Problem 6
Question
Expand the quotients in Exercises \(1-8\) by partial fractions. $$ \frac{z}{z^{3}-z^{2}-6 z} $$
Step-by-Step Solution
Verified Answer
The partial fraction expansion is \( \frac{1}{z-3} - \frac{1}{z+2} \).
1Step 1: Factor the Denominator
First, we need to factor the denominator of the fraction \( z^3 - z^2 - 6z \). Start by factoring out the greatest common factor, \( z \), which gives us \( z(z^2 - z - 6) \). Next, factor the quadratic \( z^2 - z - 6 \), which can be factored into \( (z - 3)(z + 2) \). Thus, the denominator now is \( z(z - 3)(z + 2) \).
2Step 2: Setup Partial Fraction Decomposition
With the denominator factored, we represent the rational expression as a sum of partial fractions: \( \frac{z}{z(z-3)(z+2)} = \frac{A}{z} + \frac{B}{z-3} + \frac{C}{z+2} \) where \( A \), \( B \), and \( C \) are constants to be determined.
3Step 3: Clear the Denominator
Multiply through by the common denominator \( z(z-3)(z+2) \) to clear the fractions: \( z = A(z-3)(z+2) + Bz(z+2) + Cz(z-3) \).
4Step 4: Expand and Collect Like Terms
Expand the right side: \( Az^2 - A + 2Az - 3Az - 6A + Bz^2 + 2Bz + Cz^2 - 3Cz \). Combine like terms: \( (A + B + C)z^2 + (2A - 3A + 2B - 3C)z - 6A = z \).
5Step 5: Setup Equations to Solve for Constants
Compare coefficients from both sides of the equation:\( A + B + C = 0 \), \( 2A + 2B - 3C = 1 \), and \( -6A = 0 \).
6Step 6: Solve the System of Equations
From \(-6A = 0\), we immediately find that \( A = 0 \). Substitute \( A = 0 \) into the other equations to get \( B + C = 0 \) and \( 2B - 3C = 1 \). Solving these gives \( B = 1 \), \( C = -1 \).
7Step 7: Write the Final Answer
Substitute the found values back into the partial fractions: \( \frac{z}{z^3 - z^2 - 6z} = \frac{0}{z} + \frac{1}{z-3} - \frac{1}{z+2} \). Simplifying, we have \( \frac{z}{z^3 - z^2 - 6z} = \frac{1}{z-3} - \frac{1}{z+2} \).
Key Concepts
Rational ExpressionsFactoringSystem of EquationsQuadratic Functions
Rational Expressions
Rational expressions are fractions where the numerator and the denominator are polynomials. It’s essential to gain a strong understanding of rational expressions, as they form the basis for many algebraic and calculus concepts. The expression given here, \( \frac{z}{z^3 - z^2 - 6z} \), is a classic example of a rational expression.
Managing rational expressions involves operations such as addition, subtraction, multiplication, and division. Moreover, simplifying these expressions often requires factoring the polynomials both in the numerator and the denominator. This simplification helps in understanding the behavior of the expression, including identifying any asymptotes or holes in the corresponding graph.
In our exercise, simplifying the rational expression through partial fraction decomposition allows us to break down the complex expression into simpler, more manageable terms.
Managing rational expressions involves operations such as addition, subtraction, multiplication, and division. Moreover, simplifying these expressions often requires factoring the polynomials both in the numerator and the denominator. This simplification helps in understanding the behavior of the expression, including identifying any asymptotes or holes in the corresponding graph.
In our exercise, simplifying the rational expression through partial fraction decomposition allows us to break down the complex expression into simpler, more manageable terms.
Factoring
Factoring is a crucial skill in dealing with rational expressions and solving algebraic equations. In this exercise, factoring was essential to simplify the denominator \( z^3 - z^2 - 6z \).
The first step was identifying the greatest common factor (GCF) among the terms, which was \( z \). Once \( z \) is factored out, the expression becomes \( z(z^2 - z - 6) \).
Next, we focus on the quadratic \( z^2 - z - 6 \). Quadratic trinomials can usually be factored into two binomials. Here, it factors into \( (z - 3)(z + 2) \).
The entire factorization of the denominator then results in \( z(z - 3)(z + 2) \), making it easier to decompose the rational expression into partial fractions.
The first step was identifying the greatest common factor (GCF) among the terms, which was \( z \). Once \( z \) is factored out, the expression becomes \( z(z^2 - z - 6) \).
Next, we focus on the quadratic \( z^2 - z - 6 \). Quadratic trinomials can usually be factored into two binomials. Here, it factors into \( (z - 3)(z + 2) \).
The entire factorization of the denominator then results in \( z(z - 3)(z + 2) \), making it easier to decompose the rational expression into partial fractions.
System of Equations
Once the rational expression is set up as a partial fraction, you need to determine unknown constants by solving a system of equations. In our case, this involves the setup:
\( \frac{z}{z(z-3)(z+2)} = \frac{A}{z} + \frac{B}{z-3} + \frac{C}{z+2} \).
To find the values of \( A \), \( B \), and \( C \), you clear the denominators and equate coefficients. This results in a system of linear equations. Here, there are three equations:
Understanding how to set up and solve these systems of equations is essential for decomposing and simplifying rational expressions accurately.
\( \frac{z}{z(z-3)(z+2)} = \frac{A}{z} + \frac{B}{z-3} + \frac{C}{z+2} \).
To find the values of \( A \), \( B \), and \( C \), you clear the denominators and equate coefficients. This results in a system of linear equations. Here, there are three equations:
- \( A + B + C = 0 \)
- \( 2A + 2B - 3C = 1 \)
- \( -6A = 0 \)
Understanding how to set up and solve these systems of equations is essential for decomposing and simplifying rational expressions accurately.
Quadratic Functions
Quadratic functions have the general form \( ax^2 + bx + c \). They play a vital role in factoring polynomial expressions, as seen in the exercise.
Quadratics can be factored into two binomials when possible, helping simplify expressions and solve equations efficiently. The part of our exercise that dealt with the expression \( z^2 - z - 6 \) involved identifying its factors, which are \( (z - 3)(z + 2) \).
Recognizing quadratic structures within larger expressions can help streamline problem-solving processes. This understanding is essential for working with polynomials, particularly in calculus and higher-level algebra. Moreover, learning how to factor quadratics aids in solving quadratic equations using methods like completing the square or using the quadratic formula.
Quadratics can be factored into two binomials when possible, helping simplify expressions and solve equations efficiently. The part of our exercise that dealt with the expression \( z^2 - z - 6 \) involved identifying its factors, which are \( (z - 3)(z + 2) \).
Recognizing quadratic structures within larger expressions can help streamline problem-solving processes. This understanding is essential for working with polynomials, particularly in calculus and higher-level algebra. Moreover, learning how to factor quadratics aids in solving quadratic equations using methods like completing the square or using the quadratic formula.
Other exercises in this chapter
Problem 6
Evaluate the integrals in Exercises \(1-14\) $$ \int_{0}^{\pi / 2} 7 \cos ^{7} t d t $$
View solution Problem 6
Evaluate the integrals in Exercises \(1-34\) without using tables. $$ \int_{-8}^{1} \frac{d x}{x^{1 / 3}} $$
View solution Problem 6
Evaluate the integrals. \(\int_{1}^{e} x^{3} \ln x d x\)
View solution Problem 6
Evaluate each integral in Exercises \(1-36\) by using a substitution to reduce it to standard form. $$ \int_{\pi / 4}^{\pi / 3} \frac{\sec ^{2} z}{\tan z} d z $
View solution