Problem 3 - Precalculus

Question

Match the rational expression \((1-6)\) with the form of the partial-fraction decomposition \((a-f)\). a. \(\frac{A}{x}+\frac{B}{x^{2}}+\frac{C x+D}{x^{2}+25}\) b. \(\frac{A}{x}+\frac{B x+C}{x^{2}+25}+\frac{D x+E}{\left(x^{2}+25\right)^{2}}\) c. \(\frac{A}{x}+\frac{B x+C}{x^{2}+25}\) d. \(\frac{A}{x}+\frac{B}{x+5}+\frac{C}{x-5}\) e. \(\frac{A}{x}+\frac{B}{x^{2}}+\frac{C x+D}{x^{2}+25}+\frac{E x+F}{\left(x^{2}+25\right)^{2}}\) f. \(\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x+5}+\frac{D}{x-5}\) $$ \frac{3 x+2}{x^{2}\left(x^{2}+25\right)} $$

Step-by-Step Solution

Answer

The given expression matches option a: \( \frac{A}{x} + \frac{B}{x^2} + \frac{Cx + D}{x^2 + 25} \).

Step 1: Identify the Components

Start by analyzing the rational expression \( \frac{3x+2}{x^2(x^2+25)} \). The denominator \( x^2(x^2+25) \) indicates that the expression has repeated linear factors and an irreducible quadratic factor. Specifically, it contains \( x^2 \) and \( x^2 + 25 \).

Step 2: Determine Partial Fraction Form

To decompose \( \frac{3x+2}{x^2(x^2+25)} \) using partial fractions, consider the structure required for each factor. For \( x^2 \), use terms of the form \( \frac{A}{x} + \frac{B}{x^2} \). For the irreducible quadratic \( x^2 + 25 \), include a term \( \frac{Cx + D}{x^2 + 25} \).

Step 3: Match with Options

Compare the structured decomposition \( \frac{A}{x} + \frac{B}{x^2} + \frac{Cx + D}{x^2 + 25} \) against the given options:- Option a: \( \frac{A}{x} + \frac{B}{x^2} + \frac{Cx + D}{x^2+25} \).- Option e also contains these components but includes additional terms that are unnecessary for this expression.Option a directly matches our partial fraction form.

Step 4: Verification

Verify by simplifying option a: \( \frac{A}{x} + \frac{B}{x^2} + \frac{Cx + D}{x^2 + 25} \). This matches the components in the original expression's denominator perfectly, confirming that the set up is correct for a complete decomposition involving all necessary parts.

Key Concepts

Rational Expressions

Rational expressions are fractions where the numerator and the denominator are polynomials. These expressions are crucial when it comes to parsing out different mathematical components in algebra and calculus. For a rational expression to be valid, the denominator cannot be zero, as division by zero is undefined.
One common problem solved using rational expressions is partial fraction decomposition, where a complex rational expression is split into simpler fractions. This enables easier integration, simplification, or solving of equations. Partial fraction decomposition only works when the degree of the numerator is less than the degree of the denominator. If not, polynomial division first converts it into a suitable form.
When looking at a rational expression like \( \frac{3x+2}{x^2(x^2+25)} \), recognizing the factors in its denominator helps determine the composition and suitability for partial fraction decomposition. The given example highlights both repeated and quadratic factors, showcasing common forms that arise within rational expressions.

Irreducible Quadratic Factors

Irreducible quadratic factors are polynomial components in the denominator that cannot be factored further using real numbers. They take the form \( ax^2 + bx + c \) where the discriminant \( b^2 - 4ac < 0 \). This condition indicates that there are no real roots, making the quadratic "irreducible."Utilizing irreducible quadratic factors in partial fraction decomposition involves expressing them with terms like \( \frac{Cx + D}{x^2 + 25} \). Unlike linear terms, these require both a linear numerator to account for the lack of simpler real solutions. This process introduces extra complexity, particularly in solving or integrating these expressions.
In the exercise \( \frac{3x+2}{x^2(x^2+25)} \), the factor \( x^2 + 25 \) is irreducible over the real numbers, exemplifying when these terms become a vital part of decomposition.

Linear Factors

Linear factors in polynomials are terms of the form \( x + a \). They are called linear because the power of \( x \) is one. Unlike irreducible quadratics, linear factors can be solved and factored easily, often representing the zeros of the polynomial.For rational expressions, partial fraction decomposition assigns fractions to each linear factor. For example, if you have \( \frac{1}{x^2} \), the decomposition will involve terms such as \( \frac{A}{x} \) and \( \frac{B}{x^2} \) due to its repeated nature. Linear factors allow the polynomial to be broken into simpler, more manageable parts.
In the problem \( \frac{3x+2}{x^2(x^2+25)} \), \( x^2 \) is a repeated linear factor, indicating that separate terms \( \frac{A}{x} \) and \( \frac{B}{x^2} \) need to be incorporated in the decomposition. Understanding these factors helps in precise and accurate mathematical manipulation.