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Rational expressions are fractions where both the numerator and the denominator are polynomials. The example given in the exercise, \( \frac{ax+b}{(x-c)^{2}} \) where \( c eq 0 \) , is a rational expression. The key to understanding these expressions lies in breaking them down into simpler parts; this is where the concept of partial fraction decomposition comes in. In the context of algebra, rational expressions are used to represent real-world problems involving rates, ratios, and proportions.
To grasp their behavior, we must first become comfortable with their basic properties, such as identifying restrictions on the variable (to avoid division by zero) and simplifying them by factoring and canceling common terms. When faced with integrals or solving equations involving rational expressions, partial fraction decomposition becomes a valuable tool, as it can turn a complex fraction into a sum of simpler fractions, each with its own unique denominator.
Bear in mind that understanding the nature of the numerator and the denominator is essential, for the decomposition relies heavily on the degree and type of polynomials involved.

A linear binomial expression is a polynomial with two terms, usually of the form \( x - c \) where \( c \) is a constant, and often represents a shift along the x-axis of the line \( x = 0 \). In the given exercise, the denominator is \( (x-c)^2 \), signifying that it is the square of a linear binomial expression. This detail is crucial when performing partial fraction decomposition because the power of the binomial affects the number of terms in the decomposition.

Role in Decomposition

When decomposing a rational expression with a denominator like \( (x-c)^2 \), we must account for the repeated linear factor by setting up terms in the decomposition with both \( (x-c) \) and \( (x-c)^2 \) in the denominators. It helps simplify the complex expression into more manageable parts by assigning different coefficients to each term in the decomposition.
The process of breaking down expressions with these repeated factors is not about simplifying the rational expression itself but preparing it for easier manipulation, whether for integration, solving, or another operation. The linear binomial's simplicity makes it very approachable for learners to start with before moving on to more complex algebraic expressions.

Once we have set up the correct form for decomposition, as demonstrated in Step 1 and Step 2 of the solution, we must then solve for the coefficients. These are the values of A and B in our particular problem. Solving for coefficients requires us to create and solve equations that stem from equating coefficients of corresponding powers of \( x \) on both sides of the equation. This part of the process can sometimes be non-intuitive and often requires careful observation and algebraic manipulation.

Techniques in Solving

To find these coefficients, one common approach is to first clear the denominators by multiplying through by a common factor, which in this case is \( (x-c)^2 \)—the least common multiple of the denominators. Then, equating each term on the left side with its matching term on the right side provides us with a system of equations that corresponds to the powers of \( x \).
In simple cases, you might directly match coefficients of like terms. In more complex scenarios, it could involve coefficients of multiple terms and would require isolating variables and substituting back and forth between the equations. The process requires careful linear algebra manipulation and understanding the relationship between the constants and variables within the expression.