Problem 19 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

\(\frac{x + 14}{(x - 4)(x + 2)} = \frac{3}{x - 4} - \frac{2}{x + 2}\).

Step 1: Factor the Denominator

First, we need to factor the quadratic expression in the denominator: \(x^2 - 2x - 8\). To factor this, we look for two numbers that multiply to -8 and add up to -2. The numbers are -4 and 2. So the factorization is \((x - 4)(x + 2)\).

Step 2: Set Up the Partial Fraction Decomposition

Since both factors \((x - 4)\) and \((x + 2)\) are linear, set up the decomposition as follows: \(\frac{x + 14}{(x - 4)(x + 2)} = \frac{A}{x - 4} + \frac{B}{x + 2}\), where \(A\) and \(B\) are constants to be determined.

Step 3: Clear the Denominator

Multiply both sides by \((x - 4)(x + 2)\) to clear the denominator, resulting in: \(x + 14 = A(x + 2) + B(x - 4)\).

Step 4: Expand and Collect Like Terms

Expand the right-hand side: \(A(x + 2) + B(x - 4) = Ax + 2A + Bx - 4B = (A + B)x + (2A - 4B)\).

Step 5: Equate Coefficients

By equating the coefficients from both sides, we get the system of equations: \(A + B = 1\) (for the \(x\) coefficients) and \(2A - 4B = 14\) (for the constant terms).

Step 6: Solve the System of Equations

Solve the equations: \(A + B = 1\) and \(2A - 4B = 14\). From the first equation, \(A = 1 - B\). Substituting into the second gives \(2(1 - B) - 4B = 14\) simplifying to \(2 - 2B - 4B = 14\), or \(-6B = 12\), we find \(B = -2\). Substitute \(B = -2\) back to find \(A = 1 - (-2) = 3\).

Step 7: Write the Partial Fraction Decomposition

Substitute \(A = 3\) and \(B = -2\) back into the partial fraction decomposition: \(\frac{x + 14}{(x - 4)(x + 2)} = \frac{3}{x - 4} - \frac{2}{x + 2}\).

Key Concepts

Quadratic Factorization

When you encounter a quadratic expression such as \(x^2 - 2x - 8\), your first step in addressing a partial fraction decomposition is to factor the quadratic expression. In essence, this involves rewriting the quadratic in a factored form, which makes it simpler to work with. To factor, find two numbers that multiply to the constant term (here, -8) and add up to the linear coefficient (here, -2). The quadratic \(x^2 - 2x - 8\) factors as \((x - 4)(x + 2)\). This step is crucial, as it simplifies the structure of the rational function, breaking it into linear denominators.

Linear Denominators

In the world of partial fractions, a linear denominator helps compartmentalize complex fractions into simpler, manageable components. Here, because the factors \((x - 4)\) and \((x + 2)\) are linear, it allows us to separate the original fraction \(\frac{x+14}{(x-4)(x+2)}\) into simpler fractions: \(\frac{A}{x - 4}\) and \(\frac{B}{x + 2}\). Linear denominators are straightforward because they contain variables raised to the first power, making the process of solving them more approachable. They simplify analysis, helping determine the constants \(A\) and \(B\) that aid in building the partial fraction decomposition.

System of Equations

Once you've set up your partial fraction decomposition, it's vital to solve for the unknown constants by forming a system of equations. After substituting \(\frac{x+14}{(x-4)(x+2)} = \frac{A}{x - 4} + \frac{B}{x + 2}\) and clearing the denominators, you get the expression: \(x + 14 = A(x + 2) + B(x - 4)\).

• This needs expansion and equating coefficients to form equations like \(A + B = 1\) and \(2A - 4B = 14\).

The solutions to these simple linear equations give values of \(A\) and \(B\) which are then used in the final expression. Solving systems of equations requires careful arithmetic, helping identify values that satisfy both the original equation and the newly formed ones.

Rational Functions

Rational functions are fractions involving polynomials, and understanding them is vital when dealing with partial fractions. In this exercise, \(\frac{x + 14}{x^2 - 2x - 8}\) describes a rational function where the numerator is linear and the denominator is a quadratic that we factored to simplify the task. Rational functions require an understanding of polynomials and their properties to handle the expressions effectively.

• The simplification into partial fractions aids in solving, integrating, and differentiating these functions.
• Partial fraction decomposition allows these functions to be expressed in simpler terms.

With insight into how these functions behave, manipulation becomes simpler, laying the foundation for more complex mathematical operations.