Learn on PengiSaxon Algebra 2Chapter 3: Lessons 21-30, Investigation 3

Lesson 28: Simplifying Rational Expressions

In Saxon Algebra 2 Lesson 28, Grade 10 students learn to simplify rational expressions by identifying excluded values, applying the Quotient of Powers Property, factoring out the GCF, using the difference of squares, and handling opposite binomials by factoring out negative one. The lesson covers monomial and polynomial numerators and denominators, including trinomials, with step-by-step examples showing how to divide out common factors. Students also apply simplified rational expressions to real-world problems such as calculating volume-to-surface-area ratios.

Section 1

πŸ“˜ Simplifying Rational Expressions

New Concept

A rational expression is simplified when there are no common factors, other than 1, in the numerator and denominator.

What’s next

Next, you'll learn the rules for simplifying these expressions, starting with the Quotient of Powers Property and factoring.

Section 2

Quotient of Powers Property

For a≠0a \neq 0, and integers mm and nn,

aman=amβˆ’n \frac{a^m}{a^n} = a^{m-n}

Excluded value is 0. Simplify:

5p5p2=5p5βˆ’2=5p3\frac{5p^5}{p^2} = 5p^{5-2} = 5p^3
Excluded value is 0. Simplify:
y3y8=y3βˆ’8=yβˆ’5=1y5\frac{y^3}{y^8} = y^{3-8} = y^{-5} = \frac{1}{y^5}

Imagine a power showdown! When dividing powers with the same base, just subtract the bottom exponent from the top one. It's a shortcut to clean up fractions with monomial terms, making big scary expressions like x9x2\frac{x^9}{x^2} much simpler. Remember to find any excluded values first, where the denominator would be zero, before you start simplifying!

Section 3

Simplifying by Dividing Out Common Factors

A rational expression is simplified when there are no common factors, other than 1, in the numerator and denominator. This is achieved by dividing out common factors.

Excluded value is -3. Simplify:

3x+96x+18=3(x+3)6(x+3)=36=12\frac{3x + 9}{6x + 18} = \frac{3(x + 3)}{6(x + 3)} = \frac{3}{6} = \frac{1}{2}
Excluded value is -2. Simplify:
5c2βˆ’20c+2=5(c2βˆ’4)c+2=5(c+2)(cβˆ’2)c+2=5(cβˆ’2)\frac{5c^2 - 20}{c + 2} = \frac{5(c^2 - 4)}{c + 2} = \frac{5(c+2)(c-2)}{c+2} = 5(c-2)

This is like a matching game for fractions! First, find any 'excluded values' that make the denominator zero. Then, factor the top and bottom expressions completely. If you spot the exact same factor on both levels, you can cancel them out. It’s the ultimate clean-up move for messy polynomial fractions, revealing a much simpler expression underneath.

Section 4

Simplifying by Factoring Out -1

To simplify expressions with opposite binomials, such as aβˆ’bbβˆ’a\frac{a-b}{b-a}, factor out a βˆ’1-1 from one of the binomials. For example, (bβˆ’a)(b-a) can be rewritten as βˆ’1(aβˆ’b)-1(a-b).

Excluded value is 4. Simplify:

5xβˆ’208βˆ’2x=5(xβˆ’4)2(4βˆ’x)=5(xβˆ’4)βˆ’2(xβˆ’4)=βˆ’52\frac{5x - 20}{8 - 2x} = \frac{5(x - 4)}{2(4 - x)} = \frac{5(x - 4)}{-2(x - 4)} = -\frac{5}{2}
Excluded value is 3. Simplify:
y2βˆ’93βˆ’y=(yβˆ’3)(y+3)βˆ’(yβˆ’3)=βˆ’(y+3)\frac{y^2-9}{3-y} = \frac{(y-3)(y+3)}{-(y-3)} = -(y+3)

Ever see factors that are almost identical but backward, like (xβˆ’5)(x-5) and (5βˆ’x)(5-x)? Don't worry! Use a clever trick by factoring out a βˆ’1-1 from one of them. This flips its signs, making it a perfect match for the other factor. Now you can cancel them out, leaving just a βˆ’1-1 behind. It's a ninja move for simplifying!

Book overview

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Chapter 3: Lessons 21-30, Investigation 3

  1. Lesson 1

    Lesson 21: Solving Systems of Equations Using the Substitution Method

  2. Lesson 2

    LAB 5: Graphing Calculator: Storing and Plotting a List of Data

  3. Lesson 3

    Lesson 22: Analyzing Continuous, Discontinuous, and Discrete Functions

  4. Lesson 4

    Lesson 23: Factoring Polynomials

  5. Lesson 5

    Lesson 24: Solving Systems of Equations Using the Elimination Method

  6. Lesson 6

    LAB 6: Graphing Calculator: Calculating 1- and 2-Variable Statistical Data

  7. Lesson 7

    Lesson 25: Finding Measures of Central Tendency and Dispersion

  8. Lesson 8

    Lesson 26: Writing the Equation of a Line

  9. Lesson 9

    Lesson 27: Connecting the Parabola with the Quadratic Function

  10. Lesson 10Current

    Lesson 28: Simplifying Rational Expressions

  11. Lesson 11

    Lesson 29: Solving Systems of Equations in Three Variables

  12. Lesson 12

    Lesson 30: Applying Transformations to the Parabola and Determining the Minimum or Maximum

  13. Lesson 13

    Investigation 3: Graphing Three Linear Equations in Three Variables

Lesson overview

Expand to review the lesson summary and core properties.

