Problem 9 - Precalculus

Question

(a) factor the denominator of the given \(\mathrm{ra}\) tional expression; (b) determine the form of the partial fraction decomposition for the given rational expression; and (c) determine the values of the constants in the partial fraction decomposition that you gave in part (b). To help you in spotting errors, use the fact that in part (c), each of the required constants turns out to be an integer. $$\frac{7 x+39}{x^{2}-x-6}$$

Step-by-Step Solution

Answer

The partial fraction decomposition is \(\frac{12}{x-3} - \frac{5}{x+2}\).

Step 1: Factor the Denominator

First, we need to factor the denominator \(x^2 - x - 6\). To do this, find two numbers that multiply to \(-6\) (the constant term) and add to \(-1\) (the coefficient of the middle term). The numbers are \(-3\) and \(2\). Thus, the factorization is \((x - 3)(x + 2)\).

Step 2: Set Up the Partial Fractions

The expression \(\frac{7x + 39}{(x - 3)(x + 2)}\) can be decomposed into partial fractions of the form:\[\frac{A}{x-3} + \frac{B}{x+2}\]where \(A\) and \(B\) are constants we need to find.

Step 3: Combine the Fractions on the Right Side

Combine the fractions on the right-hand side to have a common denominator:\[\frac{A(x+2) + B(x-3)}{(x-3)(x+2)}\]\[= \frac{Ax + 2A + Bx - 3B}{(x-3)(x+2)}\]\[= \frac{(A+B)x + (2A - 3B)}{(x-3)(x+2)}\]This must equal the initial expression \(\frac{7x + 39}{(x - 3)(x + 2)}\).

Step 4: Set Up and Solve the System of Equations

By equating coefficients from \(\frac{(A+B)x + (2A - 3B)}{(x-3)(x+2)}\) and \(\frac{7x + 39}{(x-3)(x+2)}\), we have:1. \(A + B = 7\)2. \(2A - 3B = 39\)Solve the system of equations. Substitute \(B = 7 - A\) into the second equation:\[2A - 3(7 - A) = 39\]\[2A - 21 + 3A = 39\]\[5A = 60\]\[A = 12\]Now substitute \(A = 12\) back into \(A + B = 7\):\[12 + B = 7\]\[B = -5\]

Step 5: Write the Partial Fraction Decomposition

Finally, substitute the values of \(A\) and \(B\) into the partial fractions setup from Step 2. Thus, the partial fraction decomposition is:\[\frac{12}{x-3} - \frac{5}{x+2}\]

Key Concepts

Factoring Polynomials

Factoring polynomials is a crucial step in solving problems involving rational expressions. To factor a polynomial, we aim to express it as a product of simpler polynomials, typically as linear factors. For the quadratic polynomial \(x^2 - x - 6\), we looked for two numbers that multiply to \(-6\) (the constant term) and add to \(-1\) (the coefficient of the middle term).

These numbers were found to be \(-3\) and \(2\), leading us directly to the factorization: \((x - 3)(x + 2)\).

This step is essential because it simplifies the denominator, allowing us to apply partial fraction decomposition effectively in subsequent steps.

Rational Expressions

Rational expressions are fractions where the numerator and the denominator are polynomials. In the given problem, the rational expression \(\frac{7x + 39}{x^2 - x - 6}\) needed to be decomposed to simpler, easy-to-manage fractions, known as partial fractions.

To achieve this for fractions like \(\frac{7x + 39}{(x-3)(x+2)}\), we rewrite it using partial fractions: \(\frac{A}{x-3} + \frac{B}{x+2}\). Here, \(A\) and \(B\) are constants that need to be determined.

The partial fraction decomposition breaks down complex rational expressions into sums of simpler ones, greatly facilitating further algebraic operations and integrations.

Systems of Equations

Systems of equations frequently appear in mathematical problems, especially when determining constants in a decomposition. In our partial fraction decomposition, we equated the numerators: \((A+B)x + (2A - 3B)\) with \(7x + 39\). This resulted in a system of two equations with two unknowns:

• \(A + B = 7\)
• \(2A - 3B = 39\)

To solve this system, we substituted \(B = 7 - A\) into the second equation, leading us to find \(A = 12\) and then substituting back to find \(B = -5\).

Systems of equations in this context are instrumental for deriving the specific values of the constants involved in the partial fraction process.