## Precalculus (6th Edition)

$$49m^{2}-4n^{2}$$
Initial equation: $$(7m+2n)(7m-2n)$$ First, we have the distribute the $7m$ located in the first set of parenthesis to the second, which involves multiplying $7m \times 7m$, and $7m \times 2n$, which will give you $49m^{2}-14mn$ Then, distribute the $2n$ located in the first set of parenthesis to the second, which using the same steps as the first distribution, gives you $14mn-4n^{2}$ Combining the products of both distributions gives you: $$49m^{2}+14mn-14mn-4n^{2}$$ To arrive at the final answer, simplify by combining like terms: $$49m^{2}-4n^{2}$$