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

Lesson 38: Dividing Polynomials Using Long Division

In this Grade 10 Saxon Algebra 2 lesson, students learn how to divide polynomials using long division, applying the same dividend-divisor-quotient-remainder structure as integer long division to expressions involving monomials, linear polynomials, and higher-degree polynomials. Students also use polynomial long division to determine whether one polynomial is a factor of another by checking for a zero remainder. The lesson includes real-world application problems such as finding the ratio of volume to surface area for rectangular prisms with polynomial dimensions.

Section 1

๐Ÿ“˜ Dividing Polynomials Using Long Division

New Concept

The result can be expressed as:

ย Dividend=Divisorร—Quotient+Remainder \ \text{Dividend} = \text{Divisor} \times \text{Quotient} + \text{Remainder}

Whatโ€™s next

Next, youโ€™ll apply this method to divide various polynomials and test if one polynomial is a factor of another.

Section 2

Polynomial long division

Property

Polynomial long division is a method for dividing a polynomial by another polynomial of the same or lower degree. The final result is expressed in the form:

Dividend=Divisorร—Quotient+Remainder\text{Dividend} = \text{Divisor} \times \text{Quotient} + \text{Remainder}

To divide (x2+7x+10)(x^2 + 7x + 10) by (x+2)(x + 2), we perform long division to find the quotient is (x+5)(x + 5) with a remainder of 00.
To divide (2x3โˆ’3x2โˆ’10x+3)(2x^3 - 3x^2 - 10x + 3) by (xโˆ’3)(x - 3), the result is a quotient of (2x2+3xโˆ’1)(2x^2 + 3x - 1) with a remainder of 00.
When dividing (x2+5x+8)(x^2 + 5x + 8) by (x+3)(x + 3), the quotient is (x+2)(x + 2) with a remainder of 22, written as (x+2)+2x+3(x+2) + \frac{2}{x+3}.

Think of this as the algebraic version of dividing 125 by 5. You divide the leading terms, multiply the result by the divisor, subtract the product, and bring down the next term. This process repeats until you have a remainder that is of a lower degree than the divisor, systematically simplifying complex polynomial fractions.

Section 3

Don't forget the zero placeholder

Property

When a dividend is missing a term for a specific power of the variable, you must insert a placeholder term with a coefficient of 0. For example, to divide (4x4โˆ’x3โˆ’11xโˆ’484)(4x^4 - x^3 - 11x - 484), you should rewrite it as 4x4โˆ’x3+0x2โˆ’11xโˆ’4844x^4 - x^3 + 0x^2 - 11x - 484.

To divide (x3โˆ’8)(x^3 - 8) by (xโˆ’2)(x - 2), rewrite the dividend as (x3+0x2+0xโˆ’8)(x^3 + 0x^2 + 0x - 8) to keep columns aligned.
To divide (3y4+2yโˆ’5)(3y^4 + 2y - 5) by (y+1)(y + 1), rewrite it as (3y4+0y3+0y2+2yโˆ’5)(3y^4 + 0y^3 + 0y^2 + 2y - 5) before starting the division.

Imagine lining up soldiers by rank. If a rank is missing, you leave a space for it, right? Using a zero placeholder like '0x20x^2' does the same thing. It keeps all your terms aligned correctly during the subtraction steps of long division, preventing you from mixing up different powers of xx and getting a chaotic result.

Section 4

Is it a factor? Check the remainder!

Property

A polynomial, such as (xโˆ’c)(x - c), is considered a factor of another polynomial if the remainder is 0 after division. If the remainder is any non-zero value, then it is not a factor.

Is (x+2)(x + 2) a factor of (2x3โˆ’x2โˆ’7x+6)(2x^3 - x^2 - 7x + 6)? Yes, because dividing gives a remainder of 00.
Is (x+3)(x + 3) a factor of (6x3โˆ’6x2โˆ’6x+6)(6x^3 - 6x^2 - 6x + 6)? No, because the division results in a non-zero remainder of 204204.

Think of factors as puzzle pieces that fit perfectly. When you divide one polynomial by another, a remainder of zero means it's a perfect fitโ€”the divisor is a factor! Any other remainder means the pieces don't align correctly, so it's not a factor. This gives you a clear 'yes' or 'no' answer every single time.

