Solved Exercises on Multiplying and Dividing Rational Expressions in Algebra 2

Master multiplying and dividing rational expressions: factoring, cancellation, domain restrictions, and advanced techniques through these 5 detailed exercises.

Solution: Exercises 1 to 3
1 Basic Multiplication
Exercise 1
Multiply: \(\frac{x+3}{x-2} \cdot \frac{x-2}{x+5}\)
Definition:

Multiplying rational expressions: \(\frac{A}{B} \cdot \frac{C}{D} = \frac{AC}{BD}\) (multiply numerators together, denominators together).

Multiplication method:
  1. Multiply numerators together
  2. Multiply denominators together
  3. Factor completely if needed
  4. Cancel common factors
  5. State domain restrictions
Multiply numerators
(x+3)(x-2)
Multiply denominators
(x-2)(x+5)
Step 1: Apply multiplication rule

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

Step 2: Identify common factors

Both numerator and denominator contain the factor \((x-2)\)

Step 3: Cancel common factors

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

Step 4: State domain restrictions

The original expression is undefined when \(x = 2\) or \(x = -5\)

After cancellation, the simplified form is still undefined at these points

\(\frac{x+3}{x-2} \cdot \frac{x-2}{x+5} = \frac{x+3}{x+5}\), where \(x \neq 2, -5\)
Final answer:

\(\frac{x+3}{x-2} \cdot \frac{x-2}{x+5} = \frac{x+3}{x+5}\), where \(x \neq 2, -5\)

Applied rules:

• Multiplication rule: \(\frac{A}{B} \cdot \frac{C}{D} = \frac{AC}{BD}\)

• Factor cancellation: Cancel identical factors in numerator and denominator

• Domain preservation: Maintain restrictions from original expressions

2 Division Conversion
Exercise 2
Divide: \(\frac{x^2 - 4}{x^2 - 9} \div \frac{x+2}{x-3}\)
Definition:

Dividing rational expressions: Convert division to multiplication by taking the reciprocal of the divisor.

Convert to multiplication
\(\frac{x^2-4}{x^2-9} \cdot \frac{x-3}{x+2}\)
Factor completely
\(\frac{(x+2)(x-2)}{(x+3)(x-3)} \cdot \frac{x-3}{x+2}\)
Step 1: Convert division to multiplication

\(\frac{x^2 - 4}{x^2 - 9} \div \frac{x+2}{x-3} = \frac{x^2 - 4}{x^2 - 9} \cdot \frac{x-3}{x+2}\)

Step 2: Factor completely

\(x^2 - 4 = (x+2)(x-2)\) (difference of squares)

\(x^2 - 9 = (x+3)(x-3)\) (difference of squares)

So: \(\frac{(x+2)(x-2)}{(x+3)(x-3)} \cdot \frac{x-3}{x+2}\)

Step 3: Multiply numerators and denominators

\(\frac{(x+2)(x-2) \cdot (x-3)}{(x+3)(x-3) \cdot (x+2)} = \frac{(x+2)(x-2)(x-3)}{(x+3)(x-3)(x+2)}\)

Step 4: Cancel common factors

Both numerator and denominator contain \((x+2)\) and \((x-3)\)

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

\(\frac{x^2 - 4}{x^2 - 9} \div \frac{x+2}{x-3} = \frac{x-2}{x+3}\), where \(x \neq \pm 3, -2\)
Final answer:

\(\frac{x^2 - 4}{x^2 - 9} \div \frac{x+2}{x-3} = \frac{x-2}{x+3}\), where \(x \neq \pm 3, -2\)

Applied rules:

• Division conversion: \(A \div B = A \cdot \frac{1}{B}\)

• Difference of squares: \(a^2 - b^2 = (a+b)(a-b)\)

• Factor cancellation: Cancel identical factors across entire numerator and denominator

3 Complex Factoring
Exercise 3
Multiply: \(\frac{x^2 - 5x + 6}{x^2 - 4} \cdot \frac{x^2 + 4x + 4}{x^2 - 3x - 10}\)
Definition:

Factoring quadratics: For \(ax^2 + bx + c\), find two numbers that multiply to \(ac\) and add to \(b\).

