Problem 4 - College Algebra with Corequisite Support

Question

You are unsure if you correctly decomposed the partial fraction correctly. Explain how you could double-check your answer.

Step-by-Step Solution

Answer

Expand terms, equate coefficients, solve, and verify by substituting back.

Step 1: Identify the Initial Decomposition

Assume you have a rational function that you've decomposed into partial fractions. Write down the initial decomposition you have obtained. For instance, let's consider a simple example: \( \frac{3x + 5}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2} \).

Step 2: Clear the Denominators

Multiply both sides of your decomposition by the original denominator to eliminate fractions. Continuing from the previous example: \( 3x + 5 = A(x+2) + B(x-1) \).

Step 3: Expand and Combine Terms

Expand the expressions on the right-hand side and then combine like terms. This gives more insight into how the numerators on both sides should equate. For example, \( 3x + 5 = Ax + 2A + Bx - B \), which simplifies to \( 3x + 5 = (A + B)x + (2A - B) \).

Step 4: Equate Coefficients

Set up equations by equating the coefficients of corresponding terms on both sides. In our example, this gives two equations: \( A + B = 3 \) (coefficient of \( x \)) and \( 2A - B = 5 \) (constant term).

Step 5: Solve the Equations

Solve the system of equations obtained from the previous step to find the values of \( A \) and \( B \). For this example, solving the equations \( A + B = 3 \) and \( 2A - B = 5 \) should give valid values for \( A \) and \( B \).

Step 6: Substitute and Verify

Substitute the values of \( A \) and \( B \) back into your initial decomposition and multiply through to check if you get the original numerator. If \( A = 2 \) and \( B = 1 \), substituting back gives \( \frac{2}{x-1} + \frac{1}{x+2} = \frac{3x + 5}{(x-1)(x+2)} \), matching the original function.

Key Concepts

Rational Functions

When delving into the world of partial fractions, it's essential to understand what rational functions are. A rational function is, simply put, a ratio of two polynomials. These functions are expressed in the form \( \frac{P(x)}{Q(x)} \), where both \( P(x) \) and \( Q(x) \) are polynomials and \( Q(x) eq 0 \).
Rational functions arise quite often in calculus and algebra, especially when dealing with integration, as they can often be simplified or broken down into partial fractions.
This breakdown makes certain mathematical operations, like integration, much more manageable. Indeed, partial fraction decomposition is a key technique used to transform complex rational expressions into simpler components.
For example, the function \( \frac{3x + 5}{(x-1)(x+2)} \) can be decomposed into simpler fractions, which are easier to handle.

Decomposition

Decomposition is a technique where a complex fraction is broken down into a sum of simpler fractions. This is particularly useful when you're trying to integrate a rational function as smaller fractions are often much easier to work with.
The process begins by expressing the rational function as a sum of fractions. Each fraction has a simpler denominator that is a part of the denominator of the original rational function. For instance, the rational function \( \frac{3x + 5}{(x-1)(x+2)} \) can be rewritten as \( \frac{A}{x-1} + \frac{B}{x+2} \).
The next step involves clearing the original denominator by multiplying through, allowing us to solve for the unknown coefficients \( A \) and \( B \). This stage is crucial as it sets the foundation for the succeeding steps.

Coefficient Comparison

Once you have multiplied out the fractions and expanded all terms, the next step is coefficient comparison. This technique involves setting up equations by comparing the coefficients of corresponding terms in the polynomial. Essentially, you're ensuring both sides of the equation are identical.
For example, take the equation from our decomposition: \( 3x + 5 = A(x+2) + B(x-1) \). After expanding, we end up with \( 3x + 5 = (A+B)x + (2A-B) \).
From this, we compare the coefficients:

• For \( x \): \( A + B = 3 \)
• For the constant term: \( 2A - B = 5 \)

These equations help us determine the values of \( A \) and \( B \), ensuring that our decomposition is correct and accurate.

System of Equations

After establishing equations through coefficient comparison, you end up with a system of equations. Solving this system is the next step to finding the unknown coefficients. A system of equations is a set of two or more equations with the same set of variables.
In our example, the system set up was:

• \( A + B = 3 \)
• \( 2A - B = 5 \)

You can solve these by various methods such as substitution, elimination, or even matrices. Here, these are fairly simple linear equations, and you can solve them by substitution or elimination. Solving the above gives \( A = 2 \) and \( B = 1 \).
Once these values are found, it's important to substitute them back into your original partial fraction equation to verify correctness. If they hold true and reconstruct your original fraction, your decomposition is correct.