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Section 1

πŸ“˜ Simplifying Rational Expressions

New Concept

A rational expression is simplified when there are no common factors, other than 1, in the numerator and denominator.

What’s next

Next, you'll learn the rules for simplifying these expressions, starting with the Quotient of Powers Property and factoring.

Section 2

Quotient of Powers Property

For a≠0a \neq 0, and integers mm and nn,

aman=amβˆ’n \frac{a^m}{a^n} = a^{m-n}

Excluded value is 0. Simplify:

5p5p2=5p5βˆ’2=5p3\frac{5p^5}{p^2} = 5p^{5-2} = 5p^3
Excluded value is 0. Simplify:
y3y8=y3βˆ’8=yβˆ’5=1y5\frac{y^3}{y^8} = y^{3-8} = y^{-5} = \frac{1}{y^5}

Imagine a power showdown! When dividing powers with the same base, just subtract the bottom exponent from the top one. It's a shortcut to clean up fractions with monomial terms, making big scary expressions like x9x2\frac{x^9}{x^2} much simpler. Remember to find any excluded values first, where the denominator would be zero, before you start simplifying!

Section 3

Simplifying by Dividing Out Common Factors

A rational expression is simplified when there are no common factors, other than 1, in the numerator and denominator. This is achieved by dividing out common factors.

Excluded value is -3. Simplify:

3x+96x+18=3(x+3)6(x+3)=36=12\frac{3x + 9}{6x + 18} = \frac{3(x + 3)}{6(x + 3)} = \frac{3}{6} = \frac{1}{2}
Excluded value is -2. Simplify:
5c2βˆ’20c+2=5(c2βˆ’4)c+2=5(c+2)(cβˆ’2)c+2=5(cβˆ’2)\frac{5c^2 - 20}{c + 2} = \frac{5(c^2 - 4)}{c + 2} = \frac{5(c+2)(c-2)}{c+2} = 5(c-2)

This is like a matching game for fractions! First, find any 'excluded values' that make the denominator zero. Then, factor the top and bottom expressions completely. If you spot the exact same factor on both levels, you can cancel them out. It’s the ultimate clean-up move for messy polynomial fractions, revealing a much simpler expression underneath.

Section 4

Simplifying by Factoring Out -1

To simplify expressions with opposite binomials, such as aβˆ’bbβˆ’a\frac{a-b}{b-a}, factor out a βˆ’1-1 from one of the binomials. For example, (bβˆ’a)(b-a) can be rewritten as βˆ’1(aβˆ’b)-1(a-b).

Excluded value is 4. Simplify:

5xβˆ’208βˆ’2x=5(xβˆ’4)2(4βˆ’x)=5(xβˆ’4)βˆ’2(xβˆ’4)=βˆ’52\frac{5x - 20}{8 - 2x} = \frac{5(x - 4)}{2(4 - x)} = \frac{5(x - 4)}{-2(x - 4)} = -\frac{5}{2}
Excluded value is 3. Simplify:
y2βˆ’93βˆ’y=(yβˆ’3)(y+3)βˆ’(yβˆ’3)=βˆ’(y+3)\frac{y^2-9}{3-y} = \frac{(y-3)(y+3)}{-(y-3)} = -(y+3)

Ever see factors that are almost identical but backward, like (xβˆ’5)(x-5) and (5βˆ’x)(5-x)? Don't worry! Use a clever trick by factoring out a βˆ’1-1 from one of them. This flips its signs, making it a perfect match for the other factor. Now you can cancel them out, leaving just a βˆ’1-1 behind. It's a ninja move for simplifying!

Book overview

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

Continue this chapter

Chapter 3: Lessons 21-30, Investigation 3

  1. Lesson 1

    Lesson 21: Solving Systems of Equations Using the Substitution Method

  2. Lesson 2

    LAB 5: Graphing Calculator: Storing and Plotting a List of Data

  3. Lesson 3

    Lesson 22: Analyzing Continuous, Discontinuous, and Discrete Functions

  4. Lesson 4

    Lesson 23: Factoring Polynomials

  5. Lesson 5

    Lesson 24: Solving Systems of Equations Using the Elimination Method

  6. Lesson 6

    LAB 6: Graphing Calculator: Calculating 1- and 2-Variable Statistical Data

  7. Lesson 7

    Lesson 25: Finding Measures of Central Tendency and Dispersion

  8. Lesson 8

    Lesson 26: Writing the Equation of a Line

  9. Lesson 9

    Lesson 27: Connecting the Parabola with the Quadratic Function

  10. Lesson 10Current

    Lesson 28: Simplifying Rational Expressions

  11. Lesson 11

    Lesson 29: Solving Systems of Equations in Three Variables

  12. Lesson 12

    Lesson 30: Applying Transformations to the Parabola and Determining the Minimum or Maximum

  13. Lesson 13

    Investigation 3: Graphing Three Linear Equations in Three Variables