## Precalculus (6th Edition) Blitzer

$a.\quad 49x^{2}-36$ $b.\quad (7x+6)(7x-6)$
$a.$ From the square with side of length $7x$, subtract 4 squares with sides of length $3$. $(7x)^{2}-4\cdot 3^{2}=49x^{2}-36$ $b.$ $49x^{2}-36$= ... a difference of squares, $(7x)^{2}-(6)^{2}$ $=(7x+6)(7x-6)$