Problem 1 - College Algebra

Question

Write the form of the partial fraction decomposition of the rational expression. It is not necessary to solve for the constants. $$\frac{11 x-10}{(x-2)(x+1)}$$

Step-by-Step Solution

Answer

The form of the partial fraction decomposition of \( \frac{11 x -10}{(x-2)(x+1)} \) is \( \frac{p}{(x-2)} + \frac{q}{(x+1)} \).

Step 1 : Identify the General Form for the Decomposition

The general form for the partial fraction decomposition of a rational function with linear factors in the denominator looks like \( \frac{p}{(x-r)} + \frac{q}{(x-s)} \), where p and q are constants, and r and s are the roots of the denominator of our fraction. For \( \frac{11 x -10}{(x-2)(x+1)} \), the roots are 2 and -1.

Step 2 : Fill In the Form

Substituting r and s into the general form gives the decomposition of \( \frac{11 x -10}{(x-2)(x+1)} \) into partial fractions which looks like this: \( \frac{p}{(x-2)} + \frac{q}{(x+1)} \).

Key Concepts

Rational Expressions

Rational expressions are fraction-like equations where both the numerator and the denominator are polynomials. For instance, in the expression \( \frac{11x - 10}{(x - 2)(x + 1)} \) we see a polynomial \( 11x - 10 \) in the numerator and the product of binomials \( (x - 2)(x + 1) \) in the denominator.

Understanding how to work with these expressions is foundational in algebra, as it allows you to simplify complex functions, solve equations, and analyze behaviors of graphs. When solving or simplifying rational expressions, it's essential to look out for values that may make the denominator zero as these would be undefined solutions or restrictions for the variable \( x \).

A key skill with rational expressions is the ability to decompose them into simpler, fractional components, known as partial fractions. This technique is particularly useful when integrating rational functions or finding limits in calculus.

Linear Factors in Denominator

The denominator of a rational expression can often be factorized into linear factors. These are first-degree polynomials that multiply together to give the original denominator. In our example, the denominator \( (x - 2)(x + 1) \) consists of the linear factors \( x - 2 \) and \( x + 1 \).

Linear factors are crucial when decomposing a rational expression into partial fractions. Each distinct linear factor will generally give rise to its own fraction in the decomposition, with the original numerator divided among these fractions.

Identifying Linear Factors

Whenever you're presented with a rational expression, identify and factor the denominator completely when possible. This action will reveal the linear factors upon which the partial fraction decomposition is based. It is these factors that help us rewrite the expression as a sum of simpler terms that are easier to integrate or evaluate.

Why Linear Factors Matter

The presence of linear factors indicates that after decomposition, each partial fraction will have a constant numerator. This is different when compared to situations where you have repeated or quadratic factors, where the numerators may be polynomials.

Constants in Partial Fractions

In the partial fraction decomposition, constants are the numerators in the fractions after the expression has been decomposed. The process breaks down a complex rational expression into a sum of simpler fractions where these constants are yet to be determined.

For the given expression \( \frac{11x - 10}{(x - 2)(x + 1)} \), the decomposition leads to a form like \( \frac{p}{(x - 2)} + \frac{q}{(x + 1)} \) where \( p \) and \( q \) are the constants that need to be found.

These constants are vital because they make sure the sum of these simpler fractions equals the original complex expression. Determining the values of these constants is a process that involves equating coefficients or using values of \( x \) that simplify the equation (commonly known as the method of 'clearing the fractions').

While our exercise doesn't require solving for \( p \) and \( q \) themselves, in actual practice, finding these constants is essential for the decomposition to be used further in calculations, such as anti-differentiation in calculus.