Rational Expressions and Their Simplification

Introduction

Key Concepts

Understanding Rational Expressions

A rational expression is a fraction where both the numerator and the denominator are polynomials. Formally, a rational expression can be written as:

**Example:** $$ \frac{2x^3 - 5x + 1}{x^2 - 4} $$ In this example, \( P(x) = 2x^3 - 5x + 1 \) and \( Q(x) = x^2 - 4 \).

Domains of Rational Expressions

The domain of a rational expression consists of all real numbers except those that make the denominator zero. To determine the domain, set the denominator equal to zero and solve for \( x \):

Simplifying Rational Expressions

Simplification involves reducing the rational expression to its lowest terms by factoring and canceling common factors in the numerator and the denominator.

**Steps to Simplify:**

1. Factor both the numerator and the denominator completely.
2. Identify and cancel common factors.
3. State the simplified expression and the restricted values.

**Example:** Simplify \( \frac{x^2 - 9}{x^2 - 6x + 9} \).

**Step 1: Factor both polynomials.** $$ x^2 - 9 = (x - 3)(x + 3) $$ $$ x^2 - 6x + 9 = (x - 3)^2 $$

**Step 2: Cancel common factors.** $$ \frac{(x - 3)(x + 3)}{(x - 3)^2} = \frac{x + 3}{x - 3}, \quad x \neq 3 $$

**Simplified Expression:** $$ \frac{x + 3}{x - 3}, \quad x \neq 3 $$

Operations with Rational Expressions

Performing operations such as addition, subtraction, multiplication, and division with rational expressions requires a solid understanding of finding common denominators and simplifying the resulting expressions.

Addition and Subtraction

To add or subtract rational expressions, find a common denominator and combine the numerators.

**Example:** $$ \frac{1}{x} + \frac{2}{x + 1} $$ **Common Denominator:** \( x(x + 1) \)

**Rewriting the Expressions:** $$ \frac{1(x + 1)}{x(x + 1)} + \frac{2x}{x(x + 1)} = \frac{x + 1 + 2x}{x(x + 1)} = \frac{3x + 1}{x(x + 1)} $$

Multiplication

Multiply the numerators together and the denominators together, then simplify.

**Example:** $$ \frac{2}{x} \times \frac{3}{x + 2} = \frac{6}{x(x + 2)} $$

Division

Multiply by the reciprocal of the divisor.

**Example:** $$ \frac{2}{x} \div \frac{3}{x + 2} = \frac{2}{x} \times \frac{x + 2}{3} = \frac{2(x + 2)}{3x} $$

Complex Fractions

A complex fraction is a fraction where the numerator, the denominator, or both contain fractions. Simplifying complex fractions involves finding a common denominator and simplifying accordingly.

**Example:** Simplify \( \frac{\frac{1}{x} + \frac{2}{y}}{\frac{3}{x} - \frac{4}{y}} \).

**Step 1: Find a common denominator for the numerator and denominator.**

Numerator: $$ \frac{y + 2x}{xy} $$ Denominator: $$ \frac{3y - 4x}{xy} $$

**Step 2: Simplify the complex fraction.** $$ \frac{\frac{y + 2x}{xy}}{\frac{3y - 4x}{xy}} = \frac{y + 2x}{3y - 4x} $$

Solving Rational Equations

To solve equations involving rational expressions, follow these steps:

1. Find the least common denominator (LCD) of all rational expressions in the equation.
2. Multiply both sides of the equation by the LCD to eliminate the denominators.
3. Simplify and solve the resulting polynomial equation.
4. Check all potential solutions in the original equation to ensure they do not make any denominators zero.

**Example:** Solve \( \frac{1}{x} + \frac{2}{x + 1} = 3 \).

**Step 1: Find the LCD.** $$ LCD = x(x + 1) $$

**Step 2: Multiply both sides by LCD.** $$ (x)(x + 1) \left( \frac{1}{x} + \frac{2}{x + 1} \right) = 3x(x + 1) $$ $$ (x + 1) + 2x = 3x(x + 1) $$

**Step 3: Simplify and solve.** $$ 3x + 1 = 3x^2 + 3x $$ $$ 3x^2 + 3x - 3x - 1 = 0 $$ $$ 3x^2 - 1 = 0 $$ $$ x^2 = \frac{1}{3} $$ $$ x = \pm \frac{\sqrt{3}}{3} $$

**Step 4: Check for restrictions.** Both solutions do not make the denominators zero, so they are valid.

Applications of Rational Expressions

Rational expressions are widely used in various fields such as engineering, physics, economics, and statistics. They model relationships where one quantity varies inversely with another.

