## Intermediate Algebra (12th Edition)

$(2r-3+s)(2r-3-s)$
$\bf{\text{Solution Outline:}}$ To factor the given expression, $4r^2-12r+9-s^2 ,$ group the first $3$ terms since these form a perfect square trinomial. Then factor the trinomial. The resulting expression becomes a difference of $2$ squares. Factor this expression using the factoring of the difference of $2$ squares. $\bf{\text{Solution Details:}}$ Grouping the first $3$ terms above results to \begin{array}{l}\require{cancel} (4r^2-12r+9)-s^2 .\end{array} The trinomial above is a perfect square trinomial. Using $a^2\pm2ab+b^2=(a\pm b)^2,$ the expression above is equivalent to \begin{array}{l}\require{cancel} (2r-3)^2-s^2 .\end{array} The expression above is a difference of $2$ squares. Using $a^2-b^2=(a+b)(a-b),$ the factored form of the expression above is \begin{array}{l}\require{cancel} [(2r-3)+s][(2r-3)-s] \\\\= (2r-3+s)(2r-3-s) .\end{array}