Answer
$ax(3x-2)(3x+7)$
Work Step by Step
$ 9ax^{3}+15ax^{2}-14ax\qquad$...factor out the common term, $ax$
$=ax(9x^{2}+15x-14)$
... Searching for two factors of $ac=-126$ whose sum is $b=15,$
we find$\qquad-6$ and $21.$
Rewrite the middle term and factor in pairs:
$=ax(9x^{2}-6x+21x-14)=$
$=ax[3x(3x-2)+7(3x-2)]$
=$ax(3x-2)(3x+7)$
