Factor. 25y^2 - 16 ______

Recognize as difference of squares: Determine the approach to factor 25y^2 - 16. We can recognize this expression as a difference of squares because it is in the form a^2 - b^2, where both 25y^2 and 16 are perfect squares.Identify perfect squares: Identify the squares in the expression 25y^2 - 16. 25y^2 can be written as (5y)^2 because 5y * 5y = 25y^2. 16 can be written as 4^2 because 4 * 4 = 16. So, 25y^2 - 16 can be rewritten as (5y)^2 - 4^2.Apply difference of squares formula: Apply the difference of squares formula to factor the expression. The difference of squares formula is a^2 - b^2 = (a - b)(a + b). Using this formula, we can factor (5y)^2 - 4^2 as (5y - 4)(5y + 4).

Home

Math Problems

Algebra 1

Factor quadratics: special cases

Full solution

Q. Factor.

\newline

25y^2 - 16

Recognize as difference of squares: Determine the approach to factor $25y^2 - 16$ . We can recognize this expression as a difference of squares because it is in the form $a^2 - b^2$ , where both $25y^2$ and $16$ are perfect squares.
Identify perfect squares: Identify the squares in the expression $25y^2 - 16$ . $25y^2$ can be written as $(5y)^2$ because $5y \times 5y = 25y^2$ . $16$ can be written as $4^2$ because $4 \times 4 = 16$ . So, $25y^2 - 16$ can be rewritten as $(5y)^2 - 4^2$ .
Apply difference of squares formula: Apply the difference of squares formula to factor the expression. $\newline$ The difference of squares formula is $a^2 - b^2 = (a - b)(a + b)$ . $\newline$ Using this formula, we can factor $(5y)^2 - 4^2$ as $(5y - 4)(5y + 4)$ .