Question

Find the partial fraction decomposition of each rational expression. $$ \frac{4 x}{2 x^{2}+3 x-2} $$

Short answer

The partial fraction decomposition is: \( \frac{4}{5} \cdot \frac{1}{2x-1} + \frac{8}{5} \cdot \frac{1}{x+2} \)

Step-by-step solution

- Factor the denominator

First, factor the quadratic expression in the denominator: \[ 2x^2 + 3x - 2 \] To do this, find two numbers that multiply to \(2 \times -2 = -4\) and add up to \(3\). These numbers are 4 and -1. So, the quadratic can be factored as: \[ 2x^2 + 4x - x - 2 = (2x - 1)(x + 2) \]

- Set up the partial fraction decomposition

Now that the denominator is factored, set up the partial fraction decomposition: \[ \frac{4x}{(2x-1)(x+2)} = \frac{A}{2x-1} + \frac{B}{x+2} \] Here, \(A\) and \(B\) are constants that need to be determined.

- Clear the fractions

Multiply both sides by the denominator \( (2x-1)(x+2) \) to clear the fractions: \[ 4x = A(x+2) + B(2x-1) \]

- Expand and collect like terms

Expand and collect like terms: \[ 4x = Ax + 2A + 2Bx - B \] Combine the like terms: \[ 4x = (A + 2B)x + (2A - B) \]

- Set up the system of equations

Compare the coefficients from both sides of the equation for \(x\) and the constant terms. This gives us the system of equations: \[ A + 2B = 4 \] \[ 2A - B = 0 \]

- Solve the system of equations

Solve the system of equations. From the second equation: \[ B = 2A \] Substitute \( B \) into the first equation: \[ A + 2(2A) = 4 \] This simplifies to: \[ 5A = 4 \] \[ A = \frac{4}{5} \] Now substitute \( A \) back into the equation for \( B \): \[ B = 2 \left( \frac{4}{5} \right) = \frac{8}{5} \]

- Write the partial fraction decomposition

Now that the values of \( A \) and \( B \) are determined, substitute them back into the partial fractions: \[ \frac{4x}{(2x-1)(x+2)} = \frac{4/5}{2x-1} + \frac{8/5}{x+2} \] This can be written as: \[ \frac{4x}{(2x-1)(x+2)} = \frac{4}{5} \cdot \frac{1}{2x-1} + \frac{8}{5} \cdot \frac{1}{x+2} \]

Key concepts

Rational Expressions

Rational expressions are fractions where both the numerator and the denominator are polynomials. In algebra, working with these expressions often involves simplifying, factoring, and performing operations like addition, subtraction, multiplication, and division. Another crucial skill is partial fraction decomposition, which breaks down a complex fraction into simpler parts. This technique is particularly useful in calculus for integrating complex rational expressions. Remember, the goal is to express the rational expression as a sum of simpler fractions that are easier to handle.

Factoring Quadratics

Factoring quadratics is a fundamental step in simplifying rational expressions and in partial fraction decomposition. A quadratic polynomial is typically in the form d(x) = ax^2 + bx + cTo factor it, we look for two numbers that multiply to the product of 'a' and 'c' and add up to 'b'. This process helps us break down the quadratic into two binomials. For example, in the given problem:2x^2 + 3x - 2We need to find two numbers that multiply to -4 (2 * -2) and add up to 3. These numbers are 4 and -1. This allows us to factor the quadratic as:(2x - 1)(x + 2)Factoring is a critical step because it makes the process of partial fraction decomposition possible by breaking the denominator into manageable parts.

System of Equations

A system of equations involves finding values for variables that satisfy multiple equations simultaneously. In partial fraction decomposition, once we have set up our partial fractions, we often need to solve for unknown constants by comparing coefficients. This results in a system of linear equations. For instance:A + 2B = 42A - B = 0Solving these equations requires methods like substitution or elimination. In the current problem, substitution is used. First, solve one of the equations for one variable (like B = 2A), then substitute back into the other equation to solve for A. After finding A, substitute back to find B. This simultaneous solving allows us to find the exact values that satisfy both equations.

Coefficients

Coefficients are the numerical or constant factors in terms of an algebraic expression. Recognizing and manipulating coefficients is essential in many algebraic processes, including the system of equations used in partial fraction decomposition. When we have:4x = Ax + 2A + 2Bx - Bwe combine like terms to get:4x = (A + 2B)x + (2A - B)Here, comparing coefficients means setting the coefficients of corresponding powers of x, and constants, equal to each other. This gives the system of equations we solve. Recognizing how coefficients relate in an equation is crucial for accurate calculations and understanding the relationships between terms in an expression. Being careful and methodical with coefficients can make complex problems much simpler.