## Intermediate Algebra (6th Edition)

$4x^2-\dfrac{1}{9}$
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*}