Problem 19 - A Graphical Approach to College Algebra

Question

Find the partial fraction decomposition for each rational expression. $$\frac{-3}{x^{2}\left(x^{2}+5\right)}$$

Step-by-Step Solution

Answer

\(\frac{-3}{x^2(x^2+5)} = \frac{-3/5}{x^2} + \frac{3/5}{x^2+5}\)

Step 1: Identify the Denominator

The denominator of the expression is \(x^2(x^2+5)\). It is a product of a repeated linear factor \(x^2\) and an irreducible quadratic \(x^2 + 5\).

Step 2: Set Up the Partial Fractions

Write the original expression as a sum of partial fractions: \[\frac{-3}{x^2(x^2+5)} = \frac{A}{x} + \frac{B}{x^2} + \frac{Cx+D}{x^2+5}\]Here, \(A\) and \(B\) are coefficients for the terms involving \(x\) and \(Cx+D\) corresponds to the quadratic factor \(x^2+5\).

Step 3: Clear the Denominator

Multiply both sides by the denominator \(x^2(x^2+5)\) to clear the fractions:\[-3 = A(x)(x^2+5) + B(x^2+5) + (Cx+D)x^2\]

Step 4: Expand and Combine Like Terms

Expand the equation on the right side:\[-3 = Ax^3 + 5Ax + Bx^2 + 5B + Cx^3 + Dx^2\]Combine like terms:\[-3 = (A + C)x^3 + (B + D)x^2 + 5Ax + 5B\]

Step 5: Set Up the System of Equations

Compare coefficients on both sides to set up equations:1. \(A + C = 0\) (Coefficient of \(x^3\))2. \(B + D = 0\) (Coefficient of \(x^2\))3. \(5A = 0\) (Coefficient of \(x\))4. \(5B = -3\) (Constant term)

Step 6: Solve the System of Equations

From equation 3, \(5A = 0\) implies \(A = 0\).Using \(5A = 0\), substitute \(A = 0\) into equation 1, \(A + C = 0\), giving \(C = 0\).From equation 4, \(5B = -3\) implies \(B = -\frac{3}{5}\).Substitute \(B = -\frac{3}{5}\) into equation 2, \(B + D = 0\), giving \(D = \frac{3}{5}\).

Step 7: Write the Partial Fraction Decomposition

The partial fraction decomposition is:\[\frac{-3}{x^2(x^2+5)} = \frac{-3/5}{x^2} + \frac{3/5}{x^2 + 5}\]

Key Concepts

Rational Expressions

A rational expression is essentially a fraction where both the numerator and the denominator are polynomial functions. When dealing with rational expressions, it is important to remember that the denominator cannot be zero, as this would make the expression undefined.
The goal of partial fraction decomposition is to express a complex rational expression into a sum of simpler fractions. This is particularly useful in calculus and algebra because it allows more complicated expressions to be integrated or manipulated more easily.
When working with rational expressions:

• Ensure the expression is in its simplest form; simplify if needed.
• Look at the denominator to identify any repeated or unique factors.
• Write each term over one of the identified factors.

This process helps in breaking down the expression into manageable parts for further algebraic manipulation or integration.

Polynomial Functions

Polynomial functions are mathematical expressions involving a sum of powers in one or more variables multiplied by coefficients. A general polynomial in one variable is written as: \[ a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0 \] where \(a_n, a_{n-1}, \ldots, a_0\) are constants.
Polynomials can have different forms based on the degree of the polynomial.

• A polynomial of degree 1 is called linear.
• Degree 2 is quadratic, degree 3 is cubic, and so on.
• If a polynomial function is arranged in descending order according to the powers, it is in its standard form.

Understanding polynomials is crucial for manipulating rational expressions. They provide the building blocks when performing operations like addition, subtraction, multiplication, division, and decomposition in partial fractions.

Algebraic Equations

Algebraic equations are statements of equality between two expressions. These involve the use of variables and constants connected by algebraic operations like addition, subtraction, multiplication, and division.
Solving algebraic equations often involves finding the value(s) of the variable(s) that satisfy the equation. In the context of partial fraction decomposition, setting up an equation is a critical step.

• Express the rational expression as a sum of fractions.
• Multiply through by the common denominator to clear fractions.
• Compare coefficients for each power of x to form a system of equations.

Once you have the system, it's a matter of solving these equations to find the unknowns that were introduced during the decomposition, like \(A, B, C, \) and \(D\) in this case.