## Intermediate Algebra (12th Edition)

$(h+k+3)(h+k-3)$
$\bf{\text{Solution Outline:}}$ To factor the given expression, $(h+k)^2-9 ,$ use the factoring of the difference of $2$ squares. $\bf{\text{Solution Details:}}$ Using $a^2-b^2=(a+b)(a-b)$ or the factoring of the difference of $2$ squares, the factored form of the expression above is \begin{array}{l}\require{cancel} [(h+k)+3][(h+k)-3] \\\\= (h+k+3)(h+k-3) .\end{array}