Problem 42 - College Algebra Essentials

Question

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

Step-by-Step Solution

Answer

The partial fraction decomposition is \(\frac{3 x-5}{x^{3}-1} = \frac{A}{x-1} + \frac{Bx + C}{x^2 + x + 1}\). The constants A, B and C need to be calculated by matching coefficients of respective powers of x on both sides of the equation.

Step 1: Factorize the denominator

The given rational function is \(\frac{3 x-5}{x^{3}-1}\). The denominator \(x^3 - 1\) can be factored using difference of cubes formula as \(x^3 - 1 = (x-1)(x^2 + x + 1)\). So, our rational function becomes: \(\frac{3 x-5}{(x-1)(x^2 + x + 1)}\)

Step 2: Decompose into partial fractions

We can rewrite this equation as a sum of two partial fractions:\(\frac{3 x-5}{(x-1)(x^2 + x + 1)} = \frac{A}{x-1} + \frac{Bx + C}{x^2 + x + 1}\) Now, to solve for the constants A, B and C, we multiply throughout by the common denominator to get rid of the fractions.

Step 3: Solve for constants

Doing as described in Step 2, we get, \(3x-5=A(x^2 + x + 1) + (Bx + C)(x-1)\). Now, by matching coefficients of respective powers of x on both sides of this equation, we can create a system of linear equations and solve for A, B and C.

Key Concepts

Rational Expressions

Rational expressions are mathematical expressions that represent the division of one polynomial by another. They appear frequently in algebra and can sometimes be quite complex. For example, in the expression \[ \frac{3x-5}{x^3-1} \]the numerator is a polynomial \(3x - 5\), and the denominator is another polynomial \(x^3 - 1\).

• The numerator and denominator are both essential to the rational expression.
• Rational expressions can be simplified, much like fractions, by factoring and reducing common factors.

Often, the goal with rational expressions is to simplify them or to perform operations such as addition, subtraction, multiplication, and division. Simplifying these expressions is the key to working smoothly with them in solving various algebraic problems. The partial fraction decomposition is a method used to break down a complex rational function into simpler parts, making it easier to analyze and integrate.

Difference of Cubes

When dealing with polynomials, one useful method is identifying and applying special factorization formulas. The difference of cubes is among these. It allows us to factor expressions of the form\[ x^3 - a^3 \]into \[ (x-a)(x^2 + ax + a^2) \].In our original problem, we have \(x^3 - 1\), which fits this formula since 1 is a cube (i.e., \(1 = 1^3\)).

• The expression \( x^3 - 1 \) is split into the product \((x - 1)(x^2 + x + 1)\).
• This factorization is critical for breaking down the rational expression into partial fractions.
• Being aware of such special formulas is incredibly helpful in math. They simplify calculations significantly.

Linear Equations

Linear equations are equations in which the highest power of the variable is one. They form the backbone of algebra and include formats like \[ ax + b = 0 \].In the context of partial fraction decomposition, after substituting presumed partial fractions into a rational expression and clearing denominators, you often get a system of linear equations.

• This system is crucial for determining unknown coefficients.
• Solving it will provide the values necessary to complete the partial fraction process.
• Techniques such as substitution or elimination can be used to find solutions.

For example, when decomposing complex expressions, you equate terms separately to form equations, giving you a linear equation system to solve for the constants.

Factoring Polynomials

Factoring polynomials is the process of breaking down a polynomial into simpler components that, when multiplied together, return the original polynomial. This skill is fundamental in algebra.When faced with a polynomial like \[ x^3 - 1\], finding its factors is the first step toward simplifying expressions and solving equations.

• In our example, \(x^3 - 1\) was factored using the difference of cubes method.
• Factoring reveals the roots or solutions of the polynomial equation when set to zero, \((x - 1)(x^2 + x + 1) = 0\).
• It's a vital skill enabling you to handle more complex algebraic tasks, especially derivative and integral calculus.

Effective polynomial factoring simplifies rational expressions and enhances understanding of their behaviors and associated equations.