Learn on PengiSaxon Algebra 2Chapter 4: Lessons 31-40, Investigation 4

Lesson 31: Multiplying and Dividing Rational Expressions

In this Grade 10 Saxon Algebra 2 lesson, students learn how to multiply and divide rational expressions by multiplying numerators and denominators, then factoring and canceling common factors to simplify. The lesson also covers dividing rational expressions by multiplying by the reciprocal and identifying values that make an expression undefined. Students practice evaluating simplified rational expressions for given variable values across a range of polynomial expressions.

Section 1

📘 Multiplying and Dividing Rational Expressions

New Concept

To divide by a rational expression, multiply by its reciprocal: ab÷cd=ab⋅dc\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c}, where bb, cc, and dd are all nonzero.

What’s next

Next, you will apply this rule, along with multiplication and factoring, to simplify complex rational expressions and solve application problems.

Section 2

Multiplying Rational Expressions

To multiply rational expressions, you multiply the numerators together and the denominators together, just like with regular fractions. The general rule is

AB⋅CD=A⋅CB⋅D \frac{A}{B} \cdot \frac{C}{D} = \frac{A \cdot C}{B \cdot D}
. After combining them, you must factor all polynomials and cancel any common factors to write the final expression in its simplest form.

Example 1:

3x2y⋅2y3x4=6x2y3x4y=6y2x2 \frac{3x^2}{y} \cdot \frac{2y^3}{x^4} = \frac{6x^2y^3}{x^4y} = \frac{6y^2}{x^2}

Example 2:
x+4x−1⋅x−15(x+4)=(x+4)(x−1)5(x−1)(x+4)=15 \frac{x+4}{x-1} \cdot \frac{x-1}{5(x+4)} = \frac{(x+4)(x-1)}{5(x-1)(x+4)} = \frac{1}{5}

Example 3:
a−2a(a+5)⋅(a2+5a)=a−2a(a+5)⋅a(a+5)1=a−2 \frac{a-2}{a(a+5)} \cdot (a^2+5a) = \frac{a-2}{a(a+5)} \cdot \frac{a(a+5)}{1} = a-2

Think of it like a team-up! The numerators join forces, and the denominators do the same. After they're combined, you look for matching factors on the top and bottom to cancel out. This process simplifies your expression significantly. It’s all about factoring first to make the multiplication and subsequent simplification an absolute breeze for everyone.

Section 3

Dividing by a Rational Expression

To divide by a rational expression, you multiply the first expression by the reciprocal of the second one. This method is often called 'keep, change, flip.' The rule is formally stated as:

ab÷cd=ab⋅dc \frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c}
, where the denominators b, c, and d cannot be equal to zero.

Example 1:

15x42y2÷3x28y3=15x42y2⋅8y33x2=120x4y36x2y2=20x2y \frac{15x^4}{2y^2} \div \frac{3x^2}{8y^3} = \frac{15x^4}{2y^2} \cdot \frac{8y^3}{3x^2} = \frac{120x^4y^3}{6x^2y^2} = 20x^2y

Example 2:
x+3x−8÷x(x+3)x−8=x+3x−8⋅x−8x(x+3)=1x \frac{x+3}{x-8} \div \frac{x(x+3)}{x-8} = \frac{x+3}{x-8} \cdot \frac{x-8}{x(x+3)} = \frac{1}{x}

Division is just multiplication in a clever disguise. Remember the 'keep, change, flip' rule: keep the first fraction, change division to multiplication, and flip the second fraction (its reciprocal). After this quick change, you simply follow the multiplication rules. Factor all the numerators and denominators completely, then cancel any common factors to find your simplified answer.

Section 4

To identify values that make an expression undefined

An expression becomes undefined when its denominator equals zero. To find these restricted values, you must inspect the denominators of the original expression, before any simplification. For a division problem

ab÷cd \frac{a}{b} \div \frac{c}{d}
, you must ensure that b, c, and d are all nonzero, as they each pose a risk of division by zero.

Example 1: For

x+1(x−5)(x+3) \frac{x+1}{(x-5)(x+3)}
, the expression is undefined when x = 5 or x = -3.
Example 2: For
xx−6÷x+2x(x−6) \frac{x}{x-6} \div \frac{x+2}{x(x-6)}
, the expression is undefined for x=6, x=-2, and x=0.

The denominator is like the ground you're standing on; dividing by zero makes it disappear! To find these forbidden values, you must check every single denominator in the original problem before you simplify or cancel anything. Even if a factor like (x-2) cancels out later, it was part of the original setup, making x=2 off-limits.

Book overview

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Chapter 4: Lessons 31-40, Investigation 4

  1. Lesson 1Current

    Lesson 31: Multiplying and Dividing Rational Expressions

    LearnPractice
  2. Lesson 2

    Lesson 32: Solving Linear Systems with Matrix Inverses (Exploration: Exploring Matrix Inverses)

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  3. Lesson 3

    Lesson 33: Applying Counting Principles

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  4. Lesson 4

    Lesson 34: Graphing Linear Equations II

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  5. Lesson 5

    Lesson 35: Solving Quadratic Equations I

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  6. Lesson 6

    Lesson 36: Using Parallel and Perpendicular Lines

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  7. Lesson 7

    Lesson 37: Adding and Subtracting Rational Expressions

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  8. Lesson 8

    Lesson 38: Dividing Polynomials Using Long Division

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  9. Lesson 9

    Lesson 39: Graphing Linear Inequalities in Two Variables

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  10. Lesson 10

    Lesson 40: Simplifying Radical Expressions

    LearnPractice
  11. Lesson 11

    Investigation 4: Understanding Cryptography

    LearnPractice

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

📘 Multiplying and Dividing Rational Expressions

New Concept

To divide by a rational expression, multiply by its reciprocal: ab÷cd=ab⋅dc\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c}, where bb, cc, and dd are all nonzero.

