## College Algebra (11th Edition)

$\dfrac{p+6q}{p+q}$
Factoring the expressions and then cancelling the common factors between the numerator and the denominator, the given expression, $\dfrac{p^2-36q^2}{p^2-12pq+36q^2}\cdot\dfrac{p^2-5pq-6q^2}{p^2+2pq+q^2} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{(p+6q)(p-6q)}{(p-6q)(p-6q)}\cdot\dfrac{(p-6q)(p+q)}{(p+q)(p+q)} \\\\= \dfrac{(p+6q)(\cancel{p-6q})}{(\cancel{p-6q})(\cancel{p-6q})}\cdot\dfrac{(\cancel{p-6q})(\cancel{p+q})}{(p+q)(\cancel{p+q})} \\\\= \dfrac{p+6q}{p+q} .\end{array}