**Examples:**

• Calculating speed: \( \text{Speed} = \frac{\text{Distance}}{\text{Time}} \)
• Economics: Marginal cost functions often involve rational expressions.
• Physics: Electrical impedance in circuits is represented using rational expressions of frequency.

Graphing Rational Expressions

Graphing rational expressions involves identifying key features such as intercepts, asymptotes, and discontinuities.

**Steps to Graph:**

1. Find the domain by determining where the denominator is zero.
2. Identify vertical asymptotes at values that make the denominator zero.
3. Find horizontal or oblique asymptotes based on the degrees of the numerator and denominator.
4. Determine the x-intercepts by setting the numerator equal to zero.
5. Find the y-intercept by evaluating the function at \( x = 0 \).
6. Plot the points and sketch the graph, considering the asymptotes and intercepts.

**Example:** Graph \( \frac{x - 1}{x^2 - 1} \).

**Step 1: Domain.** \( x^2 - 1 = 0 \Rightarrow x = \pm1 \). So, \( x \neq 1 \) and \( x \neq -1 \).

**Step 2: Vertical Asymptotes.** At \( x = 1 \) and \( x = -1 \).

**Step 3: Horizontal Asymptote.** Degree of numerator (1)

**Step 4: x-intercept.** Set \( x - 1 = 0 \Rightarrow x = 1 \). However, \( x = 1 \) is excluded from the domain, so there is no x-intercept.

**Step 5: y-intercept.** Evaluate at \( x = 0 \): $$ \frac{0 - 1}{0^2 - 1} = \frac{-1}{-1} = 1 $$

**Graph:** The graph has vertical asymptotes at \( x = 1 \) and \( x = -1 \), a horizontal asymptote at \( y = 0 \), and a y-intercept at \( (0, 1) \).

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to break down complex rational expressions into simpler fractions, making integration and equation solving more manageable.

**Steps for Partial Fraction Decomposition:**

1. Ensure the degree of the numerator is less than the degree of the denominator. If not, perform polynomial long division.
2. Factor the denominator into irreducible factors.
3. Express the rational expression as a sum of fractions with unknown coefficients.
4. Multiply through by the common denominator to eliminate the fractions.
5. Solve the resulting system of equations to find the unknown coefficients.

**Example:** Decompose \( \frac{2x + 3}{(x + 1)(x - 2)} \).

**Step 1: The degree of the numerator (1) is less than the degree of the denominator (2). Proceed.**

**Step 2: Denominator is already factored.**

**Step 3: Express as partial fractions.** $$ \frac{2x + 3}{(x + 1)(x - 2)} = \frac{A}{x + 1} + \frac{B}{x - 2} $$

**Step 4: Multiply through by \( (x + 1)(x - 2) \).** $$ 2x + 3 = A(x - 2) + B(x + 1) $$

**Step 5: Solve for \( A \) and \( B \).**

Let \( x = 2 \): $$ 2(2) + 3 = A(0) + B(3) \Rightarrow 7 = 3B \Rightarrow B = \frac{7}{3} $$

Let \( x = -1 \): $$ 2(-1) + 3 = A(-3) + B(0) \Rightarrow 1 = -3A \Rightarrow A = -\frac{1}{3} $$

**Final Decomposition:** $$ \frac{2x + 3}{(x + 1)(x - 2)} = -\frac{1}{3(x + 1)} + \frac{7}{3(x - 2)} $$

Exponents and Powers in Rational Expressions

Understanding how to manipulate exponents and powers within rational expressions is essential for simplifying and solving equations.

