Can I get help factoring 18m^3n + 3m^2n^2 – 6mn^3?

first of all take out the terms common in each term
this wud give u.

3mn (6m"2 + mn - 2n"2)
now there are two possible ways.........

simpler one:

again take n'2 common

this gives

3mn'3 ( 6 (m/n)"2 + (m/n)" - 2)

let m/n = y
then we get

3mn'3 (6y'2 + y - 2)

factorising we get

3mn'3 (3y + 2)(2y - 1)

substituting back the value of y,

we get final answer as

3mn (3m +2n)(2m - n)

take the common factor
which is 3mn
so 3mn(6m^2+mn-2n^2) and deal with the brackets as a quadratic
3mn(3m+2n)(2m-n)

18m^3n + 3m^2n^2 - 6mn^3
= 3(6m^3n + m^2n^2 - 2mn^3)
= 3mn(6m^2 + mn - 2n^2)
= 3mn(6m^2 + 4mn - 3mn - 2n^2)
= 3mn[(6m^2 + 4mn) - (3mn + 2n^2)]
= 3mn[2m(3m + 2n) - n(3m + 2n)]
= 2mn(3m + 2n)(2m - n)

18m^3n + 3m^2n^2 – 6mn^3
(3mn is the GCF.)So,

= 3mn(6m^2+mn-2n^2). <-----Answer.

find whats common in all 3 parts. i see that all coefficients are multiples of 3 and all have 1 n and 1 m
so:

(3nm)(6m^2+mn+2n^2) i hope thats right

3mn(6m^2+mn+2n^2)