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

$a^6(3a+4)(3a+2)$
Factoring the $GCF= a^6 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 18a^7+8a^6+9a^8 \\\\= 9a^8+18a^7+8a^6 \\\\= a^6(9a^2+18a+8) .\end{array} Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} a^6(9a^2+18a+8) \end{array} has $ac= 9(8)=72$ and $b= 18 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ 12,6 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} a^6(9a^2+12a+6a+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} a^6[(9a^2+12a)+(6a+8)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} a^6[3a(3a+4)+2(3a+4)] .\end{array} Factoring the $GCF= (3a+4)$ of the entire expression above results to \begin{array}{l}\require{cancel} a^6[(3a+4)(3a+2)] \\\\= a^6(3a+4)(3a+2) .\end{array}