Factor each polynomial completely. See Example 6. 25s⁴ - 9t²

Verified step by step guidance

1. Recognize that the given expression \$25s^{4} - 9t^{2}\( is a difference of squares because it can be written as \)(5s^{2})^{2} - (3t)^{2}$.
2. Apply the difference of squares formula: \(a^{2} - b^{2} = (a - b)(a + b)\), where \(a = 5s^{2}\) and \(b = 3t\).
3. Rewrite the expression as \((5s^{2} - 3t)(5s^{2} + 3t)\) after applying the difference of squares factorization.
4. Check if either factor can be factored further. Notice that \$5s^{2} - 3t\( and \)5s^{2} + 3t$ are not difference or sum of squares or any other common factorable forms.
5. Conclude that the complete factorization of the polynomial is \((5s^{2} - 3t)(5s^{2} + 3t)\).

Key Concepts

Difference of Squares

The difference of squares is a factoring technique used when an expression is in the form a² - b². It factors into (a - b)(a + b). Recognizing this pattern helps simplify polynomials like 25s⁴ - 9t² by identifying perfect squares.

Factoring Higher Powers

When variables have exponents greater than 2, such as s⁴, it can be helpful to rewrite them as powers squared (e.g., s⁴ = (s²)²). This allows the use of difference of squares or other factoring methods on more complex terms.

Prime Factorization of Coefficients

Breaking down numerical coefficients into their prime factors helps identify perfect squares and simplifies the factoring process. For example, 25 and 9 are perfect squares (5² and 3²), which is essential for applying the difference of squares method.