Answer
$(4q+7)^{2}$
Work Step by Step
$16q^{2}+56q+49=$
...write first and last terms as squares.
$=(4q)^{2}+56q+(7)^{2}$
...does middle term equal $2ab$?
$56q=2\cdot(4q)\cdot(7) ?$
yes.
$=(4q)^{2}+2\cdot(4q)\cdot(7)+(7)^{2}$
...write as the square of a binomial.
$=(4q+7)^{2}$
