What is

The expression $a^2 – b^2$ is known as the difference of squares. This is a common algebraic identity that can be factored in a specific way.

Understanding the Difference of Squares

Definition

The difference of squares refers to the subtraction of one squared term from another squared term. Mathematically, it is written as:
$a^2 – b^2$

Factoring the Difference of Squares

One of the most useful properties of the difference of squares is that it can be factored into the product of two binomials. The formula for factoring $a^2 – b^2$ is:
$a^2 – b^2 = (a + b)(a – b)$

Proof of the Formula

Let’s verify this formula by expanding the right-hand side:

$(a + b)(a – b)$

Using the distributive property (also known as the FOIL method for binomials), we get:

$(a + b)(a – b) = a(a – b) + b(a – b)$

Expanding each term separately, we obtain:

$= a^2 – ab + ab – b^2$

Notice that the $-ab$ and $+ab$ terms cancel each other out, leaving us with:

$= a^2 – b^2$

This confirms that:

$a^2 – b^2 = (a + b)(a – b)$

Examples to Illustrate

Example 1

Let’s factor $9 – 4$

First, recognize that $9$ and $4$ are perfect squares:

$9 = 3^2$ and $4 = 2^2$

So, $9 – 4$ can be written as:

$3^2 – 2^2$

Using the difference of squares formula:

$3^2 – 2^2 = (3 + 2)(3 – 2) = 5 times 1 = 5$

Example 2

Factor $x^2 – 25$

Recognize that $x^2$ and $25$ are perfect squares:

$x^2$ and $25 = 5^2$

So, $x^2 – 25$ can be written as:

$x^2 – 5^2$

Using the difference of squares formula:

$x^2 – 5^2 = (x + 5)(x – 5)$

Example 3

Factor $4y^2 – 81$

Recognize that $4y^2$ and $81$ are perfect squares:

$4y^2 = (2y)^2$ and $81 = 9^2$

So, $4y^2 – 81$ can be written as:

$(2y)^2 – 9^2$

Using the difference of squares formula:

$(2y)^2 – 9^2 = (2y + 9)(2y – 9)$

Why is This Important?

Understanding how to factor the difference of squares is crucial for solving various algebraic problems. It simplifies expressions and equations, making them easier to work with. This technique is widely used in algebra, calculus, and even in real-world applications such as physics and engineering.

Conclusion

The difference of squares, $a^2 – b^2$, is a fundamental algebraic identity that can be factored into $(a + b)(a – b)$. This property is not only a powerful tool in simplifying mathematical expressions but also serves as a stepping stone for more advanced topics in mathematics.

Citations

  1. 1. Khan Academy – Factoring
  2. 2. Math is Fun – Factoring
  3. 3. Purplemath – Factoring Polynomials

Related

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H + HO2 → O2 + H2 k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O2 k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) H + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-5 s^-1) φ

Table 1 Reactions, rate constants and activation energies used in the model* No. Reaction kopt (M⁻¹ s⁻¹) 1 OH + H₂ → H + H₂O 3.74 x 10⁷ 2 OH + HO₂ → HO₂ + OH⁻ 5 x 10⁹ 3 OH + H₂O₂ → HO₂ + H₂O 3.8 x 10⁷ 4 OH + O₂ → O₂ + OH 9.96 x 10⁹ 5 OH + HO₂ → O₂ + H₂O 7.1 x 10⁹ 6 OH + OH → H₂O₂ 5.3 x 10⁹ 7 OH + e⁻aq → OH⁻ 3 x 10¹⁰ 8 H + O₂ → HO₂ 2.0 x 10¹⁰ 9 H + HO₂ → H₂O₂ 2.0 x 10¹⁰ 10 H + H₂O₂ → OH + H₂O 3.44 x 10⁷ 11 H + OH → H₂O 1.4 x 10¹⁰ 12 H + H → H₂ 1.94 x 10¹⁰ 13 e⁻aq + O₂ → O₂⁻ 1.9 x 10¹⁰ 14 e⁻aq + O₂ → HO₂⁻ + OH⁻ 1.3 x 10¹⁰ 15 e⁻aq + HO₂ 2.0 x 10¹⁰ 16 e⁻aq + H₂O₂ 1.1 x 10¹⁰ 17 e⁻aq + HO₂ → OH + OH⁻ 1.3 x 10¹⁰ 18 e⁻aq + H⁺ → H 2.3 x 10¹⁰ 19 e⁻aq + e⁻aq → H₂ + OH⁻ + OH⁻ 2.5 x 10⁹ 20 HO₂ + O₂ → O₂ + HO₂ 1.3 x 10⁹ 21 HO₂ + HO₂ → O₂ + H₂O₂ 8.3 x 10⁵ 22 HO₂ + HO₂ → O₂ + OH + H₂O 3.7 23 HO₂ + HO₂ → O₂ + O₂ + OH + H₂O 7 x 10⁵ s⁻¹ 24 H⁺ + O₂⁻ → HO₂ 4.5 x 10¹⁰ 25 H⁺ + O₂⁻ → O₂ 2.0 x 10¹⁰ 26 H⁺ + OH⁻ 1.4 x 10¹¹ 27 H⁺ + HO₂⁻ 2 x 10¹⁰ 28 H₂O₂ → HO₂ + H⁺ + OH⁻ 2.5 x 10⁻⁵ s⁻¹ 29 H₂O₂ → H⁺ + OH⁻ 1.4 x 10⁻⁷ s⁻¹ 30 O₂ + O₂ → O₂ + HO₂ + OH⁻ 0.3 31 O₂ + H₂O₂ → O₂ + OH + OH 16 32

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H2O + O → 2 OH k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) OH + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-8 s^-1) φ