Problem 8 - A Graphical Approach to Precalculus with Limits

Question

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

Step-by-Step Solution

Answer

The partial fraction decomposition is \(\frac{-1}{x} + \frac{4}{x+1}\).

Step 1: Identify the Denominator

The first step in finding the partial fraction decomposition is to identify the factors in the denominator of the given rational expression. In this case, the denominator is \(x(x+1)\), which consists of two linear factors: \(x\) and \(x+1\).

Step 2: Set Up the Partial Fraction Form

Based on the linear factors identified, set up the partial fraction form of the expression. Each factor becomes a separate fraction. Thus, we can write: \(\frac{3x-1}{x(x+1)} = \frac{A}{x} + \frac{B}{x+1}\), where \(A\) and \(B\) are constants to be determined.

Step 3: Clear the Denominator

Multiply both sides of the equation by the common denominator \(x(x+1)\) to eliminate the denominators. This yields: \(3x-1 = A(x+1) + Bx\).

Step 4: Expand and Simplify

Expand the expressions on the right-hand side: \(A(x+1) = Ax + A\) and \(Bx = Bx\). Combine these to obtain \(Ax + A + Bx\). Simplify to get \((A+B)x + A\).

Step 5: Create Equations by Comparing Coefficients

Equate the coefficients from both sides of the equation based on the polynomial terms. From \(3x-1 = (A+B)x + A\), we have two equations: \(A + B = 3\) for the \(x\)-terms and \(A = -1\) for the constant terms.

Step 6: Solve the System of Equations

Using the equations \(A + B = 3\) and \(A = -1\), substitute \(A = -1\) into the first equation to find \(B\): \(-1 + B = 3\). Solving for \(B\) gives \(B = 4\).

Step 7: Write the Final Decomposition

Substitute the values of \(A\) and \(B\) back into the partial fraction form: \(\frac{3x-1}{x(x+1)} = \frac{-1}{x} + \frac{4}{x+1}\). This is the partial fraction decomposition.

Key Concepts

Rational Expressions

Rational expressions are fractions where the numerator and the denominator are both polynomials. The study of rational expressions is crucial in algebra because they often appear in various mathematical contexts, like calculus and engineering. In essence, a rational expression takes the form \(\frac{P(x)}{Q(x)}\), where both \(P(x)\) and \(Q(x)\) are polynomials and \(Q(x) eq 0\).
Understanding how to simplify and manipulate these expressions is key. This can include operations like addition, subtraction, multiplication, and division, along with processes like factoring and expanding.
Partial fraction decomposition, as seen in the exercise, is another vital technique that aims to express a complicated rational expression as a sum of simpler fractions. This is especially useful in integration. By breaking down complex fractions into their basic components, calculus becomes a much smoother process.

Linear Factors

Linear factors are the building blocks of many polynomial expressions. A linear factor refers to a polynomial of degree one, typically expressed in the form \(ax + b\), where \(a\) and \(b\) are constants. Recognizing linear factors is a crucial step in partial fraction decomposition.
When you decompose a rational expression into partial fractions, the presence of linear factors in the denominator determines how you set up the partial fraction form. For example, in our exercise, the denominator \(x(x+1)\) includes the linear factors \(x\) and \(x+1\). These factors form the basis for the fractions \(\frac{A}{x}\) and \(\frac{B}{x+1}\).
Understanding the role of linear factors helps one write rational expressions in a simpler, more manageable way. This is pivotal in solving integrals and other mathematical problems that involve rational expressions.

System of Equations

Solving a system of equations is an essential algebraic skill that involves finding the values of variables that satisfy multiple equations simultaneously. This process is integral to partial fraction decomposition.
Once you express a rational expression as a sum of partial fractions, you often wind up with equations that need solving. These equations arise from equating coefficients of like terms from the expanded form of the equation. In the original step-by-step solution, we got two equations: \(A + B = 3\) and \(A = -1\).
By solving these, we find the values for \(A\) and \(B\). This is typically done through substitution or elimination methods, both of which are fundamental techniques for solving systems of equations. Mastery of these techniques helps in effectively breaking down complex expressions into simpler, solvable parts.