Problem 16
Question
Write the partial fraction decomposition of each rational expression. $$\frac{x}{x^{2}+2 x-3}$$
Step-by-Step Solution
Verified Answer
The partial fraction decomposition of the given rational function is \(\frac{1}{x-1} - \frac{1}{x+3}\)
1Step 1: Factorize the denominator
The denominator \(x^{2}+2x-3\) can be expressed as the product of two binomials, \(x+3\) and \(x-1\). Therefore, the equation becomes \(\frac{x}{(x+3)(x-1)}\)
2Step 2: Decompose into simple fractions
The aim is to express the rational function in the form of \(\frac{A}{x+3} + \frac{B}{x-1}\). Multiply both sides by the common denominator \((x+3)(x-1)\) to get rid of the fractions. Thus the equation becomes \(x = A(x-1) + B(x+3)\)
3Step 3: Solving for A and B
Equation \(x=A(x-1)+B(x+3)\) is a polynomial equation, where for any value of x, both sides should be equal, hence you can put various values of x to get equations in terms of A and B. By choosing values for x that will eliminate either A or B, we can get two equations. Put x=1, we get \(A=1\) and put x=-3, we get \(B=-1\). Thus, the partial fraction decomposition of the given rational function is \(\frac{1}{x-1} - \frac{1}{x+3}\)
Key Concepts
Rational ExpressionFactoring PolynomialsSolving Linear EquationsBinomial Products
Rational Expression
A rational expression is a fraction where both the numerator and the denominator are polynomials. Just like regular fractions, rational expressions can be simplified or transformed to make them easier to work with. In this problem, we start with the rational expression \( \frac{x}{x^2 + 2x - 3} \). This expression has a polynomial of higher degree in the denominator compared to the numerator. To simplify such expressions further, we often use a technique called "partial fraction decomposition." This involves breaking the fraction into simpler parts that are easier to manage and understand. Rational expressions can tell us about the behavior of a function, including its discontinuities or points where the expression is undefined.
Factoring Polynomials
Factoring polynomials is a key step in simplifying rational expressions. By breaking down the polynomial into simpler, multiplied factors, you can often find simpler expressions for analysis. In this solution, the denominator \( x^2 + 2x - 3 \) is factored into two binomial parts \((x+3)(x-1)\).
To factor a quadratic polynomial like this, you need to find two numbers that multiply to the constant term (-3) and add up to the coefficient of the linear term (2). Here, those numbers are 3 and -1.
To factor a quadratic polynomial like this, you need to find two numbers that multiply to the constant term (-3) and add up to the coefficient of the linear term (2). Here, those numbers are 3 and -1.
- This method is straightforward for quadratic polynomials and serves as a foundation for more complex splittings in mathematical expressions.
- Each factor corresponds to potential values that make the denominator zero, revealing important characteristics of the original expression such as holes or vertical asymptotes on a graph.
Solving Linear Equations
Solving linear equations is integral in finding the partial fraction decomposition. Here, to determine the constants \( A \) and \( B \) in the decomposition \( \frac{A}{x+3} + \frac{B}{x-1} \), we set up an equation with these unknowns after clearing denominators: \( x = A(x-1) + B(x+3) \).
By substituting suitable values of \( x \) that simplify the equation, we isolate and solve for the constants.
By substituting suitable values of \( x \) that simplify the equation, we isolate and solve for the constants.
- When \( x = 1 \), the equation simplifies such that \( A \) is sole and equal to 1.
- When \( x = -3 \), \( B \) becomes isolated and equals -1.
Binomial Products
Binomial products occur when two binomials are multiplied. Understanding how these work is critical in both decomposition and factoring. In this exercise, the polynomial \( x^2 + 2x - 3 \) is the result of the binomial product \( (x+3)(x-1) \).
Binomial multiplication uses the distributive property, which involves multiplying each term in one binomial by every term in the other.
Binomial multiplication uses the distributive property, which involves multiplying each term in one binomial by every term in the other.
- You get terms like \( x \cdot x = x^2 \), \( x \cdot (-1) = -x \), and similar calculations.
- Combining like terms helps to achieve the expanded form \( x^2 + 2x - 3 \).
Other exercises in this chapter
Problem 15
Solve each system in Exercises \(5-18\). $$ \left\\{\begin{array}{ccc} x+y & = & -4 \\ y-z & = & 1 \\ 2 x+y+3 z & = & -21 \end{array}\right. $$
View solution Problem 16
a. A student earns \(\$ 10\) per hour for tutoring and \(\$ 7\) per hour as a teacher's aide. Let \(x=\) the number of hours each week spent tutoring and let \(
View solution Problem 16
In Exercises 1–26, graph each inequality. $$x^{2}+y^{2}>36$$
View solution Problem 16
Solve each system in Exercises \(5-18\). $$ \left\\{\begin{array}{l} x+y=4 \\ x+z=4 \\ y+z=4 \end{array}\right. $$
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