Problem 8
Question
Write the form of the partial fraction decomposition of the rational expression. It is not necessary to solve for the constants. $$\frac{7 x^{2}-9 x+3}{\left(x^{2}+7\right)^{2}}$$
Step-by-Step Solution
Verified Answer
The partial fraction decomposition of the given rational expression is \( \frac{A}{x^{2} + 7} + \frac{Bx+C}{(x^{2} + 7)^2} \)
1Step 1: Identify the form of the partial fractions
First, identify the type of factors in the denominator. The denominator \( (x^{2}+7)^2 \) is a square of a binomial factor. This corresponds to the pattern of partial fraction \(\frac{A}{x^{2}+7} + \frac{Bx+C}{(x^{2}+7)^2} \). A, B and C are the constants.
2Step 2: Write the partial fraction decomposition
Without finding the values of constants, write the partial fraction decomposition for the given expression. The partial fractions form of \( \frac{7x^{2} - 9x + 3}{(x^{2}+7)^2} \) is \( \frac{A}{x^{2} + 7} + \frac{Bx+C}{(x^{2} + 7)^2} \). This form is easier to integrate and simplify than the original function.
Key Concepts
Understanding Rational ExpressionsExploring Denominator FactorsRole of Constants in Fractions
Understanding Rational Expressions
A rational expression is essentially a fraction where the numerator and denominator are both polynomials. These expressions are vital in algebra because they extend the basic concept of fractions to expressions involving variables.
While a simple fraction might be \( \frac{3}{4} \), a rational expression can look like \( \frac{7x^2 - 9x + 3}{x^2 + 7} \).
While a simple fraction might be \( \frac{3}{4} \), a rational expression can look like \( \frac{7x^2 - 9x + 3}{x^2 + 7} \).
- The numerator is any polynomial, such as \( 7x^2 - 9x + 3 \).
- The denominator is also a polynomial, such as \( (x^2 + 7)^2 \).
- Rational expressions can be analyzed, simplified, and decomposed using algebraic techniques.
Exploring Denominator Factors
Understanding the denominator's factors is crucial for partial fraction decomposition. In our expression\( \frac{7x^2 - 9x + 3}{(x^2 + 7)^2} \), the denominator \( (x^2 + 7)^2 \) is particularly noteworthy.
- The expression \( x^2 + 7 \) is a binomial, and here, it is squared. Such a factor is termed as non-reducible quadratic because it doesn't factor down into linear factors over real numbers.
- For partial fraction decomposition, consider the original factor and its powers separately. In this case, we view both \( x^2 + 7 \) and \( (x^2 + 7)^2 \).
- When denominators involve squares, we account for each power in the decomposition. This approach helps in simplifying and working with the expressions.
Role of Constants in Fractions
In partial fraction decomposition, introducing constants allows us to break down complicated fractions into simpler parts. The goal is to express one large rational expression as a sum of simpler ones.
For the expression \( \frac{7x^2 - 9x + 3}{(x^2 + 7)^2} \), we use constants A, B, and C to set up the decomposition as:\[ \frac{A}{x^2 + 7} + \frac{Bx + C}{(x^2+7)^2} \]
For the expression \( \frac{7x^2 - 9x + 3}{(x^2 + 7)^2} \), we use constants A, B, and C to set up the decomposition as:\[ \frac{A}{x^2 + 7} + \frac{Bx + C}{(x^2+7)^2} \]
- Each constant (A, B, C) represents a number or expression yet to be determined. They are crucial in adapting the decomposition to fit the original expression perfectly when recombined.
- These constants are placeholders we solve for in a full equation, allowing simplification and integration of the expression more easily.
- Identifying and working with these constants is a key step in both solving and understanding the behavior of rational expressions.
Other exercises in this chapter
Problem 7
Solve each system in Exercises \(5-18\). $$ \left\\{\begin{array}{c} 4 x-y+2 z=11 \\ x+2 y-z=-1 \\ 2 x+2 y-3 z=-1 \end{array}\right. $$
View solution Problem 8
In Exercises 5–14, an objective function and a system of linear inequalities representing constraints are given. a. Graph the system of inequalities representin
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In Exercises 1–26, graph each inequality. $$y>3 x+2$$
View solution Problem 8
Solve each system by the substitution method. \(\left\\{\begin{array}{l}2 x-3 y=-13 \\ y=2 x+7\end{array}\right.\)
View solution