#### Answer

$n^{2}+7n-98$

#### Work Step by Step

To find the product of $(n+14)$ and $(n-7)$, we need to use the distributive property in which each term of the first polynomial is multiplied to each term of the second polynomial:
$(n+14)(n-7)$
=$n(n-7)+14(n-7)$
=$n^{2}-7n+14n-98$
=$n^{2}+7n-98$