91Ó°ÊÓ

In the world of algebra, we often encounter rational expressions. These expressions are fractions where both the numerator and the denominator are polynomials. A rational expression looks like this: \(\frac{P(x)}{Q(x)}\), where \(P(x)\) and \(Q(x)\) are polynomials, and \(Q(x)\) is not equal to zero. Understanding rational expressions is crucial, especially when working on partial fraction decomposition.

To break down a complex fraction into simpler parts, partial fraction decomposition comes into play. Here, we express a single rational expression as the sum of two or more simpler rational expressions. For example, our main exercise involves rewriting the fraction \(\frac{x^2 + x - 31}{(x+3)^2 (x-2)}\) into a simpler form: \(\frac{A}{x+3} + \frac{B}{(x+3)^2} + \frac{C}{x-2}\).

By doing so, future problems, such as integration, become more manageable. The process starts by setting up our rational expression in terms of unknown constants (like A, B, and C). We then solve these constants using algebraic techniques.

Key things to remember:

• Polynomials in the numerator and denominator
• The denominator cannot be zero
• Decomposition simplifies complex fractions

A critical step in partial fraction decomposition is solving for the constants using a system of equations. A system of equations involves multiple equations that share the same set of unknowns. Solving these equations simultaneously allows us to find the values for the unknowns.

In our exercise, after expressing the fraction in terms of partial fractions, we equate and simplify the numerator to get an equation. This leads to:

• For \(x^2\): \(A + C = 1\)
• For \(x\): \(A + 6C = 1\)
• Constant term: \(-6A - 2B + 9C = -31\)

By comparing the coefficients of like terms, we form a system of three equations with variables A, B, and C. We solve this system step-by-step:

1. From equations 1 and 2, we subtract (A + C) from (A + 6C):
5C = 0, hence, C = 0.
2. Substituting C = 0 into A + C = 1, we get A = 1.
3. Finally, substituting A = 1 and C = 0 into the third equation, we find B:
-6(1) - 2B + 9(0) = -31
-6 - 2B = -31
-2B = -25
B = \(\frac{25}{2}\)

Solving systems of equations helps us find the exact constants needed for the partial fraction decomposition. It's a fundamental math skill used in algebra and beyond.

Polynomial division is a technique used in algebra that resembles long division but applies to polynomials. While it wasn't directly used in our main exercise, understanding it is crucial, as it often appears in rational expression problems. Here’s a brief rundown of how polynomial division works:

Given polynomials \(P(x)\) and \(D(x)\), where \(D(x)\) ≠ 0, you can divide \(P(x)\) by \(D(x)\) to get a quotient \(Q(x)\) and a remainder \(R(x)\). The result looks like this:
\(P(x) = D(x) \frac{Q(x)}{R(x)} + R(x)\).

Here’s a simple step-by-step of the division process:

• Write down the polynomials \(P(x)\) and \(D(x)\).
• Divide the leading term of the dividend by the leading term of the divisor; this gives you the first term of the quotient.
• Multiply the entire divisor by this term and subtract the result from the dividend.
• Repeat the process with the new polynomial until the remainder has a lower degree than the divisor.

Polynomial division is essential because it often sets the stage for other operations, like partial fraction decomposition.
By understanding how polynomials divide, you can simplify complex expressions and solve equations more efficiently.
Use this concept alongside rational expressions and systems of equations to tackle a wide variety of algebra problems!