Problem 26 - Precalculus: Mathematical for Calculus

Question

Find the partial fraction decomposition of the rational function. $$\frac{3 x^{2}+5 x-13}{(3 x+2)\left(x^{2}-4 x+4\right)}$$

Step-by-Step Solution

Answer

The decomposition is \(\frac{1}{x-2} + \frac{1}{(x-2)^2}\).

Step 1: Identify the Denominator Factors

First, identify and factor the denominator of the given rational function, if possible. Here, the denominator is already given as \((3x + 2)(x^2 - 4x + 4)\). Notice that \(x^2 - 4x + 4\) can be further factored as \((x-2)^2\). So, the complete factorization of the denominator is \((3x + 2)(x - 2)^2\).

Step 2: Set Up the Partial Fraction Decomposition

Write down the structure for the partial fraction decomposition based on the factors of the denominator. For \((3x + 2)\), use a simple linear fraction \(\frac{A}{3x + 2}\). For each power of \((x - 2)\), use corresponding fractions. Hence, the decomposition is \(\frac{A}{3x+2} + \frac{B}{x-2} + \frac{C}{(x-2)^2}\).

Step 3: Eliminate the Denominator

Multiply both sides of the equation by \((3x + 2)(x - 2)^2\) to clear the denominators, which gives:\[3x^2 + 5x - 13 = A(x - 2)^2 + B(3x + 2)(x - 2) + C(3x + 2)\]

Step 4: Expand and Simplify the Equation

Expand the right-hand side:\[A(x^2 - 4x + 4) + B(3x^2 - 6x + 2x - 4) + C(3x + 2)\]Simplify to:\[Ax^2 - 4Ax + 4A + 3Bx^2 - 4Bx - 6B + 3Cx + 2C\]Group terms to get:\[(A + 3B)x^2 + (-4A - 4B + 3C)x + (4A - 6B + 2C)\]

Step 5: Solve for Coefficients A, B, C

Match the coefficients from both sides of the equation:- For \(x^2\) term: \(A + 3B = 3\)- For \(x\) term: \(-4A - 4B + 3C = 5\)- For the constant term: \(4A - 6B + 2C = -13\)Solve this system of equations to find \(A\), \(B\), and \(C\).

Step 6: Compute A, B, and C

Solving the equations:1. From \(A + 3B = 3\), express \(A\) as \(A = 3 - 3B\).2. Substitute into \(-4A - 4B + 3C = 5\): \[-4(3 - 3B) - 4B + 3C = 5 \ -12 + 12B - 4B + 3C = 5 \ 8B + 3C = 17\]3. Substitute \(A = 3 - 3B\) into the third equation: \[4(3 - 3B) - 6B + 2C = -13 \ 12 - 12B - 6B + 2C = -13 \ -18B + 2C = -25\]Solve \(8B + 3C = 17\) and \(-18B + 2C = -25\) simultaneously to find \(B = 1\), \(C = 1\). Substituting into \(A + 3B = 3\), we find \(A = 0\).

Step 7: Write the Final Partial Fraction Decomposition

Substitute \(A = 0\), \(B = 1\), \(C = 1\) back into the decomposition:\[\frac{3x^2 + 5x - 13}{(3x + 2)(x - 2)^2} = \frac{0}{3x + 2} + \frac{1}{x - 2} + \frac{1}{(x - 2)^2}\]Thus, the partial fraction decomposition is:\[\frac{1}{x - 2} + \frac{1}{(x - 2)^2}\]

Key Concepts

Understanding Rational Functions

A rational function is a type of mathematical expression represented as a ratio of two polynomials. Think of it as a fraction where both the numerator and the denominator are polynomials. Rational functions are widely used in algebra and calculus to describe various types of real-world phenomena. These functions can model behaviors such as rates, curves, and financial predictions.

Imagine a rational function like our example:\[ \frac{3x^2 + 5x - 13}{(3x + 2)(x - 2)^2} \]
Here, the numerator is \(3x^2 + 5x - 13\) and the denominator is \((3x + 2)(x - 2)^2\). The complexity of the denominator impacts how the function behaves across different ranges of \(x\). The process of partial fraction decomposition can break down complex rational functions into simpler parts, making them easier to integrate or differentiate.

• Polynomials in the numerator and denominator determine the degree of the rational function.
• Complexity from the denominator influences asymptotic behavior of the function.

When dealing with rational functions, always ensure they are expressed in their simplest form by factoring wherever possible.

The Role of Polynomial Division

Polynomial division is an essential technique in simplifying rational functions and conducting partial fraction decompositions. It involves dividing a polynomial (the dividend) by another polynomial (the divisor) to find a quotient and sometimes a remainder, much like in arithmetic division.

Consider your polynomial of interest as a dividend and part of your function's denominator as the divisor. This process helps reveal hidden factors or simplifies the function for further analysis. In our original example, no direct division was needed as the denominator was already factorized.

• Division offers insight into the behavior of polynomials in expressions.
• Essential for determining the formulation of partial fraction decomposition.

Understanding polynomial division sets the stage for breaking down complex rational expressions into manageable components.

Decoding Algebraic Fractions

Algebraic fractions, essentially rational functions, are expressions involving a single fraction made up of polynomials. These expressions obey the same rules as ordinary fractions but require handling more complex manipulations due to polynomial numerators and denominators.

The process of partial fraction decomposition involves expressing a complicated algebraic fraction as a sum of simpler ones, which particularly helps in integrating the function or solving equations linked to it. For example, in our problem:\[ \frac{3x^2 + 5x - 13}{(3x + 2)(x - 2)^2} \]
We decompose it into simpler fractions like \( \frac{1}{x - 2} + \frac{1}{(x - 2)^2} \), which allows us to handle each term individually.

• Simplifies complex fractions for differential equations and integration.
• Helps find partial fraction decomposition for rational expressions.

Mastering algebraic fractions can greatly enhance ease in solving advanced mathematical problems and improving understanding of function behaviors.