Answer
$9(k+1)(7k-3)$
Work Step by Step
Substitute $u=3k-1:$
$7u^{2}+26u-8=...$
When factoring $ax^{2}+bx+c$, we search for factors of $ac$ whose sum is $b,$
and, if we find them, we rewrite $bx$ and proceed to factor in groups.
Here, factors of $(7)\times(-8)=-56$ that add to $+26$ are ....$+28$ and $-2$
$7u^{2}+26u-8=7u^{2}+28u-2u-8=$
$=7u(u+4)-2(u+4)$
$=(u+4)(7u-2)$
... bring back k
$=[(3k-1)+4][7(3k-1)-2]$
$=(3k+3)(21k-7-2)$
$=(3k+3)(21k-9)$
... factor out 3 from the first, and 3 from the second parentheses:
$=3(k+1)\cdot 3(7k-3)$
$=9(k+1)(7k-3)$