**Rules to Remember:**

• $$ \frac{a^m}{a^n} = a^{m-n} $$
• $$ (a^m)^n = a^{mn} $$
• $$ a^{-n} = \frac{1}{a^n} $$

**Example:** Simplify \( \frac{x^5}{x^2} \times x^{-1} \).

$$ \frac{x^5}{x^2} \times x^{-1} = x^{5-2} \times x^{-1} = x^{3} \times x^{-1} = x^{3-1} = x^{2} $$

Common Mistakes to Avoid

• Forget to consider the domain restrictions when simplifying.
• Fail to factor polynomials completely before canceling.
• Cancel terms incorrectly, especially when dealing with polynomials.
• Neglect to check for extraneous solutions when solving rational equations.

Advanced Concepts

Polynomial Long Division

Polynomial long division is a method used to divide a polynomial by another polynomial of lower or equal degree. It is particularly useful when simplifying rational expressions where the degree of the numerator is greater than or equal to the degree of the denominator.

**Example:** Divide \( 2x^3 + 3x^2 - x + 5 \) by \( x - 2 \).

**Steps:**

1. Arrange both polynomials in descending order of degree.
2. Divide the leading term of the numerator by the leading term of the denominator.
3. Multiply the entire denominator by the result and subtract from the numerator.
4. Repeat until the degree of the remainder is less than the degree of the denominator.

**Calculation:**

• First term: \( \frac{2x^3}{x} = 2x^2 \)
• Multiply \( 2x^2 \) by \( x - 2 \): \( 2x^3 - 4x^2 \)
• Subtract: \( (2x^3 + 3x^2) - (2x^3 - 4x^2) = 7x^2 \)
• Next term: \( \frac{7x^2}{x} = 7x \)
• Multiply \( 7x \) by \( x - 2 \): \( 7x^2 - 14x \)
• Subtract: \( (7x^2 - x) - (7x^2 - 14x) = 13x \)
• Next term: \( \frac{13x}{x} = 13 \)
• Multiply \( 13 \) by \( x - 2 \): \( 13x - 26 \)
• Subtract: \( (13x + 5) - (13x - 26) = 31 \)

**Result:** $$ 2x^2 + 7x + 13 + \frac{31}{x - 2} $$

Deeper into Partial Fractions

Partial fraction decomposition becomes increasingly complex when dealing with repeated or irreducible quadratic factors.

Repeated Linear Factors

When the denominator contains repeated linear factors, the partial fractions must account for each multiplicity.

**Example:** Decompose \( \frac{5x + 6}{(x + 1)^2} \).

**Form:** $$ \frac{5x + 6}{(x + 1)^2} = \frac{A}{x + 1} + \frac{B}{(x + 1)^2} $$

**Solving:** Multiply through by \( (x + 1)^2 \): $$ 5x + 6 = A(x + 1) + B $$ Set \( x = -1 \): $$ 5(-1) + 6 = A(0) + B \Rightarrow B = 1 $$ Expand and equate coefficients: $$ 5x + 6 = A x + A + 1 $$ $$ 5x + 6 = A x + (A + 1) $$

Equate coefficients: $$ A = 5 $$ $$ A + 1 = 6 \Rightarrow 5 + 1 = 6 \quad \text{(consistent)} $$

**Final Decomposition:** $$ \frac{5x + 6}{(x + 1)^2} = \frac{5}{x + 1} + \frac{1}{(x + 1)^2} $$

Irreducible Quadratic Factors

When the denominator includes irreducible quadratic factors, the partial fractions take the form of linear expressions in the numerator.

**Example:** Decompose \( \frac{2x + 3}{x(x^2 + 1)} \).

**Form:** $$ \frac{2x + 3}{x(x^2 + 1)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1} $$

**Solving:** Multiply through by \( x(x^2 + 1) \): $$ 2x + 3 = A(x^2 + 1) + (Bx + C)x $$ $$ 2x + 3 = A x^2 + A + Bx^2 + Cx $$ $$ 2x + 3 = (A + B)x^2 + Cx + A $$>

Equate coefficients:

• For \( x^2 \): \( A + B = 0 \)
• For \( x \): \( C = 2 \)
• Constant term: \( A = 3 \)

From \( A = 3 \): $$ 3 + B = 0 \Rightarrow B = -3 $$

**Final Decomposition:** $$ \frac{2x + 3}{x(x^2 + 1)} = \frac{3}{x} + \frac{-3x + 2}{x^2 + 1} $$

Cancellation of Higher-Degree Polynomials

When simplifying rational expressions with higher-degree polynomials, it is essential to factor completely and look for common factors that may cancel out.

