Problem 13 - Algebra and Trigonometry

Question

Find the partial fraction decomposition of the rational function. \(\frac{5}{(x-1)(x+4)}\)

Step-by-Step Solution

Answer

\(\frac{5}{(x-1)(x+4)} = \frac{1}{x-1} - \frac{1}{x+4}\)

Step 1: Understanding the Denominator

Recognize that the denominator is a product of two distinct linear factors, \((x-1)\) and \((x+4)\). Whenever we see such distinct linear factors, we can decompose the fraction into two separate fractions.

Step 2: Set Up the Decomposition

Assume the partial fraction decomposition has the form: \[ \frac{5}{(x-1)(x+4)} = \frac{A}{(x-1)} + \frac{B}{(x+4)} \]Here, \(A\) and \(B\) are constants that we need to solve for.

Step 3: Clear the Denominator

Multiply both sides by the denominator \((x-1)(x+4)\) to clear the fractions: \[ 5 = A(x+4) + B(x-1) \]

Step 4: Expand and Rearrange

Expand the right-hand side: \[ Ax + 4A + Bx - B \]Combine like terms to get: \[ (A + B)x + (4A - B) \]

Step 5: Set Coefficients Equal

Set the coefficients from both sides equal to each other:- For the \(x\) terms: \(A + B = 0\)- For the constant terms: \(4A - B = 5\)

Step 6: Solve the System of Equations

Solve the system:1. \(A + B = 0\)2. \(4A - B = 5\)From the first equation, \(B = -A\).Substitute \(B = -A\) into the second equation: \[ 4A - (-A) = 5 \]\[ 5A = 5 \]\[ A = 1 \]Now, using \(B = -A\), we find \(B = -1\).

Step 7: Write the Partial Fraction Decomposition

Substitute \(A = 1\) and \(B = -1\) into the decomposition:\[ \frac{5}{(x-1)(x+4)} = \frac{1}{x-1} - \frac{1}{x+4} \]

Key Concepts

Rational Function

A rational function is essentially a fraction where both the numerator and the denominator are polynomials. These functions look like this:

• The numerator is a polynomial, for example, a simple constant like 5 or something more complex like \(3x^2 + 2x + 1\).
• The denominator is also a polynomial, like \((x-1)(x+4)\) in our example.

The beauty of rational functions is in their ability to model real-world situations very flexibly. Partial fraction decomposition is a tool we use to break down more complex rational functions into simpler pieces, making them easier to handle.

Linear Factors

In algebra, factors are numbers or expressions that you multiply together to get another number or expression. For linear factors, we're particularly interested in expressions that look like \(x+c\), where \(c\) is a constant.
This means the expression can be graphed as a straight line. The distinct linear factors in our function are \((x-1)\) and \((x+4)\).
Recognizing these factors helps us break down the rational function into simpler parts that are much easier to work with in calculus and algebra.

System of Equations

A system of equations is a set of equations with multiple variables. The goal is to find values for the variables that satisfy all the equations at once. In partial fraction decomposition, a system of equations emerges when we equate the coefficients on both sides of the equation.
For our rational function:

• We have two equations: \(A + B = 0\) and \(4A - B = 5\).
• These equations come from matching the coefficients of like terms (the \(x\) terms and constant terms) in our expressions.

Solving these equations simultaneously allows us to find the constants that complete the partial fraction decomposition.

Constants Solving

In partial fraction decomposition, finding the constants is crucial. These constants \(A\) and \(B\) help express the original rational function as a sum of simpler fractions.
We determined the constants by solving a system of equations:

• From \(A + B = 0\), we found \(B = -A\).
• Plugging \(B = -A\) into \(4A - B = 5\), we solved and found \(A = 1\) and consequently \(B = -1\).

These values finish the partial fraction decomposition, informing us that \(\frac{5}{(x-1)(x+4)} = \frac{1}{x-1} - \frac{1}{x+4}\). This simplification makes further calculations, like integration, much more manageable.