Problem 37 - Algebra and Trigonometry

Question

\(13-44=\) Find the partial fraction decomposition of the rational function. $$ \frac{x-3}{x^{3}+3 x} $$

Step-by-Step Solution

Answer

The partial fraction decomposition is \(\frac{-1}{x} + \frac{x + 1}{x^2 + 3}\).

Step 1: Factor the Denominator

The first step in partial fraction decomposition is to factor the denominator completely. The denominator is \(x^3 + 3x\). It can be factored as follows:\[x^3 + 3x = x(x^2 + 3) = x(x + \sqrt{3}i)(x - \sqrt{3}i)\]since \(x^2 + 3\) does not have real roots.

Step 2: Set Up Partial Fraction Decomposition

Based on the factors of the denominator, set up the partial fraction decomposition. For the denominator \(x(x + \sqrt{3}i)(x - \sqrt{3}i)\), the partial fraction decomposition is:\[\frac{x-3}{x(x + \sqrt{3}i)(x - \sqrt{3}i)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 3}\]This is because \(x^2 + 3\) remains an irreducible quadratic factor.

Step 3: Multiply Through by the Common Denominator

To solve for the constants \(A\), \(B\), and \(C\), multiply both sides by the common denominator \(x(x + \sqrt{3}i)(x - \sqrt{3}i)\):\[x - 3 = A(x^2 + 3) + (Bx + C)x\]

Step 4: Expand and Combine Terms

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

Step 5: Solve for Constants

Equate the coefficients from both sides of the equation:1. \(A + B = 0\) (since there is no \(x^2\) term on the left)2. \(C = 1\) (coefficient of \(x\))3. \(3A = -3\) (constant term)Solve these equations:From 3, \(A = -1\).Substitute \(A = -1\) into 1, we get \(-1 + B = 0\) which means \(B = 1\).So, \(A = -1\), \(B = 1\), \(C = 1\).

Step 6: Write the Final Partial Fraction Decomposition

Substitute the values of \(A\), \(B\), and \(C\) into the partial fraction decomposition:\[\frac{x-3}{x(x^2 + 3)} = \frac{-1}{x} + \frac{x + 1}{x^2 + 3}\]This is the partial fraction decomposition of the given rational function.

Key Concepts

Rational Function

A rational function is defined as a fraction in which both the numerator and the denominator are polynomials. In this particular exercise, the rational function is given by \(\frac{x-3}{x^3 + 3x}\). Rational functions are fundamental in calculus and algebra as they are used to describe a wide array of real-world phenomena.
Understanding the structure of rational functions helps in simplifying complex algebraic expressions.

• The numerator of our given function is \(x-3\).
• The denominator is \(x^3 + 3x\).

By finding its partial fraction decomposition, we can express it as a sum of simpler fractions, making it easier to perform integration or other operations.

Factoring Polynomials

Factoring polynomials involves expressing a polynomial as a product of its factors. This step is crucial in partial fraction decomposition. Let’s look at the given polynomial in the denominator, \(x^3 + 3x\).
The first step in factoring this polynomial is to find common factors. Here, \(x\) is a common factor. Thus:

• \(x^3 + 3x = x(x^2 + 3)\)
• \(x^2 + 3\) is an irreducible quadratic, meaning it cannot be factored further using real numbers.

Factoring transforms the original expression into an easier form necessary for developing a partial fraction decomposition.

Complex Numbers

Complex numbers come into play when real numbers are inadequate for factoring polynomials. In our exercise, the polynomial \(x^2 + 3\) has no real roots. Instead, it can be factored using complex numbers as \(x + \sqrt{3}i\) and \(x - \sqrt{3}i\).
Complex numbers are expressed in the form \(a + bi\), where \(i\) is the imaginary unit with the property that \(i^2 = -1\). These numbers allow us to extend our number system to include solutions for equations that have no real solutions.
For practical purposes, partial fraction decomposition will often express irreducible quadratics as simple fractions rather than complex factors.

Coefficient Comparison

Coefficient comparison is a method used to determine the unknowns in partial fraction decomposition. Once we expand our expression and combine like terms, we compare coefficients of the same powers of \(x\) on each side of the equation.
In this exercise, the goal was to find constants \(A\), \(B\), and \(C\). We expand the right side and equate it to the left side, combining like terms:

• \(A + B = 0\) (no \(x^2\) term on the left)
• \(C = 1\) (coefficient of \(x\))
• \(3A = -3\) (constant term)

Solving these equations, we identified that \(A = -1\), \(B = 1\), and \(C = 1\). Using this technique is essential for determining the values needed to complete the decomposition.