Problem 6 - Algebra for College Students

Question

Find the partial fraction decomposition for each rational expression. See answers below. \(\frac{-2 x-8}{10 x^{2}-x-2}\)

Step-by-Step Solution

Answer

\(\frac{1}{2x - 1} - \frac{3}{5x + 2}\)

Step 1: Identify the Form of Denominator

First, we need to identify the form of the given denominator. The denominator is \(10x^2 - x - 2\). Since it is a quadratic expression, we attempt to factor it, which is necessary for partial fraction decomposition.

Step 2: Factor the Denominator

Factor the quadratic denominator \(10x^2 - x - 2\). We look for two numbers that multiply to \( 10 \times -2 = -20 \) and add to \(-1\). The correct pair is \(-5\) and \(4\). So, we rewrite the quadratic as \(10x^2 - 5x + 4x - 2\), which can be factored into \((2x - 1)(5x + 2)\).

Step 3: Set Up the Partial Fractions

With the factored form of the denominator, \((2x - 1)(5x + 2)\), we set up the partial fractions as \(\frac{-2x - 8}{10x^2 - x - 2} = \frac{A}{2x - 1} + \frac{B}{5x + 2}\), where \(A\) and \(B\) are constants to be determined.

Step 4: Clear the Denominator and Formulate Equations

Multiply both sides of the equation by the original quadratic denominator to clear the fractions. This gives us \(-2x - 8 = A(5x + 2) + B(2x - 1)\). Now expand and combine like terms: \(-2x - 8 = (5A + 2B)x + (2A - B)\).

Step 5: Equate Coefficients to Solve for A and B

Equate the coefficients of like terms (terms with \(x\) and the constant terms) on both sides to derive a system of equations: \(5A + 2B = -2\) and \(2A - B = -8\). Solve this system to find \(A = 1\) and \(B = -3\).

Step 6: Write the Final Decomposition

Substitute \(A\) and \(B\) back into the partial fractions. The partial fraction decomposition is \(\frac{1}{2x - 1} - \frac{3}{5x + 2}\).

Key Concepts

Rational Expressions

Rational expressions are fractions where both the numerator and the denominator are polynomials. In the context of mathematics, these expressions provide a way to represent complex relationships and functions using polynomial basis. For example, a simple rational expression might look like \( \frac{x^2 + 3x + 2}{x - 1} \). This encompasses various parts to consider:

• Numerator Polynomial: The polynomial at the top of the fraction.
• Denominator Polynomial: The polynomial at the bottom, which must not be zero, as division by zero is undefined.

Understanding rational expressions is crucial to simplifying expressions and solving equations involving fractions. It involves operations like addition, subtraction, multiplication, division, and factoring. In the given problem, \( \frac{-2x - 8}{10x^2 - x - 2} \), recognizing that this is a rational expression is the first step in the decomposition process. By decomposing these into simpler fractions, calculations and integrations become more manageable.

Factoring Quadratic Expressions

Factoring quadratic expressions involves rewriting them as products of simpler factors, often in the form of two binomials. It's a method frequently used to simplify expressions and solve equations. Consider a standard quadratic expression of the form \( ax^2 + bx + c \).

• Identify Suitable Numbers: Look for pairs of numbers that multiply to \( a \cdot c \) and add up to \( b \).
• Rewrite and Group: Rewrite the middle term using these numbers to facilitate factoring by grouping.
• Factor by Grouping: Separate and group terms to find their common factors and form binomials.

In the example \( 10x^2 - x - 2 \), you look for numbers that multiply to \(-20\) and add to \(-1\). This results in splitting the middle term, allowing it to factor into \((2x - 1)(5x + 2)\), setting the stage for partial fraction decomposition.

System of Equations

A system of equations is a set of multiple equations that require simultaneous solutions. For partial fraction decomposition, systems of equations help us determine the unknown constants in the numerators of our partial fractions. These constants, such as \( A \) and \( B \) in our example, are crucial in expressing the original fraction correctly.

• Setting up Equations: After factoring, you create partial fractions and multiply through by the original denominator to get rid of fractions, yielding an equation involving \( A \) and \( B \).
• Equating Coefficients: Compare the coefficients of like terms (both variable and constant terms) from both sides of the equation to form your system of equations.
• Solving the System: Use substitution or elimination methods to solve the system and find values for the unknowns.

In our example, the equations \( 5A + 2B = -2 \) and \( 2A - B = -8 \) must be solved to determine \( A=1 \) and \( B=-3 \). This process allows us to decompose the fraction accurately.

Constants in Decomposition

The constants in decomposition are the unknown variables that we solve for in the fractions resulting from partial fraction decomposition. These constants determine the exact value of each partial fraction in the decomposition process.

• Determining Constants: After setting up the partial fractions, multiply both sides by the common denominator to clear the fractions.
• Solving the System: Use the system of equations formed by equating the coefficients of similar terms to solve for these constants.
• Importance: Correctly identifying these constants ensures the integrity of the original rational expression is maintained through decomposition.

In our example, the process involved determining \( A \) and \( B \) to be \( 1 \) and \( -3 \) respectively, resulting in the expression \( \frac{1}{2x - 1} - \frac{3}{5x + 2} \). These constants are essential for the correct representation of the initial rational expression as a sum of simpler fractions.