Factor numerators
\(\frac{(x-2)(x-3)}{x^2-4} \cdot \frac{(x+2)^2}{x^2-3x-10}\)
Factor denominators
\(\frac{(x-2)(x-3)}{(x+2)(x-2)} \cdot \frac{(x+2)^2}{(x-5)(x+2)}\)
Step 1: Factor numerator of first fraction

\(x^2 - 5x + 6\): Find two numbers that multiply to 6 and add to -5

Those numbers are -2 and -3: \((x-2)(x-3)\)

Step 2: Factor denominator of first fraction

\(x^2 - 4 = (x+2)(x-2)\) (difference of squares)

Step 3: Factor numerator of second fraction

\(x^2 + 4x + 4 = (x+2)^2\) (perfect square trinomial)

Step 4: Factor denominator of second fraction

\(x^2 - 3x - 10\): Find two numbers that multiply to -10 and add to -3

Those numbers are -5 and 2: \((x-5)(x+2)\)

Step 5: Rewrite with factored forms

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

Step 6: Multiply numerators and denominators

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

Step 7: Cancel common factors

Numerator has: \((x-2)\), \((x-3)\), \((x+2)^2\)

Denominator has: \((x+2)\), \((x-2)\), \((x-5)\), \((x+2)\)

Cancel one \((x-2)\) and two \((x+2)\) factors: \(\frac{(x-3)(x+2)}{(x-5)}\)

\(\frac{(x-3)(x+2)}{x-5} = \frac{x^2 - x - 6}{x-5}\), where \(x \neq \pm 2, -2, 5\)
Final answer:

\(\frac{x^2 - 5x + 6}{x^2 - 4} \cdot \frac{x^2 + 4x + 4}{x^2 - 3x - 10} = \frac{x^2 - x - 6}{x-5}\), where \(x \neq \pm 2, 5\)

Applied rules:

• Quadratic factoring: Find two numbers that multiply to c and add to b

• Perfect square trinomial: \(a^2 + 2ab + b^2 = (a+b)^2\)

• Difference of squares: \(a^2 - b^2 = (a+b)(a-b)\)

• Systematic cancellation: Count factors to ensure complete cancellation

Key Concepts: Definitions, Rules, and Methods
\(\frac{A}{B} \cdot \frac{C}{D} = \frac{AC}{BD}, \quad \frac{A}{B} \div \frac{C}{D} = \frac{A}{B} \cdot \frac{D}{C}\)
Fundamental Operations
Key definitions:

Rational Expression: A ratio of two polynomials where the denominator is not zero.

Multiplication: Multiply numerators together and denominators together.

Division: Convert to multiplication by taking the reciprocal of the divisor.

Domain: The set of all real numbers for which the expression is defined.

Multiplication methodology:
  1. Factor completely: Factor all numerators and denominators
  2. Multiply: Multiply all numerators together, all denominators together
  3. Cancel common factors: Cancel identical factors in numerator and denominator
  4. State domain restrictions: Identify values that make original denominators zero
Division methodology:
  1. Convert to multiplication: Take reciprocal of the divisor
  2. Factor completely: Factor all polynomials
  3. Multiply: Follow multiplication procedure
  4. State domain restrictions: Account for all original denominators and divisor
Tip 1: Always factor completely before multiplying to identify cancellation opportunities.
Tip 2: For division, convert to multiplication immediately to avoid confusion.
Tip 3: Count factors systematically to ensure complete cancellation.
Tip 4: Always state domain restrictions from the original expressions.
Common errors: Forgetting to factor, cancelling terms instead of factors, missing domain restrictions.
Exam preparation: Practice factoring patterns, memorize operation rules, verify results.
Essential rules to remember:

• Multiplication: \(\frac{A}{B} \cdot \frac{C}{D} = \frac{AC}{BD}\)

• Division: \(\frac{A}{B} \div \frac{C}{D} = \frac{A}{B} \cdot \frac{D}{C}\)

• Factor first: Always factor completely before multiplying

• Cancel factors only: Cancel identical polynomial factors

• Preserve domain: State restrictions from original expressions

\(a^2 - b^2 = (a+b)(a-b)\)
Difference of Squares
\(a^2 \pm 2ab + b^2 = (a \pm b)^2\)
Perfect Square Trinomial
\(ax^2 + bx + c = (x+p)(x+q) \text{ where } pq=c, p+q=b\)
Quadratic Factoring
Solution: Exercises 4 to 5
4 Advanced Multiplication
Exercise 4
Multiply: \(\frac{x^3 - 8}{x^2 + 2x + 1} \cdot \frac{x^2 - 1}{x^2 + 2x + 4}\)
Definition:

Special factoring patterns: Difference of cubes, perfect square trinomials, and difference of squares.

