## College Algebra (11th Edition)

$9q^2+30q+25-p^2$
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, $[(3q+5)-p][(3q+5)+p] ,$ is \begin{array}{l}\require{cancel} (3q+5)^2-(p)^2 \\\\= (3q+5)^2-p^2 .\end{array} Using $(a\pm b)^2=a^2\pm2ab+b^2$ or the square of a binomial, the expression, $(3q+5)^2-p^2 ,$ is equivalent to \begin{array}{l}\require{cancel} (3q)+2(3q)(5)+(5)^2-p^2 \\\\= 9q^2+30q+25-p^2 .\end{array}