#### Answer

$(x+2)(4x+3)(4x-3)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
Group the terms of the given expression, $
16x^3+32x^2-9x-18
,$ such that the factored form of the groupings will result to a factor that is common to the entire expression. Then, factor the $GCF$ in each group. and factor the $GCF$ of the entire expression. Finally, factor the resulting difference of $2$ squares.
$\bf{\text{Solution Details:}}$
Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to
\begin{array}{l}\require{cancel}
(16x^3+32x^2)-(9x+18)
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
16x^2(x+2)-9(x+2)
.\end{array}
Factoring the $GCF=
(x+2)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(x+2)(16x^2-9)
.\end{array}
The expressions $
16x^2
$ and $
9
$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $
16x^2-9
,$ 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+2)[(4x)^2-(3)^2]
\\\\=
(x+2)[(4x+3)(4x-3)]
\\\\=
(x+2)(4x+3)(4x-3)
.\end{array}