91Ó°ÊÓ

Rational expressions are fractions in which the numerator and/or the denominator are polynomials. They are similar to numerical fractions, but instead of numbers, you often deal with variables. In the exercise, the rational expression is \(\frac{-2x+6}{x^2-2x+1}\). The numerator is \(-2x+6\), and the denominator is \(x^2-2x+1\).

Rational expressions can appear complex, but they can be simplified or decomposed into simpler parts, much like numerical fractions. It's important to ensure that the denominator is not equal to zero, as this would make the expression undefined. Understanding how to work with rational expressions provides a stronger foundation for solving algebraic equations and inequalities.

Factoring involves expressing a polynomial as a product of its simpler polynomials. In our exercise, both the numerator \(-2x + 6\) and the denominator \(x^2 - 2x + 1\) needed to be factored.

To factor the numerator, we take out the common factor of \(-2\), giving \(-2(x - 3)\). The denominator can be factored as a perfect square trinomial, resulting in \((x - 1)^2\).

When factoring polynomials, consider the following steps:

• Identify common factors in all terms
• Look for special patterns like squares or cubes
• Apply techniques like grouping or the distributive property

Factoring polynomials simplifies rational expressions and is a crucial step in finding their partial fraction decomposition.

In partial fraction decomposition of rational expressions, the numerator often needs constants. In the exercise, these constants are represented by \(A\) and \(B\).

The goal is to rewrite the decomposed fractions in such a way that, when added together, they equal the original rational expression. The expression \(\frac{A}{x-1} + \frac{B}{(x-1)^2}\) sets up this scenario. Here, \(A\) and \(B\) are coefficients that must be determined by solving equations derived from equating the original and decomposed expressions.

Finding constants involves:

• Equating the original expression to its decomposed form
• Expanding and simplifying both sides
• Solving linear equations to determine each constant

This process often needs substituting specific values of \(x\) and solving for the unknowns, ultimately allowing you to express the rational expression in its simplest, decomposed form.