Problem 1 - 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. \(\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x+5}+\frac{D}{x-5}\) $$\frac{3 x+2}{x\left(x^{2}-25\right)}$$

Step-by-Step Solution

Answer

The rational expression matches with option d.

Step 1: Identify Denominators

Review the given rational expression \(\frac{3x+2}{x(x^2-25)}\). The denominator can be factored as \(x(x-5)(x+5)\). This indicates simple linear factors \(x-5\) and \(x+5\), as well as a linear factor \(x\).

Step 2: Determine Decomposition Form

From the factorization \(x(x-5)(x+5)\), the partial fraction decomposition will be composed of terms corresponding to \(\frac{A}{x}\), \(\frac{B}{x-5}\), and \(\frac{C}{x+5}\). Thus, the correct form is \(\frac{A}{x} + \frac{B}{x-5} + \frac{C}{x+5}\).

Step 3: Match With Provided Forms

Match the determined form \(\frac{A}{x} + \frac{B}{x-5} + \frac{C}{x+5}\) with the provided options. Check option (d): \(\frac{A}{x} + \frac{B}{x+5} + \frac{C}{x-5}\), which fits the required form.

Key Concepts

Rational Expressions

Rational expressions are a fundamental concept in algebra, involving fractions where the numerator and denominator are polynomials. By understanding rational expressions, you can simplify complex algebraic problems, especially when dealing with calculus or algebraic functions.

• These expressions are typically in the form \( \frac{P(x)}{Q(x)} \), where both \( P(x) \) and \( Q(x) \) are polynomials.
• You must ensure that the denominator \( Q(x) eq 0 \) since division by zero is undefined.

To simplify or manipulate rational expressions, you can often factor them. This process reveals how the expression can be broken into simpler components, making it easier to understand and solve. Identifying these components is crucial when performing operations like addition, subtraction, and in our case, partial fraction decomposition.

Factoring Polynomials

Factoring is a technique used to break down polynomials into simpler components, which are the product of polynomials called factors. Recognizing how to factor polynomials is key in algebra, particularly when dealing with rational expressions.

• A polynomial like \( x^2 - 25 \) can be factored using the difference of squares. The expression becomes \((x - 5)(x + 5)\).
• Another example is \( x^2 - 4x \) which factors into \( x(x - 4) \).

Factoring simplifies expressions and reveals relationships between the numerator and the denominator, crucial for identifying and decomposing rational expressions accurately. It helps determine what terms are present in a possible partial fraction decomposition. When faced with a polynomial, look for patterns such as:
- Difference of squares.- Perfect square trinomials.- Common factors.These patterns simplify breaking down polynomials into manageable parts, paving the way for other processes like decomposition.

Linear Factors

Linear factors are polynomial expressions of the first degree and play a vital role in simplifying rational expressions through partial fraction decomposition. Each linear factor corresponds to a root of the polynomial, which is crucial in breaking down expressions into simpler terms.

• For example, in the expression \( \frac{3x+2}{x(x^2-25)} \), the components \( x-5 \) and \( x+5 \) are the linear factors.
• Identifying these linear factors helps in setting up the partial fraction decomposition.

Linear factors are simple because they represent straight-line graphs and involve no squared or higher-degree terms, which makes them manageable. In the process of partial fraction decomposition, each linear factor corresponds to a term like \( \frac{A}{x} \), where \( A \) is a constant determined through solving the equation. The ability to express a complex rational expression as a sum of simpler fractions is crucial in calculus and other areas of mathematics where integration and differentiation are involved.