## Intermediate Algebra (12th Edition)

$9t^2-\dfrac{25}{16}$
Using $(a+b)(a-b)=a^2-b^2$ or the product of the sum and difference of like terms, the given expression, $\left(3t-\dfrac{5}{4}\right)\left(3t+\dfrac{5}{4}\right) ,$ is equivalent to \begin{array}{l}\require{cancel} (3t)^2-\left(\dfrac{5}{4}\right)^2 \\\\= 9t^2-\dfrac{25}{16} .\end{array}