Learn on PengiCalifornia Reveal Math, Algebra 1Unit 9: Polynomials

9-7 Factoring Special Products

In Lesson 9-7 of California Reveal Math Algebra 1, Grade 9 students learn to factor two types of special polynomial products: differences of two squares using the pattern a² − b² = (a + b)(a − b), and perfect square trinomials using the patterns (a + b)² and (a − b)². The lesson includes recognizing when repeated factoring is needed and identifying whether a trinomial meets all three conditions to qualify as a perfect square trinomial.

Section 1

Difference of Two Squares Pattern

Property

a2−b2=(a+b)(a−b)a^2 - b^2 = (a + b)(a - b). This pattern applies whenever you subtract one perfect square from another. This simple but powerful rule allows you to quickly factor expressions that have only two terms separated by a minus sign, where both terms are perfect squares. It is a fundamental pattern in algebra.

Examples

Factor x2−81x^2 - 81: (x+9)(x−9)(x + 9)(x - 9)
Factor 16x2−90016x^2 - 900: (4x+30)(4x−30)(4x + 30)(4x - 30)
Factor 25y2−4z225y^2 - 4z^2: (5y+2z)(5y−2z)(5y + 2z)(5y - 2z)

Explanation

This is the 'opposites attract' of algebra! When a perfect square is fighting a another perfect square (a2−b2a^2 - b^2), they break up into two nearly identical binomials but with opposite signs: (a+b)(a+b) and (a−b)(a-b). When you multiply them back, the middle terms cancel out perfectly, leaving you right where you started. It's clean and simple!

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).

Section 3

Perfect Square Trinomial

Property: a2+2ab+b2=(a+b)2a^2 + 2ab + b^2 = (a + b)^2 and a2−2ab+b2=(a−b)2a^2 - 2ab + b^2 = (a - b)^2. A trinomial is a perfect square if its first and last terms are perfect squares, and the middle term is exactly twice the product of their square roots. Recognizing this pattern is a major shortcut.

Factor x2−14x+49x^2 - 14x + 49: (x−7)2(x - 7)^2
Factor 9m2+24mn+16n29m^2 + 24mn + 16n^2: (3m+4n)2(3m + 4n)^2
Factor y2+20y+100y^2 + 20y + 100: (y+10)2(y + 10)^2

Spotting a perfect square trinomial is like finding a secret passage in a video game. If you see two perfect squares at the ends (a2a^2 and b2b^2) and the middle term is just 2ab2ab (or −2ab-2ab), you can skip all the hard work and jump straight to the answer: (a+b)2(a+b)^2 or (a−b)2(a-b)^2.

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

Difference of Two Squares Pattern

Property

a2−b2=(a+b)(a−b)a^2 - b^2 = (a + b)(a - b). This pattern applies whenever you subtract one perfect square from another. This simple but powerful rule allows you to quickly factor expressions that have only two terms separated by a minus sign, where both terms are perfect squares. It is a fundamental pattern in algebra.

Examples

Factor x2−81x^2 - 81: (x+9)(x−9)(x + 9)(x - 9)
Factor 16x2−90016x^2 - 900: (4x+30)(4x−30)(4x + 30)(4x - 30)
Factor 25y2−4z225y^2 - 4z^2: (5y+2z)(5y−2z)(5y + 2z)(5y - 2z)

Explanation

This is the 'opposites attract' of algebra! When a perfect square is fighting a another perfect square (a2−b2a^2 - b^2), they break up into two nearly identical binomials but with opposite signs: (a+b)(a+b) and (a−b)(a-b). When you multiply them back, the middle terms cancel out perfectly, leaving you right where you started. It's clean and simple!

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).

Section 3

Perfect Square Trinomial

Property: a2+2ab+b2=(a+b)2a^2 + 2ab + b^2 = (a + b)^2 and a2−2ab+b2=(a−b)2a^2 - 2ab + b^2 = (a - b)^2. A trinomial is a perfect square if its first and last terms are perfect squares, and the middle term is exactly twice the product of their square roots. Recognizing this pattern is a major shortcut.

Factor x2−14x+49x^2 - 14x + 49: (x−7)2(x - 7)^2
Factor 9m2+24mn+16n29m^2 + 24mn + 16n^2: (3m+4n)2(3m + 4n)^2
Factor y2+20y+100y^2 + 20y + 100: (y+10)2(y + 10)^2

Spotting a perfect square trinomial is like finding a secret passage in a video game. If you see two perfect squares at the ends (a2a^2 and b2b^2) and the middle term is just 2ab2ab (or −2ab-2ab), you can skip all the hard work and jump straight to the answer: (a+b)2(a+b)^2 or (a−b)2(a-b)^2.