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

$2(5a-9)(a+1)$
Factoring the $GCF= 2 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 10a^2-8a-18 \\\\= 2(5a^2-4a-9) .\end{array} Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} 2(5a^2-4a-9) \end{array} has $ac= 5(-9)=-45$ and $b= -4 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ -9,5 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 2(5a^2-9a+5a-9) .\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[(5a^2-9a)+(5a-9)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 2[a(5a-9)+(5a-9)] .\end{array} Factoring the $GCF= (5a-9)$ of the entire expression above results to \begin{array}{l}\require{cancel} 2[(5a-9)(a+1)] \\\\= 2(5a-9)(a+1) .\end{array}