Question: Factor. \((w + f)^2 - 64\) \((w + f)^2 - 64 = \boxed{\quad}\) (Factor completely.)

Factor.

\((w + f)^2 - 64\)

\((w + f)^2 - 64 = \boxed{\quad}\)

(Factor completely.)

Solution

To factor the expression \((w + f)^2 - 64\), we can use the difference of squares formula: \[ a^2 - b^2 = (a - b)(a + b) \] Here, \(a = (w + f)\) and \(b = 8\) since \(8^2 = 64\). Apply the formula: \[ \begin{align*} (w + f)^2 - 64 &= ((w + f) - 8)((w + f) + 8) \end{align*} \] So, the completely factored form is: \[ ((w + f) - 8)((w + f) + 8) \]