Problem 6 - A Graphical Approach to College Algebra

Question

Find the partial fraction decomposition for each rational expression. $$\frac{5 x-3}{(x+1)(x-3)}$$

Step-by-Step Solution

Answer

\(\frac{5x-3}{(x+1)(x-3)} = \frac{2}{x+1} + \frac{3}{x-3}\)

Step 1: Set Up the Partial Fraction Decomposition

Write the given rational expression as a sum of two fractions, where each denominator is a factor of the original denominator: \[\frac{5x - 3}{(x+1)(x-3)} = \frac{A}{x+1} + \frac{B}{x-3}\]Here, \( A \) and \( B \) are constants that we need to find.

Step 2: Clear the Denominator

Multiply every term by the common denominator \((x+1)(x-3)\) to eliminate the denominators:\[5x - 3 = A(x-3) + B(x+1)\]

Step 3: Expand and Combine Like Terms

Expand the right side of the equation:\[5x - 3 = Ax - 3A + Bx + B\]Combine like terms:\[5x - 3 = (A + B)x + (-3A + B)\]

Step 4: Set Up the System of Equations

To find \(A\) and \(B\), equate the coefficients of like terms on both sides of the equation:1. Coefficient of \(x\): \( A + B = 5 \)2. Constant term: \( -3A + B = -3 \)

Step 5: Solve the System of Equations

Solve the system of equations obtained:1. \( A + B = 5 \)2. \( -3A + B = -3 \)First, solve for \( B \) in terms of \( A \) using the first equation:\[ B = 5 - A \]Substitute \( B = 5 - A \) into the second equation:\[-3A + (5 - A) = -3\]Simplify and solve for \( A \):\[-3A + 5 - A = -3 \-4A + 5 = -3 \-4A = -3 - 5 \-4A = -8 \A = 2\]Substitute \( A = 2 \) back into \( B = 5 - A \):\[B = 5 - 2 = 3\]

Step 6: Write the Partial Fraction Decomposition

Substitute the values of \( A \) and \( B \) back into the partial fraction decomposition:\[\frac{5x - 3}{(x+1)(x-3)} = \frac{2}{x+1} + \frac{3}{x-3}\]

Key Concepts

Rational Expressions

In algebra, a rational expression is similar to a fraction, but instead of integers, the numerator and the denominator are polynomials. An example of a rational expression is \( \frac{5x - 3}{(x+1)(x-3)} \). This expression has a polynomial numerator, \( 5x - 3 \), and a polynomial denominator, \( (x+1)(x-3) \). Rational expressions are versatile and play a crucial role in many algebraic operations, including addition, subtraction, and especially in partial fraction decomposition.

Partial fraction decomposition helps us express complex rational expressions as a sum of simpler fractions. This is particularly useful for integration in calculus and solving complex algebraic equations. To decompose a rational expression, we equate it with a sum of fractions whose denominators are factors of the original denominator.

Denominators

The denominator of a rational expression is key in determining how we can break down the expression into simpler parts. In our original expression, which is \( \frac{5x - 3}{(x+1)(x-3)} \), the denominator consists of two linear factors: \( x+1 \) and \( x-3 \).

To decompose this rational expression, these factors form the denominators of the simpler fractions we are looking for. Each fraction in the decomposition will have one of these as its denominator. Our goal is to express \( \frac{5x - 3}{(x+1)(x-3)} \) as \( \frac{A}{x+1} + \frac{B}{x-3} \), where \( A \) and \( B \) are constants we need to determine.

• Each factor from the original denominator is used in the setup of the decomposition.
• This method only works when the degree of the numerator is less than the degree of the denominator.

System of Equations

In the process of finding the constants \( A \) and \( B \), we set up a system of equations. By equating the original expression with the decomposed form and clearing the denominator, we get \[ 5x - 3 = A(x-3) + B(x+1) \]. After expanding and simplifying, we have \[ 5x - 3 = (A + B)x + (-3A + B) \].

This equating step is crucial because it transforms our problem into one involving systems of equations. Two equations are formed:

• For the \( x \)-coefficients: \( A + B = 5 \)
• For the constant terms: \( -3A + B = -3 \)

We solve these equations simultaneously to find the values of \( A \) and \( B \), which satisfy both conditions.

Coefficient Matching

Coefficient matching is a technique used to solve for unknowns in algebra. Once we have our expressions \( A(x-3) + B(x+1) \) expanded to \( (A+B)x + (-3A+B) \), we compare these terms to the corresponding coefficients in the original expression \( 5x - 3 \).

To employ coefficient matching, we equate:

• The coefficients of \( x \) on both sides to form one equation: \( A + B = 5 \)
• The constant terms to form another equation: \( -3A + B = -3 \)

This practice allows us to split one complex equation into two simpler ones, known as a system of equations. These equations are then solved to find the constants \( A = 2 \) and \( B = 3 \), leading us to the partial fraction decomposition.