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 the given rational expression is \(A/(x-1) + B/(x+3)\) where A and B are constants.
1Step 1: Assign Variables for Constants
Assign variables, say A and B, for the constants in the numerators of the simpler fractions that are to be found.
2Step 2: Writing the Form of the Partial Fractions
The expression can be rewritten in the form \(A/(x - 1) + B/(x + 3)\), where A and B are constants.
Key Concepts
Rational ExpressionConstantsNumeratorsPartial Fractions
Rational Expression
A rational expression is simply a fraction where both the numerator and the denominator are polynomials. The expression \(\frac{5x+7}{(x-1)(x+3)}\) is an example of a rational expression. Here, you have:
- A polynomial in the numerator: \(5x + 7\).
- A polynomial in the denominator: \((x - 1)(x + 3)\).
Constants
Constants in mathematical expressions are values that do not change. In the process of partial fraction decomposition, you will often assign constants to unknown values. For the expression \(\frac{5x+7}{(x-1)(x+3)}\), we use constants like \(A\) and \(B\) to stand in for numbers we have not yet calculated.
By assigning these constants, we can express a complex rational expression as a sum of simpler fractions. The beauty of constants is that they provide placeholders, allowing us to work through algebraic manipulations without needing to know their exact values right away.
By assigning these constants, we can express a complex rational expression as a sum of simpler fractions. The beauty of constants is that they provide placeholders, allowing us to work through algebraic manipulations without needing to know their exact values right away.
- In this exercise, the goal is not to find the values of \(A\) and \(B\), but to set up the necessary form for solving.
- This involves assuming \(A\) and \(B\) will have specific values that, when solved, would satisfy the original expression.
Numerators
The numerators are the top parts of fractions. In the partial fraction decomposition of the rational expression, each fraction has its own numerator.
In the exercise provided, the rational expression \(\frac{5x+7}{(x-1)(x+3)}\) is rewritten as two separate fractions: \(\frac{A}{x-1}\) and \(\frac{B}{x+3}\).
In the exercise provided, the rational expression \(\frac{5x+7}{(x-1)(x+3)}\) is rewritten as two separate fractions: \(\frac{A}{x-1}\) and \(\frac{B}{x+3}\).
- Here, \(A\) and \(B\) are crucial as they become the numerators of each partial fraction.
- The numerators \(A\) and \(B\) help us express how each fraction can add up to reconstruct the original rational expression.
Partial Fractions
Partial fraction decomposition is a method used to break down complex rational expressions into simpler, more manageable pieces called partial fractions. This technique greatly simplifies operations like integration or algebraic manipulation.
- For the rational expression \(\frac{5x+7}{(x-1)(x+3)}\), decomposing it into its partial fractions means writing it as \(\frac{A}{x-1} + \frac{B}{x+3}\).
- The goal of partial fraction decomposition is to make a complicated rational expression easier to handle.
Other exercises in this chapter
Problem 2
Graph each inequality. $$3 x-6 y \leq 12$$
View solution Problem 2
Determine if the given ordered triple is a solution of the system. $$ \begin{aligned} &(5,-3,-2)\\\ &\left\\{\begin{aligned} x+y+z &=0 \\ x+2 y-3 z &=5 \\ 3 x+4
View solution Problem 2
Determine whether the given ordered pair is a solution of the system. \((-3,5)\) \(\left\\{\begin{array}{l}{9 x+7 y=8} \\ {8 x-9 y=-69}\end{array}\right.\)
View solution Problem 3
Solve each system by the substitution method. $$\left\\{\begin{array}{l}{x+y=2} \\\\{y=x^{2}-4 x+4}\end{array}\right.$$
View solution