Question

Are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. Find the partial fraction decomposition of \(\frac{3 x}{x^{3}-1}\)

Short answer

\( \frac{1}{x - 1} + \frac{-x + 1}{x^2 + x + 1} \)

Step-by-step solution

- Factor the denominator

First, factor the denominator. We need to factorize the cubic polynomial denominator, which is given as \[ x^3 - 1 \]This can be factored using the difference of cubes formula: \[ x^3 - 1 = (x - 1)(x^2 + x + 1) \]

- Setup the partial fraction decomposition

Express the rational function in the form of a partial fraction decomposition: \[ \frac{3x}{x^3 - 1} = \frac{3x}{(x - 1)(x^2 + x + 1)} \]The partial fractions can be written as: \[ \frac{3x}{(x - 1)(x^2 + x + 1)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + x + 1} \]

- Multiply through by the common denominator

Eliminate the denominators by multiplying through by \[ (x - 1)(x^2 + x + 1) \]This gives us: \[ 3x = A(x^2 + x + 1) + (Bx + C)(x - 1) \]

- Distribute and combine like terms

Distribute and simplify the equation: \[ 3x = A(x^2 + x + 1) + Bx(x - 1) + C(x - 1) \]Which expands to: \[ 3x = Ax^2 + Ax + A + Bx^2 - Bx + Cx - C \]Combine like terms: \[ 3x = (A + B)x^2 + (A - B + C)x + (A - C) \]

- Solve for the coefficients

Compare the coefficients from both sides of the equation: \[ 0x^2 + 3x + 0 = (A + B)x^2 + (A - B + C)x + (A - C) \]Matching coefficients gives us the system of equations: 1. \( A + B = 0 \)2. \( A - B + C = 3 \)3. \( A - C = 0 \)From equation 1: \( B = -A \)From equation 3: \( A = C \)Substitute these into equation 2: \[ A - (-A) + A = 3 \]Solve for A: \[ 3A = 3 \ A = 1 \]Therefore, \( A = 1 \), \( B = -1 \), and \( C = 1 \).

- Write the partial fraction decomposition

Substitute the values of A, B, and C back into the partial fractions: \[ \frac{3x}{(x - 1)(x^2 + x + 1)} = \frac{1}{x - 1} + \frac{-x + 1}{x^2 + x + 1} \]Therefore, the partial fraction decomposition is: \[ \frac{1}{x - 1} + \frac{-x + 1}{x^2 + x + 1} \]

Key concepts

algebraic fractions

Algebraic fractions are fractions where the numerator and the denominator are polynomials. Understanding and manipulating these fractions is a key skill in algebra. They are similar to numerical fractions but involve variables. Simplifying algebraic fractions often requires factoring polynomials to cancel out common factors.

For example, in the exercise, the given fraction is \( \frac{3x}{x^3 - 1} \). Here, both the numerator and the denominator are polynomials. The goal is to break this fraction down into simpler parts, known as a partial fraction decomposition.

• This process helps in integrating or differentiating algebraic expressions more easily.
• It’s particularly useful when dealing with rational functions or solving complex algebraic equations.

In our problem, we start by factoring the denominator. Once factored, we can express the given fraction as a sum of simpler fractions.

cubic polynomials

Cubic polynomials are polynomials of degree three, meaning the highest exponent of the variable is three. These polynomials can be expressed in the general form \( ax^3 + bx^2 + cx + d \). In this exercise, the cubic polynomial is \( x^3 - 1 \).

Cubic polynomials can often be factored using special formulas such as the difference of cubes. For example, \( x^3 - 1 \) can be factored as: \[ x^3 - 1 = (x - 1)(x^2 + x + 1) \]

• Factorizing these polynomials is crucial for simplifying algebraic fractions.
• It also helps in solving polynomial equations and finding roots more effectively.

The factorization reveals simpler components of the polynomial, which is critical for decomposition into partial fractions.

difference of cubes

The difference of cubes formula is a handy tool in algebra for factorizing expressions of the form \( a^3 - b^3 \). The formula is expressed as follows: \[a^3 - b^3 = (a - b)(a^2 + ab + b^2) \]

In the exercise, we have \( x^3 - 1 \). Here, \( a = x \) and \( b = 1 \), so using the difference of cubes formula, we factorize it to: \[ x^3 - 1 = (x - 1)(x^2 + x + 1) \]

• Understanding this formula simplifies the process of breaking down and handling cubic polynomials.
• It is particularly useful in partial fraction decomposition to get simpler fractions that can be added or integrated easily.

Applying this factorization in our problem allows us to set up our partial fractions and move forward with solving the decomposition.

system of equations

A system of equations is a set of equations with multiple variables that you solve simultaneously. These systems are crucial in finding the unknown coefficients in partial fraction decomposition.

In the exercise, we derived the system of equations:

• 1. \( A + B = 0 \)
• 2. \( A - B + C = 3 \)
• 3. \( A - C = 0 \)

By solving these equations step-by-step, we determine the values for A, B, and C. This involves:

• Substituting one equation into another to reduce the number of variables.
• Solving for one variable and then back-substituting to find the others.

For example, from equation 1, we get \( B = -A \). Substituting A and B in equation 2 helps us find C. Ultimately, solving these equations gives us the coefficients needed for the partial fractions, completing the decomposition process.