## Intermediate Algebra (6th Edition)

$16x^2-1$
Using $(a+b)(a-b)=a^2-b^2$ or the product of the sum and difference of like terms, the expression, $(4x+1)(4x-1)$, is equivalent to \begin{array}{l} (4x)^2-(1)^2 \\\\= 16x^2-1 .\end{array}