Problem 1 - Precalculus: Functions and Graphs

Question

Find the partial fraction decomposition. \(\frac{8 x-1}{(x-2)(x+3)}\)

Step-by-Step Solution

Answer

\(\frac{8x-1}{(x-2)(x+3)} = \frac{3}{x-2} + \frac{5}{x+3}\).

Step 1: Identify the Partial Fraction Setup

Since the denominator is the product of two distinct linear factors \(x - 2\) and \(x + 3\), the form of the partial fraction decomposition will be \( \frac{A}{x-2} + \frac{B}{x+3} \). We need to find the values of \(A\) and \(B\).

Step 2: Set Up the Equation

Express the original fraction as the sum of the partial fractions: \(\frac{8x-1}{(x-2)(x+3)} = \frac{A}{x-2} + \frac{B}{x+3}.\)Multiply through by \( (x-2)(x+3) \) to eliminate the denominators, resulting in:\(8x - 1 = A(x+3) + B(x-2).\)

Step 3: Expand and Rearrange

Expand the right-hand side: \(A(x + 3) = Ax + 3A\) and \(B(x - 2) = Bx - 2B\). Combine these to get: \(Ax + 3A + Bx - 2B = 8x - 1.\) \Reorganize it to group the \('x'\) terms: \( (A + B)x + (3A - 2B) = 8x - 1. \)

Step 4: Equate Coefficients

To satisfy the equation for all \(x\), we equate the coefficients of like terms. From \( (A + B)x = 8x \), we have:\(A + B = 8.\)From \(3A - 2B = -1\), we equate the constant terms:\(3A - 2B = -1.\)

Step 5: Solve the System of Equations

Use the equations \(A + B = 8\) and \(3A - 2B = -1\) to find \(A\) and \(B\).Multiply the first equation by 2: \(2A + 2B = 16\).Add to the second equation: \(3A - 2B = -1\).This gives \(5A = 15\), so \(A = 3\).Substitute \(A = 3\) back into \(A + B = 8\): \(3 + B = 8\), resulting in \(B = 5\).

Step 6: Write the Partial Fraction Decomposition

With \(A = 3\) and \(B = 5\), substitute back into the partial fractions: \\[ \frac{8x-1}{(x-2)(x+3)} = \frac{3}{x-2} + \frac{5}{x+3}. \]

Key Concepts

Linear Factors

Linear factors play a crucial role in partial fraction decomposition. In our original exercise, the denominator of the rational expression is a product of two distinct linear factors: \(x-2\) and \(x+3\). Linear factors are expressions of the form \(ax + b\) where \(a\) and \(b\) are constants. They are "linear" because their highest degree of \(x\) is 1.

When dealing with linear factors in partial fraction decomposition, each factor contributes to its own term in the decomposition. For our case, this results in terms like \(\frac{A}{x-2}\) and \(\frac{B}{x+3}\). Our goal is to determine the coefficients \(A\) and \(B\) that make the decomposition equal to the original expression. Understanding linear factors is the first step to breaking down complex expressions into simpler components that are easier to manage.

System of Equations

A system of equations is used to find the unknown coefficients in partial fraction decomposition. Once we express the complex fraction as a sum of simpler fractions, we need to eliminate the denominators by multiplying through with the product of the linear factors. This creates a new equation involving the numerators alone.

In our example, we equate: \(8x - 1 = A(x+3) + B(x-2)\).
This results in:

• \(A + B = 8\)
• \(3A - 2B = -1\)

Here, we have a system of equations based on the coefficients of like terms. Solving these equations simultaneously gives us the values of \(A\) and \(B\). This step is essential for completing the decomposition correctly.

Coefficients Matching

Coefficients matching is key in deriving accurate values for the unknowns in the decomposition. Once we expand the partial fractions into a single expression that equals the numerator of the original expression, we compare coefficients of like terms (both \(x\)-terms and constant terms).

In the case of our original expression:
\((A + B)x + (3A - 2B) = 8x - 1\), by matching coefficients, we create equations:

• \(A + B = 8\) from the \(x\) terms
• \(3A - 2B = -1\) from the constant terms

This technique allows us to find distinct values for \(A\) and \(B\) by simplifying the comparison into manageable linear equations. Coefficients matching is a powerful tool for resolving the individual components of rational expressions.

Rational Expressions

Rational expressions are fractions where both the numerator and the denominator are polynomials. In our exercise, the expression \(\frac{8x-1}{(x-2)(x+3)}\) is a rational expression. Rational expressions can often be simplified using techniques such as factoring, which can help in decomposition tasks.

Partial fraction decomposition itself is a method applied to rational expressions to rewrite them as a sum of simpler fractions. This simplification is particularly useful in calculus for integration or solving differential equations.

Understanding rational expressions includes knowing how to identify the polynomial components and manipulate them through processes like factoring and expanding. This knowledge is essential for effective decomposition and solving related mathematical problems.