Factoring 9x^2-1 Using Algebraic Techniques

Factoring 9x^2-1 can be accomplished using a well-known algebraic identity, the difference of squares. This identity can help us break down the expression into simpler factors, making it easier to work with and understand.

Using the Difference of Squares Identity

The difference of squares identity is given by:

$$a^2 - b^2 (a b)(a - b)$$

Let's apply this identity to the expression x2-1 within the larger expression 9x^2-1. First, we recognize that 9x^2 is a perfect square:

[9x^2 (3x)^2]

Step-by-Step Factorization

Determine the perfect squares in the expression: [9x^2 - 1 (3x)^2 - 1^2] Apply the difference of squares identity: [(3x)^2 - 1^2 (3x 1)(3x - 1)]

Thus, the factorization of 9x^2-1 is:

[9x^2 - 1 (3x 1)(3x - 1)]

Examples and Further Insights

Let's consider a more complex example, such as 9992-1. Applying the same principles, we can factor this expression as follows:

[999^2 - 1 999 - 1 cdot (999 1) 998 cdot 1000] Factor 1000 into its prime factors: [1000 10^3 2^3 cdot 5^3] Factor 998 into its prime components: [998 2 cdot 499] Given that 499 is prime, the complete factorization is: [999^2 - 1 2^4 cdot 5^3 cdot 499]

Conclusion

By using the difference of squares identity, we can effectively factor expressions like 9x^2-1 and 999^2-1. This method is not only useful for simplifying algebraic expressions, but it also helps in solving more complex problems involving factorization and simplification.

Key Takeaways

The difference of squares identity is a powerful tool in algebra for simplifying expressions. It can be applied to both simple and complex expressions, breaking them down into more manageable parts. Understanding the prime factorization of numbers is crucial for complete factorization.

By leveraging these algebraic techniques, students and mathematicians can solve a wide range of problems more efficiently and effectively.