Problem 20 - Precalculus with Limits

Question

Writing the Partial Fraction Decomposition. Write the partial fraction decomposition of the rational expression. Check your result algebraically. $$\frac{5}{x^{2}+x-6}$$

Step-by-Step Solution

Answer

The partial fraction decomposition of the given rational expression is \(\frac{1}{x-2} - \frac{1}{x+3}\).

Step 1: Factorize the denominator

First, we need to factorize the denominator \(x^{2}+x-6\). By setting it equal to zero and solving for x, we get: \(x^{2}+x-6 = 0\),] this factorizes to (x-2)(x+3) = 0. Hence, \(x^{2}+x-6\) can be written as \(x-2)(x+3)\).

Step 2: Set up the Partial Fraction Decomposition

Next, we are going to set up the partial fractions. The given expression \(\frac{5}{x^{2}+x-6}\) can be decomposed into \(\frac{A}{x-2} + \frac{B}{x+3}\) where A and B are constants to be determined.

Step 3: Solving for Constants A and B

Multiply through by the denominator \(x^{2}+x-6\) to get rid of that from the expressions. Hence, we have \(5 = A(x+3) + B(x-2)\). Now, to find values of variables A and B, we can equate coefficients or set values for x that will result in simple equations for A and B. For example, setting \(x = 2\) gives \(5 = 5A\), hence, \(A = 1\). Similarly, setting \(x = -3\) gives \(5 = {-5B}\), hence, \(B = -1\).

Step 4: Write the Partial Fraction Decomposition

Now we can replace A and B in the decomposition, we have the partial fraction decomposition as \(\frac{1}{x-2} - \frac{1}{x+3}\).

Step 5: Verify the result

To ensure that the result is correct, add up these fractions, which should result in the original fraction. The common denominator for adding the fractions \(\frac{1}{x-2}\) and \(-\frac{1}{x+3}\) is \((x-2)(x+3)\), which results in the original expression, confirming that the partial decompositions are correct.

Key Concepts

Rational Expression

A rational expression is a fraction wherein both the numerator and the denominator are polynomials. These expressions can hold values over real numbers, except for values that make the denominator zero. This means that care needs to be taken around these values, as they are considered restrictions for the expression.

When faced with exercises involving rational expressions, you're usually asked to simplify them or break them down into simpler pieces. This is where partial fraction decomposition comes handy, especially when the denominator is factorable. Rational expressions provide a great way to understand how different algebraic structures can interact with each other through multiplication and division.

Factoring Quadratic Expressions

Factoring is a vital skill in algebra, especially when working with quadratic expressions. A quadratic expression, such as \(x^2 + x - 6\), can often be factored into a product of two linear expressions. This is crucial because factored forms can simplify further operations like decompositions or simplifications.

To factor a quadratic expression, look for two numbers that multiply to the constant term, in this case, -6, and add up to the middle coefficient, 1. For \(x^2 + x - 6\), these numbers are 3 and -2, leading to the factors \((x+3)(x-2)\). Factoring helps break down complex expressions into more manageable parts, paving the way for decomposition in partial fractions.

Algebraic Solutions

Solving algebraic equations involves finding unknown values that satisfy the equations. With partial fraction decomposition, the goal is to find constants that allow us to express a complex rational expression as a sum of simpler fractions.

In our exercise, after rewriting the expression \(\frac{5}{x^2+x-6}\) as \(\frac{A}{x-2} + \frac{B}{x+3}\), we need values for A and B. One effective method is to eliminate the denominator by multiplying through by \((x^2+x-6)\), resulting in an equation without fractions. Setting strategic values for \(x\), such as the roots, helps isolate A and B. Through this process, you find that \(A = 1\) and \(B = -1\), which completes the partial fraction decomposition.

Verification of Decomposition

After determining the partial fractions, verification ensures that no mistakes were made during decomposition. This step involves recombining the fractions \(\frac{1}{x-2} - \frac{1}{x+3}\) back into a single fraction.

By finding the common denominator \((x-2)(x+3)\) and recombining, you should return to the original rational expression \(\frac{5}{x^2+x-6}\). This not only confirms the accuracy of the decomposition but also reinforces the understanding of how factorization and simplifying expressions work together in algebra.