How do you multiply # (2m^2 + n)(3n^2 + 6mn - m^2)#?

1 Answer
Jul 14, 2015

Answer:

#(2m^2 + n)(3n^2 + 6mn -m^2) = – 2 m^4 + 12 m^3n + 6m^2n^2 – m^2n + 6mn^2 + 3n^3#

Explanation:

We use the distributive property of multiplication.

#(2m^2 + n)(3n^2 + 6mn - m^2) = 2m^2(3n^2 + 6mn - m^2) + n(3n^2 + 6mn - m^2)#

#= (6m^2n^2 + 12 m^3n – 2 m^4) + (3n^3 + 6mn^2 – m^2n)#

Now we remove parentheses and arrange the terms in descending powers of #m# and ascending powers of #n#.

#(2m^2 + n)(3n^2 + 6mn -m^2) = – 2 m^4 + 12 m^3n + 6m^2n^2 –m^2n + 6mn^2 + 3n^3#