How would i factor m^2(3n+1)^2-4m^2?

Question

I do not understand how take out the difference of two squares and when the signs should stay the same and they should change. Any help yould be great thank you.

Answer 1

The general form of a difference of squares is a²-b² and this always factors into (a+b)(a-b). In the problem you have the first thing you should notice is that you can pull out a m² leaving you with m²((3n+1)²-4) and you still have a difference of squares. the a is (3n+1) and the b is 2. so it factors as m²((3n+1)+2)((3n+1)-2) which works out to m²(3n+3)(3n-1)

Answer 2

You have a common factor of m^2, so if you take that out you get: m^2((3n+1)^2 - 4). If you clean up the right-hand expression, you get 9n^2 + 6n -3, which obviously has a factor 3; if you move that outside, you have:
3m^2(3n^2 + 2n -1). But the right side is factorable into (3n-1)(n+1), so the final factorization is 3m^2(3n-1)(n+1).

Answer 3

m^2(3n+1)^2-4m^2
= m^2[(3n+1)^2-4]
= m^2(3n+3)(3n-1), used the difference of two squares.

Answer 4

Slightly different from other answers:
m^2(3n + 1)^2 - 4m^2

= m^2(9n^2 + 6n + 1) - 4m^2
= 9m^2 n^2 + 6n + 1 - 4m^2
= (9m^2 n^2 - 4m^2) + 6n + 1
= m^2(9n^2 - 4) + 6n + 1
= m^2 ((3n)^2 - 2^2) + 6n + 1
= m^2 (3n + 2)(3n - 2) + 6n + 1

hth

Answer 5

=m^2((3n+1)^2-4)

Answer 6

As (m(3n+1)+2m) *(m(3n+1) -2m)