## Precalculus (6th Edition)

$9(k+1)(7k-3)$
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: $=3(k+1)\cdot 3(7k-3)$ $=9(k+1)(7k-3)$