Learn on PengienVision, Algebra 1Chapter 7: Polynomials and Factoring

Lesson 7: Factoring Special Cases

In this Grade 11 enVision Algebra 1 lesson, students learn to factor special cases including perfect-square trinomials using the patterns a² + 2ab + b² = (a + b)² and a² − 2ab + b² = (a − b)², as well as the difference of two squares pattern a² − b² = (a + b)(a − b). Students also practice factoring out a greatest common factor before applying these special-case patterns to multi-term expressions.

Section 1

Factor Perfect Square Trinomials

Property

If aa and bb are real numbers, the perfect square trinomials pattern is as follows:

a2+2ab+b2=(a+b)2a^2 + 2ab + b^2 = (a+b)^2
a2−2ab+b2=(a−b)2a^2 - 2ab + b^2 = (a-b)^2
To use this pattern, first verify that the trinomial fits.
Check if the first term is a perfect square (a2a^2) and the last term is a perfect square (b2b^2).
Then, check if the middle term is twice their product (2ab2ab).
If it matches, write the square of the binomial (a+b)2(a+b)^2 or (a−b)2(a-b)^2.

Examples

  • To factor 25x2+30x+925x^2 + 30x + 9, recognize it as (5x)2+2(5x)(3)+32(5x)^2 + 2(5x)(3) + 3^2. This fits the pattern a2+2ab+b2a^2+2ab+b^2, so the factored form is (5x+3)2(5x+3)^2.
  • To factor 49y2−42y+949y^2 - 42y + 9, identify it as (7y)2−2(7y)(3)+32(7y)^2 - 2(7y)(3) + 3^2. This matches the pattern a2−2ab+b2a^2-2ab+b^2, so the factored form is (7y−3)2(7y-3)^2.

Section 2

Factor Differences of Squares

Property

If aa and bb are real numbers, a difference of squares factors to a product of conjugates using the following pattern:

a2−b2=(a−b)(a+b)a^2 - b^2 = (a-b)(a+b)

To use this pattern, ensure you have a binomial where two perfect squares are being subtracted.
Write each term as a square, (a)2−(b)2(a)^2 - (b)^2, then write the product of the conjugates, (a−b)(a+b)(a-b)(a+b).
Note that a sum of squares, a2+b2a^2+b^2, is prime and cannot be factored.

Examples

  • To factor 9x2−259x^2 - 25, rewrite the expression as a difference of squares, (3x)2−52(3x)^2 - 5^2. This factors into the product of conjugates (3x−5)(3x+5)(3x-5)(3x+5).
  • To factor 16a2−81b216a^2 - 81b^2, identify the terms as (4a)2(4a)^2 and (9b)2(9b)^2. The factored form is the product of their conjugates, (4a−9b)(4a+9b)(4a-9b)(4a+9b).

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Chapter 7: Polynomials and Factoring

  1. Lesson 1

    Lesson 1: Adding and Subtracting Polynomials

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  2. Lesson 2

    Lesson 2: Multiplying Polynomials

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  3. Lesson 3

    Lesson 3: Multiplying Special Cases

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  4. Lesson 4

    Lesson 4: Factoring Polynomials

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  5. Lesson 5

    Lesson 5: Factoring x² + bx + c

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  6. Lesson 6

    Lesson 6: Factoring ax² + bx + c

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  7. Lesson 7Current

    Lesson 7: Factoring Special Cases

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Lesson overview

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Section 1

Factor Perfect Square Trinomials

Property

If aa and bb are real numbers, the perfect square trinomials pattern is as follows:

a2+2ab+b2=(a+b)2a^2 + 2ab + b^2 = (a+b)^2
a2−2ab+b2=(a−b)2a^2 - 2ab + b^2 = (a-b)^2
To use this pattern, first verify that the trinomial fits.
Check if the first term is a perfect square (a2a^2) and the last term is a perfect square (b2b^2).
Then, check if the middle term is twice their product (2ab2ab).
If it matches, write the square of the binomial (a+b)2(a+b)^2 or (a−b)2(a-b)^2.

Examples

  • To factor 25x2+30x+925x^2 + 30x + 9, recognize it as (5x)2+2(5x)(3)+32(5x)^2 + 2(5x)(3) + 3^2. This fits the pattern a2+2ab+b2a^2+2ab+b^2, so the factored form is (5x+3)2(5x+3)^2.
  • To factor 49y2−42y+949y^2 - 42y + 9, identify it as (7y)2−2(7y)(3)+32(7y)^2 - 2(7y)(3) + 3^2. This matches the pattern a2−2ab+b2a^2-2ab+b^2, so the factored form is (7y−3)2(7y-3)^2.

Section 2

Factor Differences of Squares

Property

If aa and bb are real numbers, a difference of squares factors to a product of conjugates using the following pattern:

a2−b2=(a−b)(a+b)a^2 - b^2 = (a-b)(a+b)

To use this pattern, ensure you have a binomial where two perfect squares are being subtracted.
Write each term as a square, (a)2−(b)2(a)^2 - (b)^2, then write the product of the conjugates, (a−b)(a+b)(a-b)(a+b).
Note that a sum of squares, a2+b2a^2+b^2, is prime and cannot be factored.

Examples

  • To factor 9x2−259x^2 - 25, rewrite the expression as a difference of squares, (3x)2−52(3x)^2 - 5^2. This factors into the product of conjugates (3x−5)(3x+5)(3x-5)(3x+5).
  • To factor 16a2−81b216a^2 - 81b^2, identify the terms as (4a)2(4a)^2 and (9b)2(9b)^2. The factored form is the product of their conjugates, (4a−9b)(4a+9b)(4a-9b)(4a+9b).

Book overview

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

Continue this chapter

Chapter 7: Polynomials and Factoring

  1. Lesson 1

    Lesson 1: Adding and Subtracting Polynomials

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  2. Lesson 2

    Lesson 2: Multiplying Polynomials

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  3. Lesson 3

    Lesson 3: Multiplying Special Cases

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  4. Lesson 4

    Lesson 4: Factoring Polynomials

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  5. Lesson 5

    Lesson 5: Factoring x² + bx + c

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  6. Lesson 6

    Lesson 6: Factoring ax² + bx + c

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  7. Lesson 7Current

    Lesson 7: Factoring Special Cases

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