Difference of Squares - Algebra-1

1. Fundamental Concepts

Definition: The difference of squares is a special case in algebra where a polynomial can be factored into two binomials. It is expressed as $$a^2 - b^2 = (a + b)(a - b)$$.
Identifying Difference of Squares: A polynomial is a difference of squares if it can be written as the square of one term minus the square of another term.
Factoring Process: To factor a difference of squares, identify the terms that are squares and apply the formula $$a^2 - b^2 = (a + b)(a - b)$$.

2. Key Concepts

Basic Rule: $$a^2 - b^2 = (a + b)(a - b)$$

Application: Used to simplify and solve polynomial equations efficiently

Special Cases: Recognizing patterns like $$x^2 - y^2$$ helps in quick factorization

3. Examples

Example 1 (Basic)

Problem: Factor $$x^2 - 9$$

Step-by-Step Solution:

Identify $$a^2$$ and $$b^2$$: Here, $$a^2 = x^2$$ and $$b^2 = 9$$. So, $$a = x$$ and $$b = 3$$.
Apply the formula: $$(x + 3)(x - 3)$$

Validation: Substitute $$x = 2$$ → Original: $$2^2 - 9 = 4 - 9 = -5$$; Simplified: $$(2 + 3)(2 - 3) = 5 \cdot (-1) = -5$$ ✓

Example 2 (Intermediate)

Problem: Factor $$16y^2 - 25z^2$$

Step-by-Step Solution:

Identify $$a^2$$ and $$b^2$$: Here, $$a^2 = 16y^2$$ and $$b^2 = 25z^2$$. So, $$a = 4y$$ and $$b = 5z$$.
Apply the formula: $$(4y + 5z)(4y - 5z)$$

Validation: Substitute $$y = 1, z = 1$$ → Original: $$16(1)^2 - 25(1)^2 = 16 - 25 = -9$$; Simplified: $$(4(1) + 5(1))(4(1) - 5(1)) = 9 \cdot (-1) = -9$$ ✓

4. Problem-Solving Techniques

Pattern Recognition: Look for expressions that fit the form $$a^2 - b^2$$.
Substitution Method: Use substitution to verify the correctness of your factored form.
Practice with Variety: Practice with different types of polynomials to recognize various forms of differences of squares.