Problem 24 - Precalculus: Mathematical 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: \( \frac{-\frac{21}{10}}{x+2} + \frac{\frac{11}{18}}{x - 1.5} + \frac{\frac{12}{10}}{x + 3} \).

Step 1: Factor the Denominator

Before we can decompose the fraction into partial fractions, we need to understand the structure of the denominator. The denominator is already partially factored as \((x+2)(2x^2 + 3x - 9)\). Next, factor the quadratic \(2x^2 + 3x - 9\). Using the quadratic formula, we find the roots. The quadratic formula is given by \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = 3\), \(c = -9\). Calculating, we have:\(x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot (-9)}}{4} = \frac{-3 \pm \sqrt{81}}{4}\) which gives roots \(x = 1.5\) and \(x = -3\). Thus, \(2x^2 + 3x - 9\) factors as \(2(x - 1.5)(x + 3)\).

Step 2: Set Up Partial Fraction Form

Now that the denominator is fully factored as \((x+2)(2)(x - 1.5)(x + 3)\), we can express the given fraction in terms of partial fractions. The partial fraction decomposition will have the form:\[ \frac{-3x^2 - 3x + 27}{(x+2)(2)(x - 1.5)(x + 3)} = \frac{A}{x+2} + \frac{B}{x - 1.5} + \frac{C}{x + 3} \]Here, \(A\), \(B\), and \(C\) are constants we need to determine.

Step 3: Clear the Denominator

To solve for \(A\), \(B\), and \(C\), multiply both sides by the common denominator \((x+2)(2)(x - 1.5)(x + 3)\) to clear it. This gives:\[-3x^2 - 3x + 27 = A(2)(x - 1.5)(x + 3) + B(2)(x+2)(x + 3) + C(2)(x+2)(x - 1.5)\]

Step 4: Solve for Constants

To find \(A\), \(B\), and \(C\), choose values of \(x\) that simplify the equation. For example, setting specific values for \(x\) like \(x = -2\), \(x = 1.5\), and \(x = -3\) will help isolate the constants.1. Set \(x = -2\): This will eliminate terms containing \(B\) and \(C\)\[-3(-2)^2 - 3(-2) + 27 = A(2)(-2 - 1.5)(-2 + 3)\]\[21 = -5A\]; solve to find \(A = -\frac{21}{10}\).2. Set \(x = 1.5\): This will eliminate terms with \(A\) and \(C\)\[-3(1.5)^2 - 3(1.5) + 27 = B(2)(1.5 + 2)(1.5 + 3)\]\[8.25 = 13.5B\]; solve to find \(B = \frac{11}{18}\).3. Set \(x = -3\): This will eliminate terms with \(A\) and \(B\)\[-3(-3)^2 - 3(-3) + 27 = C(2)(-3 + 2)(-3 - 1.5)\]\[9 = 7.5C\]; solve to find \(C = \frac{12}{10}\).

Step 5: Write the Decomposition

Finally, substitute the values of \(A\), \(B\), and \(C\) into the partial fraction decomposition:\[ \frac{-3x^2 - 3x + 27}{(x+2)(2x^2 + 3x - 9)} = \frac{-\frac{21}{10}}{x+2} + \frac{\frac{11}{18}}{x - 1.5} + \frac{\frac{12}{10}}{x + 3} \]

Key Concepts

Quadratic Formula

The quadratic formula is a method used to find the roots of a quadratic equation. It is derived from the process of solving the general quadratic equation, which is in the form \(ax^2 + bx + c = 0\). The formula itself is given by:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula is invaluable when factoring quadratics directly is challenging or impossible. By substituting the coefficients \(a\), \(b\), and \(c\) from the quadratic equation into this formula, you can calculate the exact solutions or roots of the equation.

Here's how to use it:

• Identify the coefficients: \(a\), \(b\), and \(c\).
• Calculate the discriminant: \(b^2 - 4ac\).
• Determine the number of roots: if the discriminant is positive, there are two distinct real roots; if zero, one real roots; if negative, roots are complex.
• Substitute \(a\), \(b\), and \(c\) into the formula to find \(x\).

In our exercise, for \(2x^2 + 3x - 9\), substituting into the quadratic formula helps us find roots at \(1.5\) and \(-3\), enabling easier factoring of the quadratic expression.

Factoring Quadratics

Factoring quadratics is an essential skill in algebra. It involves expressing a quadratic equation in the form \(ax^2 + bx + c\) as a product of its linear factors. This kind of expression is often easier to work with, especially when dealing with problems involving simplification, solving equations, or integration of rational functions.

There are several techniques for factoring quadratics:

• Factoring by grouping: This is useful when the quadratic is a perfect square or easily seen as such after arranging terms.
• Using the quadratic formula: The quadratic formula can help find the roots, and with these roots, you can rewrite the quadratic as a factor of linear binomials.
• Trial and error: Attempting common factors and simple binomial tests helps, especially for integers.

In our exercise, factoring the quadratic \(2x^2 + 3x - 9\) was achieved using the roots found via the quadratic formula. With roots \(1.5\) and \(-3\), the expression can be rewritten as the factored form \(2(x - 1.5)(x + 3)\), simplifying the process of partial fraction decomposition.

Rational Functions

Rational functions are ratios of two polynomials, expressed as \( \frac{P(x)}{Q(x)} \), where both \(P(x)\) and \(Q(x)\) are polynomials, and \(Q(x) eq 0\). These functions can characterize diverse behavior based on their structure, making them a crucial concept in mathematics, particularly in calculus.

The decomposition of rational functions into partial fractions is often necessary for integration, allowing more straightforward calculation of integrals involving these functions.

• Identify each polynomial: The numerator \(P(x)\) and the already factored denominator \(Q(x)\).
• Factor the denominator: If it's not already in factors, use strategies like the quadratic formula to aid the process.
• Set up partial fractions: Express the rational function as a sum of simpler fractions with unknown constants over each part of the factored denominator.

In the given exercise, the rational function \( \frac{-3x^2 - 3x + 27}{(x+2)(2x^2 + 3x - 9)} \) is decomposed by expressing it in terms of its partial fractions after fully factoring the denominator. Each part contributes to simplification or solving integration problems involving the function.