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 1

    Lesson 31: Multiplying and Dividing Rational Expressions

  2. Lesson 2

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

  3. Lesson 3

    Lesson 33: Applying Counting Principles

  4. Lesson 4

    Lesson 34: Graphing Linear Equations II

  5. Lesson 5

    Lesson 35: Solving Quadratic Equations I

  6. Lesson 6

    Lesson 36: Using Parallel and Perpendicular Lines

  7. Lesson 7

    Lesson 37: Adding and Subtracting Rational Expressions

  8. Lesson 8Current

    Lesson 38: Dividing Polynomials Using Long Division

  9. Lesson 9

    Lesson 39: Graphing Linear Inequalities in Two Variables

  10. Lesson 10

    Lesson 40: Simplifying Radical Expressions

  11. Lesson 11

    Investigation 4: Understanding Cryptography

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

๐Ÿ“˜ Dividing Polynomials Using Long Division

New Concept

The result can be expressed as:

ย Dividend=Divisorร—Quotient+Remainder \ \text{Dividend} = \text{Divisor} \times \text{Quotient} + \text{Remainder}

Whatโ€™s next

Next, youโ€™ll apply this method to divide various polynomials and test if one polynomial is a factor of another.

Section 2

Polynomial long division

Property

Polynomial long division is a method for dividing a polynomial by another polynomial of the same or lower degree. The final result is expressed in the form:

Dividend=Divisorร—Quotient+Remainder\text{Dividend} = \text{Divisor} \times \text{Quotient} + \text{Remainder}

To divide (x2+7x+10)(x^2 + 7x + 10) by (x+2)(x + 2), we perform long division to find the quotient is (x+5)(x + 5) with a remainder of 00.
To divide (2x3โˆ’3x2โˆ’10x+3)(2x^3 - 3x^2 - 10x + 3) by (xโˆ’3)(x - 3), the result is a quotient of (2x2+3xโˆ’1)(2x^2 + 3x - 1) with a remainder of 00.
When dividing (x2+5x+8)(x^2 + 5x + 8) by (x+3)(x + 3), the quotient is (x+2)(x + 2) with a remainder of 22, written as (x+2)+2x+3(x+2) + \frac{2}{x+3}.

Think of this as the algebraic version of dividing 125 by 5. You divide the leading terms, multiply the result by the divisor, subtract the product, and bring down the next term. This process repeats until you have a remainder that is of a lower degree than the divisor, systematically simplifying complex polynomial fractions.

Section 3

Don't forget the zero placeholder

Property

When a dividend is missing a term for a specific power of the variable, you must insert a placeholder term with a coefficient of 0. For example, to divide (4x4โˆ’x3โˆ’11xโˆ’484)(4x^4 - x^3 - 11x - 484), you should rewrite it as 4x4โˆ’x3+0x2โˆ’11xโˆ’4844x^4 - x^3 + 0x^2 - 11x - 484.

To divide (x3โˆ’8)(x^3 - 8) by (xโˆ’2)(x - 2), rewrite the dividend as (x3+0x2+0xโˆ’8)(x^3 + 0x^2 + 0x - 8) to keep columns aligned.
To divide (3y4+2yโˆ’5)(3y^4 + 2y - 5) by (y+1)(y + 1), rewrite it as (3y4+0y3+0y2+2yโˆ’5)(3y^4 + 0y^3 + 0y^2 + 2y - 5) before starting the division.

Imagine lining up soldiers by rank. If a rank is missing, you leave a space for it, right? Using a zero placeholder like '0x20x^2' does the same thing. It keeps all your terms aligned correctly during the subtraction steps of long division, preventing you from mixing up different powers of xx and getting a chaotic result.

Section 4

Is it a factor? Check the remainder!

Property

A polynomial, such as (xโˆ’c)(x - c), is considered a factor of another polynomial if the remainder is 0 after division. If the remainder is any non-zero value, then it is not a factor.

Is (x+2)(x + 2) a factor of (2x3โˆ’x2โˆ’7x+6)(2x^3 - x^2 - 7x + 6)? Yes, because dividing gives a remainder of 00.
Is (x+3)(x + 3) a factor of (6x3โˆ’6x2โˆ’6x+6)(6x^3 - 6x^2 - 6x + 6)? No, because the division results in a non-zero remainder of 204204.

Think of factors as puzzle pieces that fit perfectly. When you divide one polynomial by another, a remainder of zero means it's a perfect fitโ€”the divisor is a factor! Any other remainder means the pieces don't align correctly, so it's not a factor. This gives you a clear 'yes' or 'no' answer every single time.

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 1

    Lesson 31: Multiplying and Dividing Rational Expressions

  2. Lesson 2

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

  3. Lesson 3

    Lesson 33: Applying Counting Principles

  4. Lesson 4

    Lesson 34: Graphing Linear Equations II

  5. Lesson 5

    Lesson 35: Solving Quadratic Equations I

  6. Lesson 6

    Lesson 36: Using Parallel and Perpendicular Lines

  7. Lesson 7

    Lesson 37: Adding and Subtracting Rational Expressions

  8. Lesson 8Current

    Lesson 38: Dividing Polynomials Using Long Division

  9. Lesson 9

    Lesson 39: Graphing Linear Inequalities in Two Variables

  10. Lesson 10

    Lesson 40: Simplifying Radical Expressions

  11. Lesson 11

    Investigation 4: Understanding Cryptography