#### Answer

$(x+3+5y)(x+3-5y)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To factor the given expression, $
x^2+6x+9-25y^2
,$ group the first $3$ terms since these form a perfect square trinomial. Factor this trinomial. The resulting expression becomes a difference of $2$ squares. Factor this using the factoring of the difference of $2$ squares.
$\bf{\text{Solution Details:}}$
Grouping the first $3$ terms, the given expression is equivalent to
\begin{array}{l}\require{cancel}
(x^2+6x+9)-25y^2
.\end{array}
To factor the trinomial expression above, note that the value of $ac$ is $
1(9)=9
$ and the value of $b$ is $
6
.$
The possible pairs of integers whose product is $ac$ are
\begin{array}{l}\require{cancel}
\{1,9\}, \{3,3\},
\\
\{-1,-9\}, \{-3,-3\}
.\end{array}
Among these pairs, the one that gives a sum of $b$ is $\{
3,3
\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
(x^2+3x+3x+9)-25y^2
.\end{array}
Grouping the first and second terms and the third and fourth terms of the trinomial, the given expression is equivalent to
\begin{array}{l}\require{cancel}
[(x^2+3x)+(3x+9)]-25y^2
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
[x(x+3)+3(x+3)]-25y^2
.\end{array}
Factoring the $GCF=
(x+3)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
[(x+3)(x+3)]-25y^2
\\\\=
(x+3)^2-25y^2
.\end{array}
The expressions $
(x+3)^2
$ and $
25y^2
$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $
(x+3)^2-25y^2
,$ is a difference of $2$ squares. Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
(x+3)^2-(5y)^2
\\\\=
[(x+3)+5y][(x+3)-5y]
\\\\=
(x+3+5y)(x+3-5y)
.\end{array}