1. **Stating the problem:** We want to understand the method used to factor expressions like $p^2 - 1^2$ and $25v^2 - 49$ using special algebraic identities. 2. **Formula used:** The key formula here is the difference of squares identity: $$a^2 - b^2 = (a + b)(a - b)$$ This means any expression that is a difference between two perfect squares can be factored into the product of the sum and difference of their square roots. 3. **Applying the formula to $p^2 - 1^2$:** - Recognize that $p^2$ and $1^2$ are perfect squares. - Using the difference of squares formula: $$p^2 - 1^2 = (p + 1)(p - 1)$$ This shows the factorization of $p^2 - 1$. 4. **Clarifying the incorrect step:** The expression $p^3 - 1^2$ is not a correct continuation from $p^2 - 1$. The difference of squares applies only to expressions of the form $a^2 - b^2$, not $a^3 - b^2$. So, $p^3 - 1^2$ is not factored by this method. 5. **Applying the formula to $25v^2 - 49$:** - Recognize $25v^2 = (5v)^2$ and $49 = 7^2$. - Using the difference of squares formula: $$25v^2 - 49 = (5v + 7)(5v - 7)$$ 6. **Summary:** - The difference of squares formula is a powerful tool to factor expressions where two perfect squares are subtracted. - It transforms $a^2 - b^2$ into $(a + b)(a - b)$. - This method does not apply to expressions like $p^3 - 1^2$. This method simplifies factoring and helps solve equations or simplify expressions efficiently.