# How do you factor 12m^2n^2p^2-mnp-6?

$3 \times {2}^{2} \times \left(m n p + \frac{2}{3}\right) \times \left(m n p - \frac{3}{4}\right)$
Making $x = m n p$ the resulting equation $12 {x}^{2} - x - 6 = 0$
has roots: $x = - \frac{2}{3}$ and $x = \frac{3}{4}$
So we can represent $12 {x}^{2} - x - 6 = 12 \left(x + \frac{2}{3}\right) \left(x - \frac{3}{4}\right)$ The final factorization is
$3 \times {2}^{2} \times \left(m n p + \frac{2}{3}\right) \times \left(m n p - \frac{3}{4}\right)$