## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

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