Chapter 5 - Polynomials and Factoring - 5.4 Factoring Perfect-Square Trinomials and Differences of Squares - 5.4 Exercise Set - Page 333: 104

$\dfrac{1}{3}(3x+1)(3x-1)$

Factoring $ \dfrac{1}{3} $, the given expression, $ 3x^2-\dfrac{1}{3} ,$ is equivalent to \begin{array}{l} \dfrac{1}{3}(9x^2-1) .\end{array} Using $x^2-y^2=(x+y)(x-y)$ or the factoring of the difference of 2 squares, then the factored form of the expression, $ \dfrac{1}{3}(9x^2-1) ,$ is \begin{array}{l} \dfrac{1}{3}(3x+1)(3x-1) .\end{array}