Problem 2 - Precalculus: Mathematics for Calculus

Question

Choose from (i)-(iv) the appropriate form for its partial fraction decomposition. $$r(x)=\frac{2 x+8}{(x-1)\left(x^{2}+4\right)}$$ (i) \(\frac{A}{x-1}+\frac{B}{x^{2}+4}\) (ii) \(\frac{A}{x-1}+\frac{B x+C}{x^{2}+4}\) (iii) \(\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{x^{2}+4}\) (iv) \(\frac{A x+B}{x-1}+\frac{C x+D}{x^{2}+4}\)

Step-by-Step Solution

Answer

Option (ii): \(\frac{A}{x-1}+\frac{Bx+C}{x^{2}+4}\).

Step 1: Recognize the Denominator Composition

Identify the structure of the denominator in the given fraction. The denominator is \((x-1)(x^2+4)\), which consists of one linear factor \(x-1\) and one irreducible quadratic factor \(x^2+4\).

Step 2: Choose Partial Fraction Decomposition Form

Based on the structure of the denominator, determine the appropriate form from the given choices. A linear factor \((x-1)\) has a numerator of the form \(A\). An irreducible quadratic factor \((x^2 + 4)\) has a numerator of the form \(Bx + C\) because of its quadratic nature. Therefore, the correct form is \(\frac{A}{x-1} + \frac{Bx + C}{x^2+4}\), which corresponds to option (ii).

Step 3: Confirm Choice Against Options

Compare your selected form \(\frac{A}{x-1} + \frac{Bx + C}{x^2+4}\) to the provided options: (i) has the wrong numerator structure for the quadratic term, (iii) adds an unnecessary term \(\frac{B}{x+2}\), and (iv) incorrectly assigns numerators for the factors. Thus, option (ii) is the correct choice.

Key Concepts

linear factor

The term "linear factor" in partial fraction decomposition refers to a factor in the denominator of the form \(x - r\), where \(r\) is a constant. This means the variable \(x\) in the factor is raised to the power of one. Linear factors are straightforward because their numerators are simple constants. For the rational function \(\frac{2x+8}{(x-1)(x^2+4)}\), the linear factor is \(x-1\). This factor dictates a numerator in the partial fraction form to be a constant \(A\). In other words, when dealing with linear factors in partial fractions, we use numerators like \(\frac{A}{x-1}\). This ensures that each part of the decomposition can separately handle simpler factors without complexity.

• Linear factors are always of the form \(x - r\).
• The numerator for a linear factor is a constant, labeled \(A\) in this case.
• Recognizing linear factors aids in correctly setting up partial fractions.

irreducible quadratic factor

An irreducible quadratic factor is a quadratic expression that cannot be factored into real linear factors. Such factors remain as quadratics over the real numbers. Within the denominator of the function \(\frac{2x+8}{(x-1)(x^2+4)}\), \(x^2+4\) is an irreducible quadratic factor because it cannot be broken down into linear components with real coefficients.With irreducible quadratic factors, the numerator of the corresponding term in the partial fraction decomposition must be a linear expression. This is represented as \(Bx + C\). The additional degrees in the numerator (compared to a simple constant) allow for flexibility needed to capture all behavior of the original fraction.

• Irreducible quadratic factors remain as such over the real number set.
• Examples include expressions like \(x^2 + 4\), which can't be simplified further using real numbers.
• Numerators over these factors are of the form \(Bx + C\) to accommodate their complexity.

numerator structure

The choice of the numerator structure in partial fraction decomposition is crucial, as it must reflect the type of factor it is associated with. The guideline is simple:- If the denominator factor is linear, the numerator is a constant like \(A\).- If the denominator factor is irreducible quadratic, the numerator takes the form \(Bx + C\).In the given problem, the function \(\frac{2x+8}{(x-1)(x^2+4)}\) was dissected based on these rules. The linear factor, \(x-1\), received a constant numerator \(A\). On the other hand, the irreducible quadratic factor, \(x^2+4\), needed a linear numerator, \(Bx + C\). This correspondence provides a simple but powerful strategy to correctly configure the decomposition according to the nature of each factor. Such precise numerator choices ensure that when added together, the partial fractions can reconstruct the original rational function.

• The structure of the numerator changes depending on the associated factor.
• Linear factors pair with constant numerators, while irreducible quadratics use linear successions \(Bx + C\).
• Correct setting of numerators allows accurate decompositions and problem solving.