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

1 Answer
May 28, 2016

Answer:

#3 times 2^2 times (m n p +2/3)times (m n p - 3/4)#

Explanation:

Making #x = m n p# the resulting equation #12 x^2-x-6=0#
has roots: #x = -2/3# and #x= 3/4#
So we can represent #12 x^2-x-6 = 12(x+2/3)(x-3/4)# The final factorization is
#3 times 2^2 times (m n p +2/3)times (m n p - 3/4)#