Problem 26 - Precalculus: Mathematics for Calculus

Question

Find the partial fraction decomposition of the rational function. $$\frac{-3 x^{2}-3 x+27}{(x+2)\left(2 x^{2}+3 x-9\right)}$$

Step-by-Step Solution

Answer

The decomposition is \(\frac{21/5}{x+2} + \frac{(-51/5)x + 162/5}{2x^2 + 3x - 9}\).

Step 1 - Determine the form of the decomposition

The denominator is factored into \((x+2)(2x^2 + 3x - 9)\). First, identify that \(2x^2 + 3x - 9\) is a quadratic factor. Set up the partial fraction decomposition as:\[ \frac{-3x^2 - 3x + 27}{(x+2)(2x^2 + 3x - 9)} = \frac{A}{x+2} + \frac{Bx + C}{2x^2 + 3x - 9} \]

Step 2 - Clear the denominators

Multiply both sides of the equation by the denominator \((x+2)(2x^2 + 3x - 9)\) to clear the fractions:\[ -3x^2 - 3x + 27 = A(2x^2 + 3x - 9) + (Bx + C)(x+2) \]

Step 3 - Expand the right side

Distribute \(A\) into the polynomial and expand \((Bx + C)(x+2)\):\[ -3x^2 - 3x + 27 = A(2x^2 + 3x - 9) + Bx^2 + 2Bx + Cx + 2C \]Simplify further:\[ -3x^2 - 3x + 27 = (2A + B)x^2 + (3A + 2B + C)x + (-9A + 2C) \]

Step 4 - Equate coefficients

Equate the coefficients of the corresponding powers of \(x\) from both sides of the equation:1. \(2A + B = -3\)2. \(3A + 2B + C = -3\)3. \(-9A + 2C = 27\)

Step 5 - Solve the system of equations

From the set of equations:1. \(2A + B = -3\)2. \(3A + 2B + C = -3\)3. \(-9A + 2C = 27\)Start by solving the system. From eq. (1), express \(B\) in terms of \(A\):\[ B = -3 - 2A \]Substitute into eq. (2) and (3):- Equation (2): \[ 3A + 2(-3 - 2A) + C = -3 \Rightarrow C = 7A + 3 \]- Equation (3): \[ -9A + 2(7A + 3) = 27 \Rightarrow -9A + 14A + 6 = 27 \Rightarrow 5A = 21 \Rightarrow A = \frac{21}{5} \]Substitute \(A\) back to find \(B\) and \(C\):- \( B = -3 - 2\left(\frac{21}{5}\right) = -\frac{51}{5} \)- \( C = 7\left(\frac{21}{5}\right) + 3 = \frac{162}{5} \)

Step 6 - Write the partial fraction decomposition

Substitute \(A\), \(B\), and \(C\) back into the partial fraction equation:\[ \frac{-3x^2 - 3x + 27}{(x+2)(2x^2 + 3x - 9)} = \frac{21/5}{x+2} + \frac{(-51/5)x + 162/5}{2x^2 + 3x - 9} \]

Key Concepts

Rational Functions

Rational functions are mathematical expressions representing the division of two polynomials. In simpler terms, they are fractions where both the numerator and the denominator are polynomials. For example, the expression \( \frac{-3x^2 - 3x + 27}{(x+2)(2x^2 + 3x - 9)} \) is a rational function because both the top part (numerator) and the bottom part (denominator) are polynomials.
Understanding how these functions work is important in mathematics, as they can model real-world situations and present complex curves in graphs. When working with rational functions, it's essential to identify the behavior of the function, like where it might become undefined or reach asymptotes.
In partial fraction decomposition, we break down a rational function into simpler fractions that are easier to handle, integrate, or differentiate.

Factoring Polynomials

Factoring polynomials is a core skill in algebra that involves expressing a polynomial as a product of simpler polynomials. This step is crucial because it often helps simplify many complex algebraic operations.
For the problem \( \frac{-3x^2 - 3x + 27}{(x+2)(2x^2 + 3x - 9)} \), the denominator is factored into \( (x+2)(2x^2 + 3x - 9) \). Factoring aims to express the polynomial in terms of its simplest components, making it easier to solve equations or simplify fractions.
To manage quadratic or higher degree polynomials, techniques such as grouping, using the quadratic formula, or special formulas for cubes and squares can be practiced. Once factored, these polynomials reveal critical points, like roots, which are essential for solving equations.

System of Equations

A system of equations involves finding values that satisfy all given equations simultaneously. In this problem, after clearing the denominators and simplifying, a system of equations arises from matching coefficients of equivalent terms on both sides of the equation.
The equation \(-3x^2 - 3x + 27 = (2A + B)x^2 + (3A + 2B + C)x + (-9A + 2C)\) leads to three distinct equations:\

• \(2A + B = -3\)
• \(3A + 2B + C = -3\)
• \(-9A + 2C = 27\)

To solve these equations, algebraic methods such as substitution or elimination are used. You start by expressing one variable in terms of another and substituting it into other equations, gradually solving for each unknown. Systems of equations are powerful tools in finding relationships between multiple variables.

Algebraic Techniques

Algebraic techniques encompass various strategies used to manipulate and solve equations, making them crucial in solving complex mathematical problems. In the partial fraction decomposition exercise, several techniques come into play.

• Distribution: Using distribution, we expand expressions like \((Bx + C)(x+2)\) to combine and equate parts of the polynomial.
• Clearing Fractions: This involves multiplying through by a common denominator to eliminate fractions, making polynomials easier to work with.
• Substitution: By substituting expressions from one equation into another, we can simplify systems of equations and reduce the number of variables.

Mastering these algebraic tools requires practice. These steps help in breaking down the rational expression into its partial fractions, ultimately simplifying the integration process and aiding in calculus operations.