What’s next

Next, you will apply this rule, along with multiplication and factoring, to simplify complex rational expressions and solve application problems.

Section 2

Multiplying Rational Expressions

To multiply rational expressions, you multiply the numerators together and the denominators together, just like with regular fractions. The general rule is

AB⋅CD=A⋅CB⋅D \frac{A}{B} \cdot \frac{C}{D} = \frac{A \cdot C}{B \cdot D}
. After combining them, you must factor all polynomials and cancel any common factors to write the final expression in its simplest form.

Example 1:

3x2y⋅2y3x4=6x2y3x4y=6y2x2 \frac{3x^2}{y} \cdot \frac{2y^3}{x^4} = \frac{6x^2y^3}{x^4y} = \frac{6y^2}{x^2}

Example 2:
x+4x−1⋅x−15(x+4)=(x+4)(x−1)5(x−1)(x+4)=15 \frac{x+4}{x-1} \cdot \frac{x-1}{5(x+4)} = \frac{(x+4)(x-1)}{5(x-1)(x+4)} = \frac{1}{5}

Example 3:
a−2a(a+5)⋅(a2+5a)=a−2a(a+5)⋅a(a+5)1=a−2 \frac{a-2}{a(a+5)} \cdot (a^2+5a) = \frac{a-2}{a(a+5)} \cdot \frac{a(a+5)}{1} = a-2

Think of it like a team-up! The numerators join forces, and the denominators do the same. After they're combined, you look for matching factors on the top and bottom to cancel out. This process simplifies your expression significantly. It’s all about factoring first to make the multiplication and subsequent simplification an absolute breeze for everyone.

Section 3

Dividing by a Rational Expression

To divide by a rational expression, you multiply the first expression by the reciprocal of the second one. This method is often called 'keep, change, flip.' The rule is formally stated as:

ab÷cd=ab⋅dc \frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c}
, where the denominators b, c, and d cannot be equal to zero.

Example 1:

15x42y2÷3x28y3=15x42y2⋅8y33x2=120x4y36x2y2=20x2y \frac{15x^4}{2y^2} \div \frac{3x^2}{8y^3} = \frac{15x^4}{2y^2} \cdot \frac{8y^3}{3x^2} = \frac{120x^4y^3}{6x^2y^2} = 20x^2y

Example 2:
x+3x−8÷x(x+3)x−8=x+3x−8⋅x−8x(x+3)=1x \frac{x+3}{x-8} \div \frac{x(x+3)}{x-8} = \frac{x+3}{x-8} \cdot \frac{x-8}{x(x+3)} = \frac{1}{x}

Division is just multiplication in a clever disguise. Remember the 'keep, change, flip' rule: keep the first fraction, change division to multiplication, and flip the second fraction (its reciprocal). After this quick change, you simply follow the multiplication rules. Factor all the numerators and denominators completely, then cancel any common factors to find your simplified answer.

Section 4

To identify values that make an expression undefined

An expression becomes undefined when its denominator equals zero. To find these restricted values, you must inspect the denominators of the original expression, before any simplification. For a division problem

ab÷cd \frac{a}{b} \div \frac{c}{d}
, you must ensure that b, c, and d are all nonzero, as they each pose a risk of division by zero.

Example 1: For

x+1(x−5)(x+3) \frac{x+1}{(x-5)(x+3)}
, the expression is undefined when x = 5 or x = -3.
Example 2: For
xx−6÷x+2x(x−6) \frac{x}{x-6} \div \frac{x+2}{x(x-6)}
, the expression is undefined for x=6, x=-2, and x=0.

The denominator is like the ground you're standing on; dividing by zero makes it disappear! To find these forbidden values, you must check every single denominator in the original problem before you simplify or cancel anything. Even if a factor like (x-2) cancels out later, it was part of the original setup, making x=2 off-limits.

Book overview

Jump across lessons in the current chapter without opening the full course modal.

Continue this chapter

Chapter 4: Lessons 31-40, Investigation 4

  1. Lesson 1Current

    Lesson 31: Multiplying and Dividing Rational Expressions

    LearnPractice
  2. Lesson 2

    Lesson 32: Solving Linear Systems with Matrix Inverses (Exploration: Exploring Matrix Inverses)

    LearnPractice
  3. Lesson 3

    Lesson 33: Applying Counting Principles

    LearnPractice
  4. Lesson 4

    Lesson 34: Graphing Linear Equations II

    LearnPractice
  5. Lesson 5

    Lesson 35: Solving Quadratic Equations I

    LearnPractice
  6. Lesson 6

    Lesson 36: Using Parallel and Perpendicular Lines

    LearnPractice
  7. Lesson 7

    Lesson 37: Adding and Subtracting Rational Expressions

    LearnPractice
  8. Lesson 8

    Lesson 38: Dividing Polynomials Using Long Division

    LearnPractice
  9. Lesson 9

    Lesson 39: Graphing Linear Inequalities in Two Variables

    LearnPractice
  10. Lesson 10

    Lesson 40: Simplifying Radical Expressions

    LearnPractice
  11. Lesson 11

    Investigation 4: Understanding Cryptography

    LearnPractice