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 partial fraction decomposition is \( \frac{-3}{x+2} + \frac{3x}{2x^2+3x-9} \).

Step 1: Factor the Denominator

First, we need to factor the denominator of the rational function. The denominator given is \((x+2)(2x^2+3x-9)\). Check if the quadratic \(2x^2+3x-9\) can be further factored. However, upon testing discriminant and factorization techniques, we find it cannot be factored nicely with real numbers. Thus, keep it as \((x+2)(2x^2+3x-9)\).

Step 2: Set Up the Partial Fraction Decomposition

Since we cannot factor \(2x^2+3x-9\) further, the partial fraction decomposition will be:\[ \frac{-3x^2-3x+27}{(x+2)(2x^2+3x-9)} = \frac{A}{x+2} + \frac{Bx+C}{2x^2+3x-9} \]where \(A\), \(B\), and \(C\) are constants to be determined.

Step 3: Multiply Through by the Denominator

Multiply both sides by \((x+2)(2x^2+3x-9)\) to eliminate the denominators:\[-3x^2-3x+27 = A(2x^2+3x-9) + (Bx+C)(x+2)\]

Step 4: Expand and Combine Like Terms

Expand the right-hand side:\[ A(2x^2+3x-9) = 2Ax^2 + 3Ax - 9A \]\[ (Bx+C)(x+2) = Bx^2 + 2Bx + Cx + 2C = Bx^2 + (2B+C)x + 2C \]Combine all terms:\[ 2Ax^2 + 3Ax - 9A + Bx^2 + (2B+C)x + 2C = (2A+B)x^2 + (3A+2B+C)x + (-9A+2C) \]

Step 5: Set Up System of Equations

Equate the coefficients of the same powers of \(x\) from both sides of the equation:For \(x^2\): \(2A + B = -3\) For \(x\):\(3A + 2B + C = -3\)For the constant term:\(-9A + 2C = 27\)

Step 6: Solve the System of Equations

Solve the above system of equations:1. From \(2A + B = -3\), express \(B\) as: \(B = -3 - 2A\).2. Substitute into the second equation: \(3A + 2(-3 - 2A) + C = -3\) \(3A - 6 - 4A + C = -3\) \(-A + C = 3\) -> \(C = 3 + A\).3. Substitute \(C = 3 + A\) into the third equation: \(-9A + 2(3 + A) = 27\) \(-9A + 6 + 2A = 27\) \(-7A + 6 = 27\) \(-7A = 21\) \(A = -3\).4. With \(A = -3\), find \(B\): \(B = -3 - 2(-3) = -3 + 6 = 3\).5. With \(A = -3\), find \(C\): \(C = 3 + (-3) = 0\).

Step 7: Write the Decomposition

Substitute \(A = -3\), \(B = 3\), and \(C = 0\) back into the partial fractions:\[ \frac{-3x^2-3x+27}{(x+2)(2x^2+3x-9)} = \frac{-3}{x+2} + \frac{3x}{2x^2+3x-9} \].

Key Concepts

Rational Function

A rational function is essentially a fraction where both the numerator and the denominator are polynomials. These types of functions are quite prevalent in mathematics because they can model a variety of real-world situations. In our specific exercise, we are working with the rational function \( \frac{-3x^2 -3x + 27}{(x+2)(2x^2 + 3x - 9)} \).
The numerator here is \(-3x^2 - 3x + 27\) and the denominator is \((x+2)(2x^2 + 3x - 9)\).

To perform partial fraction decomposition, we focus on breaking down this rational function into simpler fractions that are easier to manage, analyze, and integrate. This process is especially useful in calculus for techniques such as integration, where dealing with simple fractions is often more straightforward than handling complex ones. The main goal is to rewrite the original function as a sum of simpler fractions, facilitating various mathematical operations.

System of Equations

A system of equations is a set of equations with multiple variables that you solve simultaneously. In partial fraction decomposition, a system of equations arises when you match coefficients from the expanded equation of your rational function to your decomposed fractions.

In this exercise, after expanding and rearranging terms, we get three linear equations:

• For \(x^2\): \(2A + B = -3\)
• For \(x\): \(3A + 2B + C = -3\)
• For constants: \(-9A + 2C = 27\)

Solving this system involves expressing one variable in terms of others and substituting back until all variables, \(A\), \(B\), and \(C\), are found. This step is crucial as it ensures the decomposition accurately represents the original rational function. Managing this system correctly allows for determining the coefficients which transform the function into its decomposed form without altering its mathematical properties.

Denominator Factorization

Denominator factorization involves breaking down the denominator of a rational function into its simplest factors. This step is crucial in partial fraction decomposition because the factorized form dictates how to set up the fractions for decomposition.

In our example, the denominator \((x+2)(2x^2+3x-9)\) is partially factorized. While \((x+2)\) is linear and cannot be reduced further, \(2x^2 + 3x - 9\) was checked for further factorization, but it was determined that it cannot be factored over the reals.

The inability to factor the quadratic further guides how the decomposition is set up: linear factors take a simple constant in the numerator, while higher degree polynomials, like the irreducible quadratic, take a linear numerator. Therefore, we write the decomposition as \( \frac{A}{x+2} + \frac{Bx+C}{2x^2+3x-9} \), allowing us to proceed with finding the unknown constants that satisfy this equation.