## Precalculus (6th Edition)

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 $9\times(-2)=-18$ that add to $+4$ are .... we can't find such a pair. $(-3)\times 6$ add to $+3,$ $(-2)\times 9$ add to $+7$, etc. No two add to $4$. We can't further factor this trinomial.