Algebra 1: Common Core (15th Edition)

Perfect square, $(9r-5)^{2}$.
$81r^{2}-90r+25=$ ...write first and last terms as squares. $=(9r)^{2}-90r+(5)^{2}$ ...does the middle term equal $2ab$? $90r=2\cdot(9r)\cdot(5)$ $=(9r)^{2}-2\cdot(9r)\cdot(5)+(5)^{2}$ ...write as the square of a binomial. $=(9r-5)^{2}$ Perfect square, $(9r-5)^{2}$.