Problem 2
Question
Write the form of the partial fraction decomposition of the rational expression. It is not necessary to solve for the constants. $$\frac{5 x+7}{(x-1)(x+3)}$$
Step-by-Step Solution
Verified Answer
The form of the partial fraction decomposition of \(\frac{5x+7}{(x-1)(x+3)}\) is \[ \frac{A}{x-1} + \frac{B}{x+3} \] where A and B are constants.
1Step 1: Identify the Denominator Factors
The denominator of the given function is \((x-1)(x+3)\). These are the factors of the denominator. Each of these factors will be the denominator of a simpler rational function in the partial fraction decomposition.
2Step 2: Write the General Form of Partial Fraction Decomposition
The general form of a partial fraction decomposition, for this case, is:\[ \frac{A}{x-1} + \frac{B}{x+3} \]where A and B are constants.
3Step 3: The Result
Our result is, therefore, the general form of the partial fraction decomposition. Note that the constants A and B are not specified in this case because the exercise only asks for the form of the decomposition, not the specific values of the constants.
Key Concepts
AlgebraRational ExpressionsPolynomial Functions
Algebra
Algebra sits at the heart of mathematical problem solving, allowing us to represent numbers and their relationships in abstract ways. One of the core tools of algebra is the manipulation of equations to find unknown values.
In the context of partial fraction decomposition, algebra is used to break down complex rational expressions into simpler fragments. Each fragment corresponds to fractions where the numerators, represented by constants like A and B in our example, are pieces of a puzzle waiting to be solved. You'll recognize that the process of finding these constants involves setting up and solving linear equations, an essential skill in algebra.
Part of the beauty of algebra is in how it systematically deconstructs and reconstructs understanding of how quantities relate to each other. So, when you're faced with a rational expression like the one in our exercise, algebra guides you step by step to find clarity and simplicity from complexity.
In the context of partial fraction decomposition, algebra is used to break down complex rational expressions into simpler fragments. Each fragment corresponds to fractions where the numerators, represented by constants like A and B in our example, are pieces of a puzzle waiting to be solved. You'll recognize that the process of finding these constants involves setting up and solving linear equations, an essential skill in algebra.
Part of the beauty of algebra is in how it systematically deconstructs and reconstructs understanding of how quantities relate to each other. So, when you're faced with a rational expression like the one in our exercise, algebra guides you step by step to find clarity and simplicity from complexity.
Rational Expressions
Rational expressions are much like fractions, but instead of integers, you have polynomials in the numerator, the denominator, or both. Understanding how to work with these expressions is critical in algebra, opening the door to solving many types of problems.
Partial fraction decomposition is a powerful tool used to work with rational expressions, especially when integrating or finding limits. It's a way of expressing a complex rational expression as a sum of simpler ones. The given exercise breaks down the expression into \(A/(x-1) + B/(x+3)\), which are fractions with linear denominators.
Recognizing how to rewrite complex expressions as a sum of simpler ones does not just make calculations easier; it also enhances comprehension by showing the structure hidden within the rational expression. For students, mastering rational expressions builds a strong foundation for calculus and other advanced topics in mathematics.
Partial fraction decomposition is a powerful tool used to work with rational expressions, especially when integrating or finding limits. It's a way of expressing a complex rational expression as a sum of simpler ones. The given exercise breaks down the expression into \(A/(x-1) + B/(x+3)\), which are fractions with linear denominators.
Recognizing how to rewrite complex expressions as a sum of simpler ones does not just make calculations easier; it also enhances comprehension by showing the structure hidden within the rational expression. For students, mastering rational expressions builds a strong foundation for calculus and other advanced topics in mathematics.
Polynomial Functions
Polynomial functions are sums of terms consisting of variables raised to whole number exponents and their coefficients. They are ubiquitous in mathematics, from simple linear functions like \(f(x) = mx + b\) to more complex quadratic equations like \(f(x) = ax^2 + bx + c\).
In the context of our textbook exercise, the denominator of our rational expression consists of two binomials \(x-1\) and \(x+3\), which are, in fact, first-degree polynomial functions. Polynomial functions often play the role of building blocks in algebraic expressions and equations. With partial fraction decomposition, we leverage the predictable behavior of polynomials to simplify the integration or differentiation of complex rational expressions.
It's essential to appreciate how polynomial functions interact when they form a part of a rational expression. When factoring or expanding these polynomials, you unlock the ability to manipulate and simplify expressions in ways that can reveal solutions to the problems at hand.
In the context of our textbook exercise, the denominator of our rational expression consists of two binomials \(x-1\) and \(x+3\), which are, in fact, first-degree polynomial functions. Polynomial functions often play the role of building blocks in algebraic expressions and equations. With partial fraction decomposition, we leverage the predictable behavior of polynomials to simplify the integration or differentiation of complex rational expressions.
It's essential to appreciate how polynomial functions interact when they form a part of a rational expression. When factoring or expanding these polynomials, you unlock the ability to manipulate and simplify expressions in ways that can reveal solutions to the problems at hand.
Other exercises in this chapter
Problem 1
Determine if the given ordered triple is a solution of the system. $$\begin{array}{rr}x+y+z= & 4 \\\x-2 y-z= & 1 \\\2 x-y-2 z= & -1\\\\(2,-1,3) &\end{array}$$
View solution Problem 2
In Exercises \(1-4,\) determine whether the given ordered pair is a solution of the system. $$ \begin{aligned} &(-3,5)\\\ &9 x+7 y=8\\\ &8 x-9 y=-69 \end{aligne
View solution Problem 2
Graph each inequality. $$ 3 x-6 y \leq 12 $$
View solution Problem 2
Solve each system by the substitution method. $$\begin{aligned} &x-y=-1\\\ &y=x^{2}+1 \end{aligned}$$
View solution