Problem 3 - Algebra for College Students

Question

Find the partial fraction decomposition for each rational expression. See answers below. \(\frac{-2 x-8}{x^{2}-1}\)

Step-by-Step Solution

Answer

\(\frac{-2x-8}{x^2-1} = \frac{-5}{x-1} + \frac{3}{x+1}\).

Step 1: Identify the Form of the Denominator

The denominator is \(x^2 - 1\), which can be factored further. Recognize that \(x^2 - 1\) is a difference of squares, hence it factors to \((x-1)(x+1)\).

Step 2: Set the Form for Partial Fractions

For the rational expression \(\frac{-2x-8}{x^2-1}\), which is now \(\frac{-2x-8}{(x-1)(x+1)}\), we set up the partial fraction decomposition as \(\frac{A}{x-1} + \frac{B}{x+1}\), where \(A\) and \(B\) are constants to be determined.

Step 3: Combine the Partial Fractions

Combine the fractions on the right: \( \frac{A}{x-1} + \frac{B}{x+1} = \frac{A(x+1) + B(x-1)}{(x-1)(x+1)}\).

Step 4: Set Equal to Original Expression

Equate the two expressions: \(\frac{-2x-8}{(x-1)(x+1)} = \frac{A(x+1) + B(x-1)}{(x-1)(x+1)}\). Since the denominators are the same, set the numerators equal: \[-2x - 8 = A(x+1) + B(x-1)\].

Step 5: Expand and Collect Like Terms

Expand the right-hand side: \[A(x+1) + B(x-1) = Ax + A + Bx - B = (A+B)x + (A-B)\].This means we equate coefficients: -2 (coefficient of \(x\)) and -8 (constant term).

Step 6: Set Up System of Equations

From comparing coefficients, we have:1. \(A + B = -2\) (equate the coefficient of \(x\))2. \(A - B = -8\) (equate the constant term).Solve for \(A\) and \(B\).

Step 7: Solve the System of Equations

Add the two equations: \[(A + B) + (A - B) = -2 + (-8)\]\[2A = -10 \Rightarrow A = -5\]Substitute \(A = -5\) into \(A + B = -2\):\[-5 + B = -2 \Rightarrow B = 3\].

Step 8: Write the Partial Fractions

Therefore, the partial fraction decomposition is:\[\frac{-2x-8}{x^2-1} = \frac{-5}{x-1} + \frac{3}{x+1}\].

Key Concepts

Rational Expressions

Rational expressions are expressions formed by the ratio of two polynomials. Much like fractions, these expressions can often be simplified and decomposed into simpler parts to make operations like addition, subtraction, and partial fraction decomposition more manageable. For example, when working with the rational expression \(\frac{-2x-8}{x^2-1}\), the goal is to break it into simpler fractions that add up to the same form as the original, using the method known as partial fraction decomposition. This often involves first factoring the denominator, hence simplifying the overall calculation. With rational expressions, understanding the role of both the numerator and the denominator is crucial, as one must be familiar with factoring, equivalence, and transformation techniques to work effectively with them.

Difference of Squares

The difference of squares is a special factoring technique that turns expressions like \(x^2 - 1\) into a product of two binomials. It follows the identity \(a^2 - b^2 = (a - b)(a + b)\). Applying this to our example's denominator, \(x^2 - 1\), we get \((x - 1)(x + 1)\). This powerful method helps with factorization, making the process of partial fraction decomposition much more straightforward. Recognizing the difference of squares quickens the simplification of complex algebraic expressions, leading to more straightforward solutions when solving equations, integrating, or performing other mathematical operations. Often once this pattern is noted, subsequent steps in solving rational expressions become much clearer.

System of Equations

When decomposing a rational expression, you often end up creating a system of equations to find unknown constants (like \(A\) and \(B\) in our partial fractions form). For the expression \(\frac{-2x-8}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}\), setting up a system is crucial for solving for \(A\) and \(B\). This involves equating the coefficients of like terms. In our scenario, we had two equations derived from this setup:

• \(A + B = -2\)
• \(A - B = -8\)

By solving this system, we found \(A = -5\) and \(B = 3\). Understanding and solving such systems of equations is essential in finding the right coefficients for decomposing rational expressions correctly. Practice with different types of these systems improves algebraic manipulation skills and enhances problem-solving strategies."}]}]}]}'