#### Answer

$2(2n-3)(n+8)$

#### Work Step by Step

Since $2$ is common to all the terms of the equation, we take it out as a common factor:
$4n^{2}+26n-48$
=$2(2n^{2}+13n-24)$
In accordance with the rules of factoring trinomials, we need to find two integers whose product is $-48$ and whose sum is $13$. A little searching determines that these numbers are $-3$ and $16$. Therefore, we can express the middle term $13n$ as $(-3n+16n)$ and proceed to factoring by grouping:
$2(2n^{2}+13n-24)$
=$2(2n^{2}-3n+16n-24)$
=$2[n(2n-3)+8(2n-3)]$
=$2[(2n-3)(n+8)]$
=$2(2n-3)(n+8)$