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Magazine Chapter 8: Problem 47
write the partial fraction decomposition of each rational expression. $$\frac{1}{x^{2}-c^{2}} \quad(c \neq 0)$$
Short Answer
Expert verified
\(\frac{1}{{x^{2}-c^{2}}} = \frac{\frac{1}{2c}}{{x+c}} - \frac{\frac{1}{2c}}{{x-c}}\)
Step by step solution
01
Factor the Denominator
The denominator of the given fraction is a difference of squares, which can be factored using the identity \(a^2 - b^2 = (a+b)(a-b)\). Hence, the denominator can be factored as \(x^2 - c^{2} = (x+c)(x-c)\). So, now we have rewritten the fraction \(\frac{1}{{x^{2}-c^{2}}}\) as \(\frac{1}{{(x+c)(x-c)}}\).
02
Set up the Partial Fractions
The form of the partial fractions will depend on the factors in the denominator. Here, we have a product of two different linear binomials. Therefore, the partial fraction decomposition will have the form \(\frac{1}{{(x+c)(x-c)}} = \frac{A}{{x+c}} + \frac{B}{{x-c}}\), where A and B are constants that we need to find.
03
Find the Values of A and B
To find A and B, we can equate the right-hand side to the left-hand side of the equation and then equate the coefficients of like terms on both sides. First, we clear the denominator on the right side by multiplying every term by \((x+c)(x-c)\) to get \(1 = A(x-c) + B(x+c)\). Choosing convenient values for x can make things simpler. Let's choose \(x=c\) and \(x=-c\). From \(x=c\), we get \(1 = 2Ac \Rightarrow A = \frac{1}{2c}\), and from \(x=-c\), we get \(1 = -2Bc \Rightarrow B = -\frac{1}{2c}\).
04
Write the Final Partial Fraction Decomposition
Our solution from step 3 gives us A and B values. We can substitute these back into our decomposition \(\frac{A}{{x+c}} + \frac{B}{{x-c}}\) to get the final partial fraction decomposition: \(\frac{1}{{(x+c)(x-c)}} = \frac{\frac{1}{2c}}{{x+c}} - \frac{\frac{1}{2c}}{{x-c}}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Rational Expression
A rational expression is a fraction where both the numerator and the denominator are polynomials. In our given problem, the rational expression is \( \frac{1}{x^{2}-c^{2}} \). This expression has a polynomial numerator, which in this case is simply 1, and a polynomial denominator \( x^2 - c^2 \).
Rational expressions can sometimes be simplified or decomposed into simpler parts, helping in resolving complex algebraic equations. This simplification process is called partial fraction decomposition, which breaks a complex fraction into simpler fractions that are easier to work with.
Rational expressions can sometimes be simplified or decomposed into simpler parts, helping in resolving complex algebraic equations. This simplification process is called partial fraction decomposition, which breaks a complex fraction into simpler fractions that are easier to work with.
Difference of Squares
The difference of squares is a common expression in algebra, characterized by the formula \( a^2 - b^2 = (a+b)(a-b) \). This formula is used to factor polynomials that are the difference of two perfect squares.
In the expression \( x^2 - c^2 \), we observe a difference of squares where \( a = x \) and \( b = c \). Using the identity, we rewrite the denominator as \( (x+c)(x-c) \). This factorization is crucial in setting up the partial fraction decomposition for the rational expression provided in the exercise.
In the expression \( x^2 - c^2 \), we observe a difference of squares where \( a = x \) and \( b = c \). Using the identity, we rewrite the denominator as \( (x+c)(x-c) \). This factorization is crucial in setting up the partial fraction decomposition for the rational expression provided in the exercise.
Linear Binomials
Linear binomials are algebraic expressions consisting of two terms, such as \( ax + b \). Linear binomials play a significant role in partial fractions and factorization.
In our example, from the factorization of \( x^2 - c^2 \), we have two linear binomials: \( (x+c) \) and \( (x-c) \). These expressions represent the factors of the original quadratic expression. Each of these binomials will be part of the decomposition setup, corresponding to separate fractions with unknown constants to be determined.
In our example, from the factorization of \( x^2 - c^2 \), we have two linear binomials: \( (x+c) \) and \( (x-c) \). These expressions represent the factors of the original quadratic expression. Each of these binomials will be part of the decomposition setup, corresponding to separate fractions with unknown constants to be determined.
- Understand that each factor represents a distinct behavior in the expression's graph.
- Linear binomials depict simple linear relationships that can be independently analyzed.
Constants A and B
In partial fraction decomposition, the expression is rewritten as a sum of simpler fractions. With the factors \( (x+c) \) and \( (x-c) \), we arrange the expression as \( \frac{A}{{x+c}} + \frac{B}{{x-c}} \). Here, \( A \) and \( B \) are constants that need to be found.
To determine these constants, you can use algebraic techniques like equating coefficients or simply substituting convenient values for the variable \( x \). In this solution, specific values for \( x \) were chosen, such as \( x=c \) and \( x=-c \), making it simple to calculate \( A = \frac{1}{2c} \) and \( B = -\frac{1}{2c} \).
To determine these constants, you can use algebraic techniques like equating coefficients or simply substituting convenient values for the variable \( x \). In this solution, specific values for \( x \) were chosen, such as \( x=c \) and \( x=-c \), making it simple to calculate \( A = \frac{1}{2c} \) and \( B = -\frac{1}{2c} \).
- Constants simplify the representation of the fraction.
- Once determined, these constants allow us to express the rational expression in its decomposed form.
- The correct values for \( A \) and \( B \) ensure that the original expression is accurately represented by the partial fractions.
