Factoring the Difference of Squares: 16y²w⁸ – 81x⁶

Factoring is a fundamental concept in algebra that involves breaking down a mathematical expression into simpler expressions that, when multiplied together, result in the original expression. In this case, we’re dealing with a binomial expression, meaning it consists of two terms: 16y²w⁸ and 81x⁶. To factor this expression, we’ll utilize a specific pattern known as the difference of squares.

Understanding the Difference of Squares

The difference of squares pattern is a useful tool for factoring expressions that follow a specific structure. It states that the difference of two perfect squares can be factored into the product of the sum and difference of their square roots. Mathematically, this can be represented as:

$a^2 – b^2 = (a + b)(a – b)$

Let’s break down this pattern:

  • a²: Represents the first term, which is a perfect square (a number obtained by squaring another number). For example, 9 is a perfect square because it’s the result of 3 squared (3² = 9).
  • b²: Represents the second term, also a perfect square.
  • (a + b): The sum of the square roots of the first and second terms.
  • (a – b): The difference of the square roots of the first and second terms.

Applying the Difference of Squares to Our Expression

Let’s analyze our expression, 16y²w⁸ – 81x⁶, to see if it fits the difference of squares pattern:

  1. Identify Perfect Squares: Both 16y²w⁸ and 81x⁶ are perfect squares. Let’s break them down:

    • 16y²w⁸: This is the square of 4yw⁴ (4yw⁴)² = 16y²w⁸
    • 81x⁶: This is the square of 9x³ (9x³)² = 81x⁶

  2. Apply the Pattern: Now that we’ve identified the perfect squares, we can apply the difference of squares pattern:

    • a² = 16y²w⁸ => a = 4yw⁴
    • b² = 81x⁶ => b = 9x³

  3. Factor the Expression: Substituting the values of a and b into the difference of squares formula, we get:

    • 16y²w⁸ – 81x⁶ = (4yw⁴ + 9x³)(4yw⁴ – 9x³)

Conclusion

Therefore, the factored form of the expression 16y²w⁸ – 81x⁶ is (4yw⁴ + 9x³)(4yw⁴ – 9x³). By recognizing the difference of squares pattern and applying the appropriate formula, we successfully factored the binomial expression into two simpler expressions. This factoring process is crucial in various algebraic manipulations, including solving equations and simplifying expressions.

Additional Examples

Let’s explore a few more examples to solidify our understanding of factoring the difference of squares:

  1. Factoring 49a² – 25b²:

    • Identify perfect squares: 49a² = (7a)² and 25b² = (5b)²
    • Apply the pattern: a = 7a and b = 5b
    • Factor: 49a² – 25b² = (7a + 5b)(7a – 5b)

  2. Factoring 100m⁴ – 64n⁶:

    • Identify perfect squares: 100m⁴ = (10m²)² and 64n⁶ = (8n³)²
    • Apply the pattern: a = 10m² and b = 8n³
    • Factor: 100m⁴ – 64n⁶ = (10m² + 8n³)(10m² – 8n³)

  3. Factoring 9x² – 1:

    • Identify perfect squares: 9x² = (3x)² and 1 = (1)²
    • Apply the pattern: a = 3x and b = 1
    • Factor: 9x² – 1 = (3x + 1)(3x – 1)

By practicing with various examples, you’ll become more comfortable recognizing the difference of squares pattern and applying it to factor different expressions. Remember, factoring is a fundamental skill in algebra, and mastering it will pave the way for solving more complex problems in mathematics.

Citations

  1. 1. Khan Academy – Factoring the Difference of Squares
  2. 2. Purplemath – Factoring the Difference of Squares
  3. 3. Math is Fun – Difference of Squares

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