Problem 66 - Precalculus with Limits

Question

Writing the Partial Fraction Decomposition , write the partial fraction decomposition of the rational expression. Then assign a value to the constant \(a\) to to check the result algebraically and graphically. $$\frac{1}{x(x+a)}$$

Step-by-Step Solution

Answer

The partial fraction decomposition of \( \frac{1}{x(x+a)} \) is \( \frac{1}{ax} - \frac{1}{a(x+a)} \).

Step 1: Express in Partial Fractions

We rewrite the expression \( \frac{1}{x(x+a)} \) as \( \frac{A}{x} + \frac{B}{x+a} \), where A and B are constants that we need to determine. This step is standard for partial fractions, where we express the given fraction in terms of simpler fractions.

Step 2: Solve for A and B

To find A and B, we equate \( \frac{1}{x(x+a)} \) back to \( \frac{A}{x} + \frac{B}{x+a} \) and solve for A and B. If we clear the denominator by multiplying the equation by \( x(x+a) \), we get 1 = A(x+a) + Bx. Setting x = 0 gives us A = 1/a. Setting x = -a gives us B = -1/a.

Step 3: Substitute A and B

We substitute A = 1/a and B = -1/a back into the expression in Step 1, yielding \( \frac{1}{ax} - \frac{1}{a(x+a)} \).

Step 4: Verify the Result

Now that we have the partial fraction decomposition, we can check our work both algebraically (by adding the two fractions and simplifying to see if we get the original expression) and graphically (by plotting the original function \( \frac{1}{x(x+a)} \) and the resulting function \( \frac{1}{ax} - \frac{1}{a(x+a)} \) and comparing the two plots. It's important to verify the answers with the original function to ensure accuracy.

Key Concepts

Rational Expressions

A rational expression is a quotient of two polynomials, similar to how a rational number is a quotient of two integers. As with fractions that have numeric denominators, the key to understanding rational expressions is knowing how to manipulate them properly.

When working with rational expressions, we often aim to simplify them, to identify their restrictions, or to rewrite them in a more useful form, such as by using partial fraction decomposition. Simplifying a rational expression involves reducing it to its lowest terms by canceling any common factors between the numerator and the denominator. Identifying restrictions is crucial because the values that make the denominator equal to zero are not part of the expression's domain.

To aid in understanding, let's consider our example \( \frac{1}{x(x+a)} \) as a rational expression. We have two polynomials where the numerator is 1 (which is also a polynomial of degree zero) and the denominator is \( x(x+a) \), a polynomial of degree two. One cannot directly simplify this expression further by factoring or canceling out terms. Therefore, the technique of partial fraction decomposition becomes useful to break this complex expression into simpler, more manageable parts.

Algebraic Identities

Algebraic identities are equations that are true for all values of the variables involved. They are foundational tools used to manipulate algebraic expressions and to simplify complex equations. Common examples include \( (a+b)^2 = a^2 + 2ab + b^2 \) and \( (a-b)(a+b) = a^2 - b^2 \) among others.

In the process of partial fraction decomposition, we leverage algebraic identities to solve for unknown constants. By setting the decomposed expression equal to the original function, we form an identity. To illustrate, from our initial problem, the identity formed is \( 1 = A(x+a) + Bx \). This equation holds for all values of \( x \), giving us a system of equations that can be solved for \( A \) and \( B \) which are later used in the reconstitution of the original expression into its partial fractions.

The key to constructing and solving these algebraic identities lies in choosing strategic values for \( x \) that simplify the process of finding the constants \( A \) and \( B \), as demonstrated in the provided step-by-step solution.

Polynomial Division

Polynomial division is a process similar to long division with numbers, where one polynomial, the dividend, is divided by another, the divisor, to find a quotient and sometimes a remainder. This operation is particularly useful in simplifying expressions and solving equations.

The method of partial fractions is closely related to polynomial division because we effectively 'divide' a complex rational expression into simpler ones. The decomposed parts often represent the result of dividing the numerator by each factor of the denominator separately. In our example, while the original expression doesn't permit polynomial division due to its numerator being of a lesser degree than the denominator, the concept still informs the restructuring of \( \frac{1}{x(x+a)} \) into its partial fractions.

Through the process we've applied to our problem, polynomial division isn't explicitly used; however, understanding that partial fraction decomposition can be seen as an extension of division can help with grasping the reason behind why the method works. It's essentially about finding expressions that, when combined, will give back the original rational expression, akin to how quotient and remainder together reconstruct the dividend in polynomial division.