## College Algebra (11th Edition)

$81r^2-18rs+s^2-4$
Using $(a+b)(a-b)=a^2-b^2$ or the special product of the sum and difference of like terms, the product of the given expression, $[(9r-s)+2][(9r-s)-2 ,$ is \begin{array}{l}\require{cancel} (9r-s)^2-(2)^2 \\\\= (9r-s)^2-4 .\end{array} Using $(a\pm b)^2=a^2\pm2ab+b^2$ or the square of a binomial, the expression, $(9r-s)^2-4 ,$ is equivalent to \begin{array}{l}\require{cancel} (9r)^2-2(9r)(s)+(s)^2-4 \\\\= 81r^2-18rs+s^2-4 .\end{array}