Problem 21 - Precalculus with Limits

Question

Writing the Partial Fraction Decomposition. Write the partial fraction decomposition of the rational expression. Check your result algebraically. $$\frac{3}{x^{2}+x-2}$$

Step-by-Step Solution

Answer

The partial fraction decomposition of \( \frac{3}{x^{2} + x - 2} \) is \( \frac{1}{x - 1} - \frac{1}{x + 2} \).

Step 1: Factorize the denominator

The first thing to be done is to factorize the denominator: \(x^{2} + x - 2\). This will break down to \((x - 1)(x + 2)\).

Step 2: Set up the general form of decomposition

According to the Partial Fraction Decomposition method, the rational expression can be broken down into the following form:\[\frac{3}{x^{2} + x - 2} = \frac{A}{x - 1} + \frac{B}{x + 2}\]. This equation is the general form of Partial Fraction Decomposition where A and B are constants which need to be found.

Step 3: Solve for A and B

To compute the values of constants A and B, multiply throughout by the denominator of the left side to clear out the fractions:\[3 = A(x + 2) + B(x - 1)\]Next, choose suitable values for x to solve for A and B. Let's take x = 1 to find A:\[3 = A(1 + 2) + B(1 - 1) \rightarrow A = 1\]Similarly, let's take x = -2 to find B:\[3 = A(-2 + 2) + B(-2 - 1) \rightarrow B = -1\]

Step 4: Write down the partial fraction decomposition

Substitute the values of A and B back into the general expression to get two separate fractions. Therefore, the partial fraction decomposition of \( \frac{3}{x^{2} + x - 2} \) is: \( \frac{1}{x - 1} - \frac{1}{x + 2} \).

Key Concepts

Factorization of Polynomials

Factorization is a key tool in algebra, where we express a polynomial as a product of its factors. Factors are polynomials of lower degrees that, when multiplied together, give the original polynomial. Understanding how polynomials factor is crucial when working with rational expressions and algebraic equations, especially when dealing with partial fraction decomposition.

Let's consider the polynomial in the denominator of our exercise, which is a quadratic polynomial given as \(x^2 + x - 2\). The polynomial can be written as a product of two first-degree polynomials. This process begins by looking for two numbers that multiply to give \(-2\) (the constant term) and add up to \(1\) (the coefficient of the x term). These numbers are \(-1\) and \(2\). Thus, the polynomial \(x^2 + x - 2\) factors into \((x - 1)(x + 2)\), which are its factors. This step is critical in the partial fraction decomposition process as it allows us to split a complex rational expression into simpler parts.

Rational Expressions

Rational expressions are fractions where the numerator and the denominator are both polynomials. They can often appear intimidating due to their complexity. However, just like numeric fractions, they can be simplified and manipulated using similar rules.

In the context of our exercise, \(\frac{3}{x^2 + x - 2}\) is a rational expression with 3 in the numerator and the polynomial \(x^2 + x - 2\) in the denominator. To work with such expressions, especially to integrate or differentiate them or to determine their limits, we often use partial fraction decomposition. This involves breaking the expression down into simpler fractional components whose denominators are the factors of the original denominator. In the given example, after factoring the polynomial, the expression was decomposed into \(\frac{1}{x - 1} - \frac{1}{x + 2}\). Each of these new rational expressions is easier to handle than the original.

Algebraic Equations

Algebraic equations are mathematical statements indicating that two expressions are equal. They have one or more variables and are useful in expressing relationships and solving problems. When solving algebraic equations, the goal is often to isolate the variable(s) and find their values.

In the partial fraction decomposition process, we set up an equation based on the original rational expression, like in our exercise: \(3 = A(x + 2) + B(x - 1)\). To find the unknowns A and B, we need to choose values for x that simplify the equation. By strategically picking x-values that correspond to the zeroes of the factors in the denominator, we eliminate one of the variables at a time. For instance, choosing \(x = 1\) nullifies the term with B, letting us solve for A directly. Similarly, selecting \(x = -2\) zeroes out the term with A, enabling us to find B. The solutions to these algebraic equations are the coefficients needed for constructing the segmented parts of the rational expression in its decomposed form.