## Precalculus (6th Edition)

$(6z + 9y^{2})(6z - 9y^{2})$
This polynomial is a difference of squares, so $a^{2} - b^{2} = (a-b)(a+b)$. In this case, $a^{2} = 36z^{2}$ so $a = 6z$ and $b^{2} = 81y^{4}$ so $b = 9y^{2}$. Therefore, $(a-b)(a+b) = (6z + 9y^{2})(6z - 9y^{2})$.