#### Answer

$(4p+q)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To factor the given expression, $
16p^2-4pq-2q^2
,$ factor first the $GCF.$ Then, find two numbers whose product is $ac$ and whose sum is $b$ in the quadratic expression $ax^2+bx+c.$ Use these $2$ numbers to decompose the middle term of the given quadratic expression and then use factoring by grouping.
$\bf{\text{Solution Details:}}$
Factoring the $GCF=
2
,$ the given expression is equivalent to
\begin{array}{l}\require{cancel}
2(8p^2-2pq-q^2)
.\end{array}
In the trinomial expression above the value of $ac$ is $
8(-1)=-8
$ and the value of $b$ is $
-2
.$
The possible pairs of integers whose product is $ac$ are
\begin{array}{l}\require{cancel}
\{1,-8\}, \{2,-4\},
\{-1,8\}, \{-2,4\}
.\end{array}
Among these pairs, the one that gives a sum of $b$ is $\{
2,-4
\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
2(8p^2+2pq-4pq-q^2)
.\end{array}
Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to
\begin{array}{l}\require{cancel}
2[(8p^2+2pq)-(4pq+q^2)]
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
2[2p(4p+q)-q(4p+q)]
.\end{array}
Factoring the $GCF=
(4p+q)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
2[(4p+q)(2p-q)]
\\\\=
2(4p+q)(2p-q)
.\end{array}
The missing factor of the given expression is $
(4p+q)
.$