Answer
$16(2t+1)(2t-1)$
Work Step by Step
Factoring the $GCF=
16
,$ the given expression, $
64t^2-16
,$ is equivalent to
\begin{align*}
16(4t^2-1)
\end{align*}
Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to
\begin{align*}
&
16[(2t)^2-(1)^2]
\\&=
16[(2t+1)(2t-1)]
\\&=
16(2t+1)(2t-1)
\end{align*}
