#### Answer

$[(t-3)^{n}+1][7(t-3)^n-2]$

#### Work Step by Step

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}