**Example:** Simplify \( \frac{x^4 - 1}{x^2 - 1} \).

**Step 1: Factor the numerator and the denominator.** $$ x^4 - 1 = (x^2)^2 - 1^2 = (x^2 - 1)(x^2 + 1) = (x - 1)(x + 1)(x^2 + 1) $$> $$ x^2 - 1 = (x - 1)(x + 1) $$>

**Step 2: Cancel common factors.** $$ \frac{(x - 1)(x + 1)(x^2 + 1)}{(x - 1)(x + 1)} = x^2 + 1, \quad x \neq \pm1 $$>

**Simplified Expression:** $$ x^2 + 1, \quad x \neq \pm1 $$>

Solving Systems of Rational Equations

When dealing with systems of rational equations, the approach involves finding a common denominator and solving the resulting system of polynomial equations.

**Example:** Solve the system: $$ \frac{1}{x} + \frac{1}{y} = 2 $$> $$ \frac{1}{x} - \frac{1}{y} = 0 $$>

**Step 1: Let \( u = \frac{1}{x} \) and \( v = \frac{1}{y} \). The system becomes:**

• \( u + v = 2 \)
• \( u - v = 0 \)

**Step 2: Solve the system.**

• Add the equations: \( 2u = 2 \Rightarrow u = 1 \)
• Substitute \( u = 1 \) into \( u - v = 0 \Rightarrow v = 1 \)

**Step 3: Find \( x \) and \( y \).**

• \( u = \frac{1}{x} = 1 \Rightarrow x = 1 \)
• \( v = \frac{1}{y} = 1 \Rightarrow y = 1 \)

**Solution:** \( x = 1 \), \( y = 1 \)

Rational Expressions in Complex Numbers

Rational expressions extend to complex numbers, where the numerator and denominator can involve complex polynomials.

**Example:** Simplify \( \frac{(x + i)}{(x^2 + 1)} \), where \( i = \sqrt{-1} \).

**Factor the denominator:** $$ x^2 + 1 = (x - i)(x + i) $$>

**Express as partial fractions:** $$ \frac{x + i}{(x - i)(x + i)} = \frac{A}{x - i} + \frac{B}{x + i} $$>

**Solve for \( A \) and \( B \):** Multiply through by \( (x - i)(x + i) \): $$ x + i = A(x + i) + B(x - i) $$>

Let \( x = i \): $$ i + i = A(2i) + B(0) \Rightarrow 2i = 2iA \Rightarrow A = 1 $$>

Let \( x = -i \): $$ -i + i = A(0) + B(-2i) \Rightarrow 0 = -2iB \Rightarrow B = 0 $$>

**Final Decomposition:** $$ \frac{x + i}{x^2 + 1} = \frac{1}{x - i} $$>

**Note:** Simplifications involving complex numbers require careful handling of imaginary units and factoring.

Rational Expressions with Radical Expressions

Rational expressions may sometimes include radical expressions. Simplifying these requires combining rational expression techniques with those for radicals.

**Example:** Simplify \( \frac{\sqrt{x}}{x} \).

$$ \frac{\sqrt{x}}{x} = \frac{x^{1/2}}{x^1} = x^{1/2 - 1} = x^{-1/2} = \frac{1}{\sqrt{x}} $$>

Asymptotic Behavior and Limits

Understanding the asymptotic behavior of rational expressions involves analyzing the limits as \( x \) approaches infinity or specific points, providing insight into the end-behavior and discontinuities of the function.

