Math Problem Statement
Solve
Solution
The problem is asking how many second-type partial fractions will be present in the formal decomposition of the expression:
Step 1: Factor the Denominator
To perform partial fraction decomposition, we first factor the denominator .
- Factor by grouping:
- Factor as a difference of squares:
- Thus, the denominator factors as:
Step 2: Set Up Partial Fraction Decomposition
The partial fraction decomposition will involve terms based on the factors of the denominator. For each linear factor and , we set up terms as follows:
- For the factor : a single fraction term .
- For the repeated factor : two terms are required, and .
Thus, the decomposition is:
Step 3: Identify Second-Type Partial Fractions
In this context, "second-type partial fractions" refers to fractions with terms of higher powers in the denominator. Here, is the only term with a higher power (specifically, ).
Conclusion
There is 1 second-type partial fraction in the decomposition.
Would you like a step-by-step breakdown of how to solve for each coefficient (A, B, and C)?
Here are five related questions:
- How do we solve for the coefficients , , and in partial fractions?
- What are the types of partial fraction decompositions for different polynomial degrees in the denominator?
- How would the decomposition change if the denominator included a quadratic irreducible factor?
- What are the general rules for setting up partial fractions with repeated factors in the denominator?
- Can partial fractions be used in integration, and how does the decomposition simplify the process?
Tip: When factoring the denominator, always look for common patterns such as grouping or difference of squares to simplify the process.
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Math Problem Analysis
Mathematical Concepts
Partial Fractions
Factorization
Polynomial Division
Formulas
Partial Fraction Decomposition Formula
Theorems
Partial Fractions Theorem
Suitable Grade Level
Grades 10-12
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