Problem 7 - Precalculus

Question

Write the form of the partial-fraction decomposition. Do not solve for the constants. $$\frac{9}{x^{2}-x-20}$$

Step-by-Step Solution

Answer

Decompose as \( \frac{A}{x-5} + \frac{B}{x+4} \).

Step 1: Factor the Denominator

First, we need to factor the quadratic expression in the denominator: \(x^2 - x - 20\). We are looking for two numbers that multiply to -20 and add to -1. These numbers are 4 and -5. So, we can rewrite the denominator as \((x - 5)(x + 4)\).

Step 2: Set Up the Partial-Fraction Decomposition

Now that we have factored the denominator, we can set up the partial-fraction decomposition. Since \((x - 5)\) and \((x + 4)\) are linear factors, the decomposition will take the form \[ \frac{9}{x^2 - x - 20} = \frac{A}{x - 5} + \frac{B}{x + 4} \] where \(A\) and \(B\) are constants that need to be determined.

Key Concepts

Factoring Quadratics

Factoring quadratics is a method used to express a quadratic expression in the form \( ax^2 + bx + c \) as a product of two binomials. This is an essential step in many algebraic procedures, particularly in calculus and algebra, because it simplifies expressions and makes them easier to work with. In the context of partial fraction decomposition, factoring the quadratic in the denominator is crucial.
Let's take the quadratic expression \( x^2 - x - 20 \) as an example. To factor it:

• Look for two numbers that multiply to the constant term, \(-20\), and add to the linear coefficient \(-1\).
• The numbers \(4\) and \(-5\) satisfy both conditions: \(4 \times (-5) = -20\) and \(4 + (-5) = -1\).
• Thus, the expression can be factored to \((x - 5)(x + 4)\).

The benefit of factoring is that it reveals the roots of the quadratic equation, making it easier to perform operations like decomposition.

Linear Factors

Linear factors are expressions of the form \( (x + a) \) or \( (x - b) \), appearing as components of a factored polynomial. Each linear factor represents a solution or root of the equation where the expression equals zero. In the context of partial fraction decomposition, identifying these factors is vital because they define the terms of the decomposition.
For instance, after factoring the quadratic \(x^2 - x - 20\) into linear factors, we obtained \((x - 5)(x + 4)\). These linear factors correspond to the roots \(x = 5\) and \(x = -4\).
Once the linear factors are identified:

• They are used to set up the partial fractions.
• Each factor becomes the denominator of a separate fraction in the decomposition, such as \( \frac{A}{x - 5} + \frac{B}{x + 4} \).

Understanding linear factors is fundamental because they simplify equations and aid in solving polynomials within various mathematical contexts.

Constants in Decomposition

When performing partial fraction decomposition, the constants represent coefficients in the partial terms. These constants are placeholders we solve for to complete the decomposition process. In cases where the factored denominator consists of linear factors, the numerator of each partial fraction is a constant.
In the example given with the expression \( \frac{9}{x^2 - x - 20} \), the partial fraction decomposition is set as:

• \( \frac{A}{x - 5} \)
• \( \frac{B}{x + 4} \)

Where \(A\) and \(B\) are constants.
The steps do not require solving for \(A\) and \(B\) yet, but typically:

• The constants are determined by equating the original expression with the decomposition and solving the resulting equations.
• This often involves plugging in values or using systems of equations.

Recognizing the role of these constants allows for the flexibility to express more complex expressions as a sum of simpler fractions, simplifying integration and other algebraic operations.