Problem 42 - College Algebra

Question

Write the partial fraction decomposition of each rational expression. $$\frac{3 x-5}{x^{3}-1}$$

Step-by-Step Solution

Answer

The factorization of the denominator leads to \((x - 1)(x^2 + x + 1)\). The given rational expression can then be expressed as a sum of partial fractions \(\frac{3 x-5}{x^3 -1} = \frac{A}{x-1} + \frac{Bx + C}{x^2 + x + 1}\). By equating coefficients on both sides and solving the resulting system of linear equations, the final values of A, B, and C are determined. The values would enable completing the fraction decomposition.

Step 1: Factorize the denominator

The denominator of the given rational expression is \(x^3 - 1\). This can be factorized as \((x - 1)(x^2 + x + 1)\) using difference of cubes rule.

Step 2: Write partial fraction decomposition

After we have the factorization of the denominator, the given rational expression can be expressed as a sum of partial fractions \(\frac{3 x-5}{x^3 -1} = \frac{A}{x-1} + \frac{Bx + C}{x^2 + x + 1}\), where A, B, and C are constants to be determined.

Step 3: Solve for A, B and C

Multiply both sides of the equation by \(x^3 - 1\) to clear the denominator: \(3x - 5 = A(x^2 + x + 1) + (Bx + C)(x - 1)\). Simplify and collect like terms. Then, set the coefficients of similar power of x equal to each other on both sides, which will result in a system of linear equations. Solve this system to find values of A, B, and C.

Key Concepts

Factorization

Factorization is an essential process in algebra. It involves breaking down an expression into a product of simpler factors. In our exercise, we have the denominator of a rational expression as \(x^3 - 1\). To find its factors, we use a special rule known as the "difference of cubes." This rule applies to expressions of the form \(a^3 - b^3\) and states that they can be factored into \((a-b)(a^2 + ab + b^2)\). For instance, in our case, we have \(x^3 - 1^3\). Therefore, it can be factorized as \((x - 1)(x^2 + x + 1)\). This process of factorization is crucial for simplifying complex expressions, solving equations, and of course, working with rational expressions like we are here.

Rational Expression

A rational expression is quite simply a fraction that has polynomials in its numerator and denominator. Just like normal fractions, they can be added, subtracted, multiplied, and divided, but they also have unique properties due to their polynomial nature. In this exercise, we have the rational expression \(\frac{3x-5}{x^3-1}\). As with any fraction, simplifying or decomposing into parts can make it easier to work with. This is where partial fraction decomposition comes in. We decompose the fraction into a sum of simpler fractions, making it easier to handle in various mathematical operations. Understanding rational expressions is a key part of algebra and higher-level math due to its broad applications.

Difference of Cubes

The difference of cubes is a specific algebraic identity used to factor certain polynomials. Considered a special factoring technique, it aids in breaking down expressions like \(a^3 - b^3\) into a product of a binomial and a trinomial. The formula is given by:

• \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\)

For our expression \(x^3 - 1\), here \(a = x\) and \(b = 1\). Using the difference of cubes, it becomes \((x-1)(x^2 + x + 1)\). It is always beneficial to remember this identity or pattern since certain polynomials can seem complex until factorized using such rules. This technique simplifies expressions and is particularly useful in calculus and algebra, assisting in integrations and solving equations.

System of Linear Equations

A system of linear equations consists of multiple equations that need to be solved together, finding the values of the variables that satisfy all the equations simultaneously. In the context of our original exercise, after decomposing the rational expression, we expressed it as \(\frac{A}{x-1} + \frac{Bx+C}{x^2+x+1}\). By clearing the denominators and aligning terms, we are left with equations based on comparing coefficients of like-terms.These equations form what we call a system of linear equations. For example, if we had terms involving \(x^2\), \(x\), and constants, we would equate coefficients from both sides for each power of \(x\), resulting in a system that might look like:

• \(Ax^2 + Bx^2 + C = Constant\)
• \(Ax + Bx + C = Constant\)
• \(PreferenceOf x = Constant\)

The goal is to solve this system to find specific values for \(A\), \(B\), and \(C\). Techniques often used are substitution, elimination, or matrix methods. Solving these is fundamental in understanding relationships in algebra and finding more complex solutions.