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Chapter 8: Problem 20

Write the partial fraction decomposition of each rational expression. $$\frac{3}{x^{2}+3 x+2}$$

Short Answer

Expert verified
The partial fraction decomposition of \( \frac{3}{x^{2}+3 x+2} \) is \( \frac{1.5}{x + 1} - \frac{1.5}{x + 2} \).

Step by step solution

01

Factor the denominator

First, factorize the quadratic polynomial in the denominator by finding numbers that multiply to 2 (the constant term) and add to 3 (the coefficient of x), which are 1 and 2. Therefore, \(x^{2}+ 3x + 2\) can be factorized to \( (x + 1)(x + 2) \). This gives us the new expression \( \frac{3}{(x + 1)(x + 2)} \).
02

Express the rational function as the sum of partial fractions

According to the partial fraction decomposition, a rational function can be expressed as the sum of simpler fractional parts. Now, write \( \frac{3}{(x + 1)(x + 2)} \) as \( \frac{A}{x + 1} + \frac{B}{x + 2} \), where A and B are constants to be determined.
03

Solve for A and B

To find the constants A and B, multiply the entire equation \( \frac{3}{(x + 1)(x + 2)} = \frac{A}{x + 1} + \frac{B}{x + 2} \) by \( (x + 1)(x + 2) \) to clear the denominator, getting 3 = A(x + 2) + B(x + 1). Now, choose the convenient values of x to solve for A and B. For example, x = -1 gives A = 1.5 and x = -2 gives B = -1.5. So, the partial fractions are \( \frac{1.5}{x + 1} - \frac{1.5}{x + 2} \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Factorizing Polynomials
Understanding how to factorize polynomials is crucial when working with algebraic expressions, especially in the context of partial fraction decomposition. Factorization is the process of breaking down a complex expression into simpler parts that, when multiplied together, give back the original expression. For instance, when we have a quadratic polynomial like \(x^2 + 3x + 2\), we look for two numbers that multiply to the constant term (here, 2) and add up to the coefficient of the linear term (here, 3).

In our example, these numbers are 1 and 2, which gives us the factors \((x+1)\) and \((x+2)\). This factorization is the first step in partial fraction decomposition and is essential for simplifying rational expressions so that they can be more easily integrated or differentiated if needed.

Why is Factorization Important?

By converting a complex polynomial into a product of simpler binomials or other polynomials, we simplify the complexity of algebraic manipulations. Factorization also helps in finding roots of the polynomial and assists in graphing functions.
Rational Expressions
A rational expression is a fraction where both the numerator and the denominator are polynomials. The expression \(\frac{3}{x^2+3x+2}\) from the given exercise is a perfect example of a rational expression. The key to dealing with rational expressions is understanding their domain (the set of all possible values of x that make the expression valid), simplifying them, and discovering their behavior as the value of x changes.

Simplification often involves factorizing the polynomials in the denominator and the numerator when possible, and then canceling out common factors. This is also the first step before proceeding with partial fraction decomposition, which we use to break down more complex rational expressions into simpler parts. This is particularly useful when you want to perform integration or need to solve equations that involve rational expressions.
Solving Algebraic Equations
Solving algebraic equations is a foundational skill in mathematics. Whether you are finding the constant terms in the partial fraction decomposition or looking for the unknown in a straightforward equation, the key steps involve isolating the variable and making the equation as simple as possible. In the context of finding the values for A and B in our partial fraction, we clear the denominators and obtain an equation without fractions, making it easier to solve for the unknowns.

Choosing strategic values for x to simplify the equation is another helpful technique. For example, if x is set equal to -1, the term involving B drops out, letting you easily solve for A. Similarly, setting x to -2 allows you to solve for B without A interfering. This trick is particularly useful for decomposing rational expressions into partial fractions and showcases a method of choosing values that simplifies the process of finding the solution to an algebraic equation.

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Most popular questions from this chapter

In this set of exercises, you will use the method of solving linear systems using matrices to study real-world problems. A grocery store carries two brands of diapers. For a certain week, the number of boxes of Brand \(A\) diapers sold was 4 more than the number of boxes of Brand B diapers sold. Brand A diapers cost \(\$ 10\) per box and Brand B diapers cost \(\$ 12\) per box. If the total revenue generated that week from the sale of diapers was \(\$ 172,\) how many of each brand did the store sell?

Criminology In 2004 , there were a total of 3.38 million car thefts and burglaries in the United States. The number of burglaries exceeded the number of car thefts by \(906,000 .\) Find the number of burglaries and the number of car thefts.

Involve positive-integer powers of a square matrix \(A . A^{2}\) is defined as the product \(A A ;\) for \(n \geq 3, A^{n}\) is defined as the product \(\left(A^{n-1}\right) A\) Find \(\left(A^{3}\right)^{-1}\) and \(\left(A^{-1}\right)^{3},\) where \(A=\left[\begin{array}{rr}-5 & -1 \\ 4 & 1\end{array}\right] .\) What do you observe?

You wish to make a 1 -pound blend of two types of coffee, Kona and Java. The Kona costs \(\$ 8\) per pound and the Java costs \(\$ 5\) per pound. The blend will sell for \(\$ 7\) per pound. (a) Let \(k\) and \(j\) denote the amounts (in pounds) of Kona and Java, respectively, that go into making a 1 -pound blend. One equation that must be satisfied by \(k\) and \(j\) is $$k+j=1$$ Both \(k\) and \(j\) must be between 0 and \(1 .\) Why? (b) Using the variables \(k\) and \(j\), write an equation that expresses the fact that the total cost of 1 pound of the blend will be \(\$ 7\) (c) Solve the system of equations from parts (a) and (b), and interpret your solution. (d) To make a 1 -pound blend of Kona and Java that costs \(\$ 7.50\) per pound, which type of coffee would you use more of? Explain without solving any equations.

Find \(A^{2}\) (the product \(A A\) ) and \(A^{3}\) (the prod\(\left.u c t\left(A^{2}\right) A\right)\). $$A=\left[\begin{array}{rr}-4 & 0 \\\0 & 3\end{array}\right]$$
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