#### Answer

$-(3x^{m}-4)(5x^m-2)$

#### Work Step by Step

Factoring the negative $GCF=
-1
,$ the given expression is equivalent to
\begin{array}{l}\require{cancel}
-15x^{2m}+26x^m-8
\\\\=
-(15x^{2m}-26x^m+8)
.\end{array}
Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{
expression
}$
\begin{array}{l}\require{cancel}
-(15x^{2m}-26x^m+8)
\end{array} has $ac=
15(8)=120
$ and $b=
-26
.$
The two numbers with a product of $c$ and a sum of $b$ are $\left\{
-20,-6
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
-(15x^{2m}-20x^m-6x^m+8)
.\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}
-[(15x^{2m}-20x^m)-(6x^m-8)]
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
-[5x^m(3x^{m}-4)-2(3x^m-4)]
.\end{array}
Factoring the $GCF=
(3x^{m}-4)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
-[(3x^{m}-4)(5x^m-2)]
\\\\=
-(3x^{m}-4)(5x^m-2)
.\end{array}