Problem 65 - Algebra and Trigonometry

Question

determine whether each statement makes sense or does not make sense, and explain your reasoning. Find the partial fraction decomposition of $$\frac{4 x^{2}+5 x-9}{x^{3}-6 x-9}$$

Step-by-Step Solution

Answer

The partial fraction decomposition of \(\frac{4x^{2}+5 x-9}{x^{3}-6 x-9}\) is the sum of the simpler fractions found in Step 4, each fraction has coefficients derived from solving the equations in Step 3.

Step 1: Factoring the Denominator

The first step is to factor the denominator \(x^{3}-6x-9\). If it can be factored, it will take the form \(x(x-a)(x-b)\), where \(a\) and \(b\) are roots of the equation \(x^2-6x-9=0\). To find the roots, solve this quadratic equation.

Step 2: Setting Up Equivalent Expressions

The original fraction is now set equal to a new expression with the discovered factors in the denominator. If the factored denominator takes the form \(x(x-a)(x-b)\), the corresponding expression would be in the form \(\frac{A}{x} + \frac{B}{x-a} + \frac{C}{x-b}\) where 'A', 'B', and 'C' are coefficients that are to be determined.

Step 3: Equating Coefficients and Solving

To find the coefficients 'A', 'B', and 'C', multiply both sides of the equation by \(x(x-a)(x-b)\) to get rid of the denominator. Now, equate coefficients on both sides of the equation to solve for 'A', 'B', and 'C'.

Step 4: Obtaining the Partial Fraction Decomposition

Insert the solved values 'A', 'B', and 'C' into the expression found in Step 2. This gives the partial fraction decomposition of the original fraction. If needed, these simpler fraction can now be easily integrated or differentiated.

Key Concepts

Factoring Polynomials

Factoring polynomials is an essential mathematical skill that assists in simplifying expressions and solving equations. When you factor a polynomial, you rewrite it as a product of its factors, which are simpler polynomials. Let's consider the polynomial given in the original exercise: \(x^3 - 6x - 9\). The first task is to see if it can be factored further. To achieve this, one should look for potential factors or roots using trial, error, or mathematical techniques such as the Rational Root Theorem. Another common method involves synthetic division if one root is known.

After finding one factor, the remaining polynomial can often be simplified by further division or by solving a quadratic equation if the remaining polynomial is of degree 2. Factoring is crucial because having the polynomial in its simplest form allows you to set up partial fraction decompositions and solve equations more straightforwardly.

Quadratic Equations

Quadratic equations are foundational in algebra. A quadratic equation typically takes the form \(ax^2 + bx + c = 0\). The solutions or roots of quadratic equations can be found using several methods:

• Factoring: This method works best when the polynomial can be easily rewritten as a product of two binomials.
• Quadratic Formula: If a quadratic equation cannot be factored quickly, the formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) becomes handy.
• Completing the Square: By rearranging and manipulating terms, you can often bring a quadratic into a form that is immediately solvable.

In the exercise's context, solving \(x^2 - 6x - 9 = 0\) might require using the quadratic formula, especially if the roots aren't immediately obvious by simple factoring. Solving this will aid in determining the complete factorization of the polynomial.

Rational Expressions

Rational expressions are fractions where the numerator and the denominator are polynomials. Simplifying these expressions often involves finding a common denominator or factoring to cancel out common terms. In the exercise, we start with the rational expression \(\frac{4x^2 + 5x - 9}{x^3 - 6x - 9}\).

Simplifying the expression through partial fraction decomposition is a process that involves:

• Factoring the denominator into its simpler parts, as discussed previously.
• Setting up the rational expression as a sum of simpler fractions, where each has one factor of the denominator as its new denominator.
• Solving for unknown coefficients to match the original expression when combined.

This decomposition is beneficial for integration, especially in calculus, where complex fractions are simplified to allow for easier manipulation. Rational expressions are omnipresent in algebra and many practical applications, making understanding their simplification and decomposition techniques invaluable in mathematics.