## Algebra 1: Common Core (15th Edition)

$3k^{2}(k+1)$
This becomes : $\frac{k^{2}+k}{5k} \times \frac{15k^{2}}{1}$ $= \frac{k(k+1)}{5k} \times \frac{15k^{2}}{1}$ $= \frac{k(k+1)}{1} \times \frac{3k}{1}$ $= 3k^{2}(k+1)$