Answer
$9(k+1)(7k-3)$
Work Step by Step
Let $z=3k-1.$ Then the given expression, $
7(3k-1)^2+26(3k-1)-8
,$ is equivalent to
\begin{array}{l}\require{cancel}
7z^2+26z-8
.\end{array}
The two numbers whose product is $ac=
7(-8)=-56
$ and whose sum is $b=
26
$ are $\{
28,-2
\}.$ Using these two numbers to decompose the middle term, then the factored form of $
7z^2+26z-8
$ is
\begin{array}{l}\require{cancel}
7z^2+28z-2z-8
\\\\=
(7z^2+28z)-(2z+8)
\\\\=
7z(z+4)-2(z+4)
\\\\=
(z+4)(7z-2)
.\end{array}
Since $z=3k-1$, then the expression $
(z+4)(7z-2)
$ is equivalent to
\begin{array}{l}\require{cancel}
(3k-1+4)(7(3k-1)-2)
\\\\=
(3k-1+4)(21k-7-2)
\\\\=
(3k+3)(21k-9)
\\\\=
3(k+1)3(7k-3)
\\\\=
9(k+1)(7k-3)
.\end{array}
