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

$[(t-3)^{n}+1][7(t-3)^n-2]$
Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} 7(t-3)^{2n}+5(t-3)^n-2 \end{array} has $ac= 7(-2)=-14$ and $b= 5 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ 7,-2 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 7(t-3)^{2n}+7(t-3)^n-2(t-3)^n-2 .\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} [7(t-3)^{2n}+7(t-3)^n]-[2(t-3)^n+2] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 7(t-3)^n[(t-3)^{n}+1]-2[(t-3)^n+1] .\end{array} Factoring the $GCF= (a^{3n}-1)$ of the entire expression above results to \begin{array}{l}\require{cancel} [(t-3)^{n}+1][7(t-3)^n-2] .\end{array}