Factor numerator
\(x^3 - 8 = (x-2)(x^2+2x+4)\)
Factor denominator
\(x^2 + 2x + 1 = (x+1)^2\)
Step 1: Factor \(x^3 - 8\) using difference of cubes

\(x^3 - 8 = x^3 - 2^3 = (x-2)(x^2 + 2x + 4)\)

Step 2: Factor \(x^2 + 2x + 1\) as perfect square trinomial

\(x^2 + 2x + 1 = (x+1)^2\)

Step 3: Factor \(x^2 - 1\) as difference of squares

\(x^2 - 1 = (x+1)(x-1)\)

Step 4: Rewrite with factored forms

\(\frac{(x-2)(x^2+2x+4)}{(x+1)^2} \cdot \frac{(x+1)(x-1)}{x^2+2x+4}\)

Step 5: Multiply numerators and denominators

\(\frac{(x-2)(x^2+2x+4)(x+1)(x-1)}{(x+1)^2(x^2+2x+4)}\)

Step 6: Cancel common factors

Numerator has: \((x-2)\), \((x^2+2x+4)\), \((x+1)\), \((x-1)\)

Denominator has: \((x+1)^2\), \((x^2+2x+4)\)

Cancel one \((x^2+2x+4)\) and one \((x+1)\): \(\frac{(x-2)(x-1)}{x+1}\)

\(\frac{(x-2)(x-1)}{x+1} = \frac{x^2 - 3x + 2}{x+1}\), where \(x \neq -1\)
Final answer:

\(\frac{x^3 - 8}{x^2 + 2x + 1} \cdot \frac{x^2 - 1}{x^2 + 2x + 4} = \frac{x^2 - 3x + 2}{x+1}\), where \(x \neq -1\)

Applied rules:

• Difference of cubes: \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\)

• Perfect square trinomial: \(a^2 + 2ab + b^2 = (a+b)^2\)

• Difference of squares: \(a^2 - b^2 = (a+b)(a-b)\)

• Systematic cancellation: Track all factors carefully

5 Complex Division
Exercise 5
Divide: \(\frac{x^2 - 6x + 9}{x^2 - x - 6} \div \frac{x^2 - 9}{x^2 + 5x + 6}\)
Definition:

Reciprocal division: Converting division to multiplication by flipping the divisor.

Convert to multiplication
\(\frac{x^2-6x+9}{x^2-x-6} \cdot \frac{x^2+5x+6}{x^2-9}\)
Factor all polynomials
\(\frac{(x-3)^2}{(x-3)(x+2)} \cdot \frac{(x+2)(x+3)}{(x+3)(x-3)}\)
Step 1: Convert division to multiplication

\(\frac{x^2 - 6x + 9}{x^2 - x - 6} \div \frac{x^2 - 9}{x^2 + 5x + 6} = \frac{x^2 - 6x + 9}{x^2 - x - 6} \cdot \frac{x^2 + 5x + 6}{x^2 - 9}\)

Step 2: Factor numerator of first fraction

\(x^2 - 6x + 9 = (x-3)^2\) (perfect square trinomial)

Step 3: Factor denominator of first fraction

\(x^2 - x - 6\): Find two numbers that multiply to -6 and add to -1

Those numbers are -3 and 2: \((x-3)(x+2)\)

Step 4: Factor numerator of second fraction

\(x^2 + 5x + 6\): Find two numbers that multiply to 6 and add to 5

Those numbers are 2 and 3: \((x+2)(x+3)\)

Step 5: Factor denominator of second fraction

\(x^2 - 9 = (x+3)(x-3)\) (difference of squares)

Step 6: Rewrite with factored forms

\(\frac{(x-3)^2}{(x-3)(x+2)} \cdot \frac{(x+2)(x+3)}{(x+3)(x-3)}\)

Step 7: Multiply numerators and denominators

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

Step 8: Cancel common factors

Numerator has: \((x-3)^2\), \((x+2)\), \((x+3)\)

Denominator has: \((x-3)\), \((x+2)\), \((x+3)\), \((x-3)\)

All factors cancel completely: \(\frac{(x-3)^2(x+2)(x+3)}{(x-3)^2(x+2)(x+3)} = 1\)

The expression simplifies to 1, where \(x \neq 3, -2, -3\)
Final answer:

\(\frac{x^2 - 6x + 9}{x^2 - x - 6} \div \frac{x^2 - 9}{x^2 + 5x + 6} = 1\), where \(x \neq 3, -2, -3\)

Applied rules:

• Division conversion: \(A \div B = A \cdot \frac{1}{B}\)

• Perfect square trinomial: \(a^2 - 2ab + b^2 = (a-b)^2\)

• Difference of squares: \(a^2 - b^2 = (a+b)(a-b)\)

• Complete cancellation: When all factors cancel, result is 1

Multiplying and Dividing Rational Expressions: Complete Guide
\(\frac{A}{B} \cdot \frac{C}{D} = \frac{AC}{BD}, \quad \frac{A}{B} \div \frac{C}{D} = \frac{A}{B} \cdot \frac{D}{C}\)
Fundamental Operations
Key definitions:

Rational Expression: A ratio of two polynomials where the denominator is not zero.

Multiplication: Multiply numerators together and denominators together.

Division: Convert to multiplication by taking the reciprocal of the divisor.

Simplified Form: When no common factors remain between numerator and denominator.