**Example:** Find the limit of \( \frac{2x^2 + 3x + 1}{x^2 - x - 2} \) as \( x \to \infty \).

**Solution:** Divide numerator and denominator by \( x^2 \): $$ \lim_{x \to \infty} \frac{2 + \frac{3}{x} + \frac{1}{x^2}}{1 - \frac{1}{x} - \frac{2}{x^2}} = \frac{2}{1} = 2 $$>

**Conclusion:** The horizontal asymptote is \( y = 2 \).

Applications in Calculus

Rational expressions play a significant role in calculus, particularly in differentiation and integration. Techniques such as partial fraction decomposition are essential for integrating rational functions.

**Example: Integration Using Partial Fractions** Integrate \( \frac{2x + 3}{(x + 1)(x - 2)} \):

From the earlier partial fraction decomposition: $$ \frac{2x + 3}{(x + 1)(x - 2)} = \frac{5}{x - 2} + \frac{-3x + 1}{x + 1} $$>

However, based on correct decomposition: $$ \frac{2x + 3}{(x + 1)(x - 2)} = \frac{5}{x - 2} + \frac{1}{x + 1} $$>

Therefore, $$ \int \frac{2x + 3}{(x + 1)(x - 2)} dx = 5 \ln|x - 2| + \ln|x + 1| + C $$>

Rational Expressions in Optimization Problems

Rational expressions are frequently used in optimization problems where the objective is to maximize or minimize a certain quantity.

**Example:** Find the dimensions that minimize the cost function \( C = \frac{2x + 3}{x^2} \).

**Solution:** To minimize \( C \), take the derivative and set it to zero: $$ C(x) = \frac{2x + 3}{x^2} = 2x^{-1} + 3x^{-2} $$> $$ C'(x) = -2x^{-2} - 6x^{-3} = -\frac{2}{x^2} - \frac{6}{x^3} $$> Set \( C'(x) = 0 \): $$ -\frac{2}{x^2} - \frac{6}{x^3} = 0 \Rightarrow \frac{2}{x^2} + \frac{6}{x^3} = 0 $$> $$ 2x + 6 = 0 \Rightarrow x = -3 $$> **Note:** Since \( x > 0 \) in most optimization problems, check endpoints or constraints.

Rational Inequalities

Rational inequalities involve expressions with polynomials in fractional form and require determining the intervals where the inequality holds true.

**Example:** Solve \( \frac{x + 1}{x - 2} > 0 \).

**Steps:**

1. Find critical points by setting numerator and denominator to zero: \( x = -1 \) and \( x = 2 \).
2. Divide the number line into intervals based on critical points: \( (-\infty, -1) \), \( (-1, 2) \), \( (2, \infty) \).
3. Test a point in each interval to determine the sign of the expression.

**Testing:**

• For \( x = -2 \): \( \frac{-1}{-4} > 0 \) (True)
• For \( x = 0 \): \( \frac{1}{-2} > 0 \) (False)
• For \( x = 3 \): \( \frac{4}{1} > 0 \) (True)

**Solution:** $$ x \in (-\infty, -1) \cup (2, \infty) $$>

Rational Expressions in Sequences and Series

In sequences and series, rational expressions can represent the general term, enabling the analysis of convergence and other properties.

**Example:** Find the general term of the sequence defined by \( a_n = \frac{2n + 1}{n^2 + n} \).

**Simplification:** $$ a_n = \frac{2n + 1}{n(n + 1)} = \frac{2n}{n(n + 1)} + \frac{1}{n(n + 1)} = \frac{2}{n + 1} + \frac{1}{n(n + 1)} $$>

Optimization Using Lagrange Multipliers

Although more advanced, rational expressions are integral to optimization problems involving constraints, where methods like Lagrange multipliers are applied.

**Example:** Maximize \( f(x, y) = \frac{x + y}{xy} \) subject to \( x + y = k \), where \( k \) is a constant.

**Approach:** Use substitution to reduce the problem to a single variable and apply calculus techniques to find extrema.

Rational Expressions in Probability and Statistics

Rational expressions model probabilities, especially in distributions and expectation calculations.

**Example:** The probability generating function for a discrete random variable \( X \) can be expressed as: $$ G_X(s) = \sum_{k=0}^{\infty} P(X = k)s^k $$> If \( G_X(s) \) is a rational function, it simplifies analysis of the distribution.

Comparison Table

Summary and Key Takeaways