \[\frac{x+14}{x^{2}-2x-8} = \frac{A}{x-4} + \frac{B}{x+2}.\] Here, \(A\) and \(B\) are constants that we need to determine.

Step 3: Clear the Denominator

Multiply both sides by the original denominator \((x - 4)(x + 2)\) to clear the fractions: \[x + 14 = A(x + 2) + B(x - 4).\]

Step 4: Expand and Rearrange

Expand the right side: \[x + 14 = Ax + 2A + Bx - 4B.\] Combine like terms to rewrite: \[x + 14 = (A + B)x + (2A - 4B).\]

Step 5: Solve for A and B

By comparing coefficients, we equate terms. For the coefficient of \(x\), we have: \[A + B = 1.\] For the constant term, we have: \[2A - 4B = 14.\] Solve these simultaneous equations for \(A\) and \(B\).

Step 6: Solve the System of Equations

The first equation is \(A + B = 1\) and the second is \(2A - 4B = 14\). From \(A + B = 1\), we find \(A = 1 - B\). Substitute into the second equation: \[2(1 - B) - 4B = 14.\] Solve for \(B\): \[2 - 2B - 4B = 14,\] which simplifies to \[-6B = 12\], giving \(B = -2\). Now substitute back to find \(A = 1 - (-2) = 3\).

Step 7: Write the Final Decomposition

Having found \(A = 3\) and \(B = -2\), the partial fraction decomposition is: \[\frac{x+14}{x^{2}-2x-8} = \frac{3}{x-4} - \frac{2}{x+2}.\]

Key Concepts

Rational Functions

Rational functions are the backbone of many algebra topics. They consist of a ratio or quotient of two polynomials. Specifically, the form \( \frac{P(x)}{Q(x)} \), where \( P(x) \) and \( Q(x) \) are polynomials, represents a rational function. In the exercise, you worked with \( \frac{x+14}{x^2 - 2x - 8} \). Here, \( x + 14 \) is your numerator (the \( P(x) \)) and \( x^2 - 2x - 8 \) is the denominator (the \( Q(x) \)).

To break down a rational function into partial fractions, you express it as a sum of simpler rational expressions. This strategy allows for easier integration or other operations in calculus. It's crucial to ensure the degree (the highest power of x) of the numerator is less than the degree of the denominator. If not, you'll need to perform polynomial division first. By understanding how to manipulate these expressions, you'll gain more control over solving complex mathematical problems efficiently.

Factoring Quadratics

Factoring quadratics is a fundamental technique in algebra. It involves expressing a quadratic expression, typically of the form \( ax^2 + bx + c \), as a product of two binomials, such as \((px + q)(rx + s)\).

In the provided problem, you started with \( x^2 - 2x - 8 \). Here, we need to find two numbers that multiply to \(-8\) (the constant term) and add to \(-2\) (the coefficient of the linear term). These numbers are \(2\) and \(-4\), leading to the factorization \((x - 4)(x + 2)\). Understanding this process allows you to decompose more complex algebraic expressions further down the line.

• Identify target numbers for multiplication and addition/subtraction.
• Factor by grouping or using the quadratic formula for more complex expressions.
• Ensure the factors correctly multiply and sum to the original coefficients.

Factoring helps simplify rational functions into manageable parts called partial fractions, paving the way for further analysis.

Simultaneous Equations

Solving simultaneous equations is another essential skill often used in algebra and calculus. This method involves finding the values of variables that satisfy multiple equations at once.

In the partial fraction decomposition task, after setting up your equation, you arrived at two simultaneous equations:

• \( A + B = 1 \)
• \( 2A - 4B = 14 \)

These equations help find the unknown constants, \( A \) and \( B \), in your partial fractions.

To solve these:

• First, express one variable in terms of the other, like \( A = 1 - B \).
• Substitute this expression into the second equation to solve for \( B \): \(-6B = 12\), giving \( B = -2 \).
• Substitute back to find \( A \), which results in \( A = 3 \).

These solutions plug right back into your partial fraction decomposition, allowing you to complete the task with ease. Mastering simultaneous equations enables you to navigate through various mathematical problems more efficiently.