Answer
$(3a-4+b)(3a-4-b)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To factor the given expression, $
9a^2-24a+16-b^2
,$ group the first $3$ terms since these form a perfect square trinomial. Then factor the trinomial. The resulting expression becomes a difference of $2$ squares. Factor this expression using the factoring of the difference of $2$ squares.
$\bf{\text{Solution Details:}}$
Grouping the first $3$ terms above results to
\begin{array}{l}\require{cancel}
(9a^2-24a+16)-b^2
.\end{array}
The trinomial above is a perfect square trinomial. Using $a^2\pm2ab+b^2=(a\pm b)^2,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
(3a-4)^2-b^2
.\end{array}
The expression above is a difference of $2$ squares. Using $a^2-b^2=(a+b)(a-b),$ the factored form of the expression above is
\begin{array}{l}\require{cancel}
[(3a-4)+b][(3a-4)-b]
\\\\=
(3a-4+b)(3a-4-b)
.\end{array}
