How do you factor this problem completely??

Question

Factor completely. m^3 – 9m^2n + 18n^2m

Answer 1

First factor an "m" out of every term, leaving you with:
m*(m^2 – 9*m*n + 18*n^2)

Now factor the part in parentheses, yielding:
m*(m - 6n)*(m - 3n)

Answer 2

First factor out a m
m(m^2 - 9mn + 18n^2)

Next find what two numbers multiply together to make 18 and add together to make -9

Hint: -6, -3

put it all together and make sure to include the n's
m(m - 3n)(m-6)

Answer 3

Factor out m.

m(m^2 – 9mn – 18n^2)

Using the quadratic formula reveals two real roots for the polynomial, so it can be factored completely.

The two roots are 6n and 3n (if you chose n as part of the coefficients instead of as the variable).

m(m – 6n)(m – 3n)

Answer 4

factor out m
m (m^2 - 9mn + 18n^2)

two numbers that multiply to 18 and add to -9

the two numbers are -3 and -6

m [ (m^2 - 3mn - 6mn + 18^2)

factor by groups
m [ (m^2 - 3mn) + (-6mn + 18n^2) ]

m [ (m (m - 3n) + -6n(m - 3n)]

m [ (m (m - 3n) - 6n(m - 3n)]

m (m - 6n) (m - 3n)

Answer 5

m(m-3n)(m-6n)