Answer
$4x^2-\dfrac{1}{9}$
Work Step by Step
Using the special product of the sum and difference of like terms, $(a+b)(a-b)=a^2-b^2$, the given expression is equivalent to
\begin{align*}
&
\left( 2x-\dfrac{1}{3} \right)\left( 2x+\dfrac{1}{3} \right)
\\\\&
(2x)^2-\left( \dfrac{1}{3} \right)^2
\\\\&
4x^2-\dfrac{1}{9}
.\end{align*}
