Problem 18 - Precalculus: Mathematics for Calculus

Question

Find the partial fraction decomposition of the rational function. $$\frac{x-12}{x^{2}-4 x}$$

Step-by-Step Solution

Answer

\( \frac{3}{x} - \frac{2}{x - 4} \).

Step 1: Factor the Denominator

The first step is to factor the denominator of the rational function. The given denominator is \(x^2 - 4x\). We can factor this by taking out the common factor \(x\). Thus, \(x^2 - 4x = x(x - 4)\).

Step 2: Set Up the Partial Fraction Decomposition

Since the denominator \(x(x - 4)\) consists of two linear factors, the partial fraction decomposition can be set up as: \( \frac{x-12}{x^2 - 4x} = \frac{A}{x} + \frac{B}{x - 4} \), where \(A\) and \(B\) are constants to be determined.

Step 3: Clear the Fractions

To find \(A\) and \(B\), clear the fractions by multiplying both sides of the equation by the denominator \(x(x - 4)\). This gives: \[ x - 12 = A(x - 4) + Bx \]

Step 4: Expand and Combine Like Terms

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

Step 5: Equate Coefficients

Equate the coefficients from both sides of the equation for terms involving \(x\) and constant terms. This yields two equations: 1. \(A + B = 1\) (for the coefficient of \(x\))2. \(-4A = -12\) (for the constant terms).

Step 6: Solve for A and B

Solve the simpler equation first: \(-4A = -12\) gives \(A = 3\). Substitute \(A = 3\) into the first equation: \(3 + B = 1\) gives \(B = -2\).

Step 7: Write the Partial Fraction Decomposition

Substitute the values of \(A\) and \(B\) back into the partial decomposition:\[ \frac{x-12}{x^2 - 4x} = \frac{3}{x} + \frac{-2}{x - 4} = \frac{3}{x} - \frac{2}{x - 4} \]

Key Concepts

Rational Function

Let's delve into the concept of a rational function. A rational function is a fraction where both the numerator and the denominator are polynomials. In simple terms, it's a ratio, just like any common fraction, but with polynomials involved. For example, in the case of our exercise, we have \( \frac{x-12}{x^2 - 4x} \) as the rational function.

Such functions are noteworthy because they can have interesting behaviors, like asymptotes, depending upon the zeroes of the denominator. When the denominator equals zero, the function becomes undefined. Exploring this helps us understand where the potential points of discontinuity and unique values for the function lie. That's why figuring out and simplifying the rational function is so crucial.

Factoring Polynomials

Factoring polynomials is a core skill in algebra that simplifies complex expressions. When we factor a polynomial, we're essentially breaking it down into simpler, multipliable components called factors. This process is like pulling apart a jigsaw puzzle to see each individual piece.

Take our exercise's step one, where the denominator \(x^2 - 4x\) is factored. By recognizing a common term \(x\), we can factor it out, arriving at \(x(x - 4)\).

Why do we go through the hassle of factoring? It simplifies the expression and paves the way for further calculations and manipulations, such as partial fraction decomposition. Plus, it's helpful for identifying zeros of polynomials, which can tell us where rational functions are undefined due to division by zero.

Linear Factors

A key part of breaking down polynomials into components is identifying linear factors. Linear factors, such as \(x-a\), represent a straight line when plotted on a graph—they're the simplest type of polynomial.

In our example, the denominator \(x(x - 4)\) post-factoring is composed of two linear factors: \(x\) and \(x - 4\). These linear factors are crucial for setting up partial fraction decomposition. Each of these factors corresponds to a part of the decomposition: \( \frac{A}{x} \) and \( \frac{B}{x - 4} \).

• Linear factors simplify complex expressions and allow for a straightforward path to solve for coefficients \(A\) and \(B\).
• They also provide insight into the graph of the rational function, indicating where intercepts and asymptotes might be.

By understanding and using linear factors, we can efficiently solve and analyze rational functions.