Answer
$(x-3+7y)(x-3-7y)$
Work Step by Step
The first three terms form a perfect square, $(a -b)^{2}=a^{2}-2ab+b^{2}$
a=x and b=3
the last term is a square of $7y$
$=( x-6)^{2}-(7y)^{2}$
now we have a difference of squares, $(a+b)(a-b)=a^{2}-b^{2}$
=$[(x-3)+7y]\cdot[(x-3)-7y]$
= $(x-3+7y)(x-3-7y)$