Complete methodology:
  1. Factor completely: Factor all numerators and denominators into irreducible polynomials
  2. For division: Convert to multiplication by taking the reciprocal of the divisor
  3. Multiply: Multiply all numerators together, all denominators together
  4. Cancel common factors: Cancel identical factors in numerator and denominator
  5. State domain restrictions: Identify all values that made original denominators zero
  6. Verify: Check that no further factoring or cancellation is possible
Tip 1: Always factor completely before multiplying to maximize cancellation opportunities.
Tip 2: For division, convert to multiplication immediately to avoid confusion.
Tip 3: Systematically count factors to ensure complete cancellation.
Tip 4: Always state domain restrictions from the original expressions.
Common errors: Forgetting to factor, cancelling terms instead of factors, missing domain restrictions, not converting division properly.
Exam preparation: Practice factoring patterns, memorize operation rules, verify results by substitution.
Essential formulas and rules:

• Multiplication: \(\frac{A}{B} \cdot \frac{C}{D} = \frac{AC}{BD}\)

• Division: \(\frac{A}{B} \div \frac{C}{D} = \frac{A}{B} \cdot \frac{D}{C}\)

• Difference of squares: \(a^2 - b^2 = (a+b)(a-b)\)

• Perfect square trinomial: \(a^2 \pm 2ab + b^2 = (a \pm b)^2\)

• Difference of cubes: \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\)

\(a^2 - b^2 = (a+b)(a-b)\)
Difference of Squares
\(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\)
Difference of Cubes
\(a^2 \pm 2ab + b^2 = (a \pm b)^2\)
Perfect Square Trinomial
Common factoring patterns:

GCF: Always factor out the greatest common factor first

Quadratic trinomials: Find two numbers that multiply to c and add to b

Grouping: For 4-term polynomials, group pairs and factor out common terms

Higher degree: May require multiple rounds of factoring

Verification techniques:
  1. Substitution: Plug in a value (not in the restricted domain) to both original and simplified forms
  2. Domain check: Verify that excluded values are properly identified
  3. Factor check: Ensure all possible factoring has been performed
Key note: The simplified form and original form are equivalent everywhere except at excluded values.
Key note: When all factors cancel completely, the result is 1.

Questions & Answers

Question: I'm confused about when to flip the second fraction in division. Is it always the divisor that gets flipped?

Answer: Yes, in division of rational expressions, you always flip (take the reciprocal of) the DIVISOR - the second fraction.

The rule is: \(\frac{A}{B} \div \frac{C}{D} = \frac{A}{B} \cdot \frac{D}{C}\)

  • \(\frac{A}{B}\) is the dividend (what you're dividing)
  • \(\frac{C}{D}\) is the divisor (what you're dividing by)
  • To divide by \(\frac{C}{D}\), multiply by its reciprocal: \(\frac{D}{C}\)

Think of it as: "dividing by a fraction is the same as multiplying by its reciprocal."

Example: \(\frac{x+1}{x-2} \div \frac{x+3}{x-1} = \frac{x+1}{x-2} \cdot \frac{x-1}{x+3}\)

Question: How do I know if I've canceled all possible factors? What if I miss some?

Answer: To ensure you've canceled all possible factors, follow this systematic approach:

  1. Factor completely: Make sure every polynomial is factored into irreducible factors
  2. List all factors: Write out every factor in the numerator and denominator
  3. Match identical factors: Pair up identical factors in numerator and denominator
  4. Cancel pairs: Remove one factor from numerator and one from denominator for each pair

Example: \(\frac{(x+2)(x-1)^2(x+3)}{(x+2)(x-1)(x+4)}\)

  • Numerator factors: \((x+2)\), \((x-1)^2\), \((x+3)\)
  • Denominator factors: \((x+2)\), \((x-1)\), \((x+4)\)
  • Common factors: One \((x+2)\) and one \((x-1)\)
  • Result: \(\frac{(x-1)(x+3)}{x+4}\)

Always double-check by ensuring no identical factors remain in both numerator and denominator.

Question: Why do we need to keep track of domain restrictions? What happens if we ignore them?

Answer: Domain restrictions are crucial because they preserve the original function's behavior and ensure mathematical correctness.

Consider: \(\frac{x^2-4}{x-2} \cdot \frac{x-1}{x^2-4}\)

  • The original expression is undefined when \(x = 2\), \(x = -2\), or \(x = 1\)
  • After factoring: \(\frac{(x+2)(x-2)}{x-2} \cdot \frac{x-1}{(x+2)(x-2)}\)
  • After cancellation: \(\frac{x-1}{x-2}\)

Even though the simplified form appears to be defined at \(x = -2\), we must remember that the original expression was undefined there. The complete answer is: \(\frac{x-1}{x-2}\), where \(x \neq \pm 2, 1\).

Ignoring domain restrictions can lead to incorrect evaluations and loss of important mathematical properties.