1. m^2 + 6n - 9n^2 - 2m
a. (m-3n)(m+3n-2)
b. (m-3n +2)(m+3n)
c. (m+3n +2) (m-3n)
d. (m + 3n)(m-3n -2)
Thanks so much everybody!!! :)
2007-01-08 05:04:50 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
First: when there's four terms > group "like" terms in parenthesis:
(m^2 - 2m) - (9n^2 + 6n)
Sec: factor both sets of parenthesis (find the least common factor from each set and multiply it by the remaining terms to get the inside expression) >
m(m - 2) - 3n(3n + 2)
(m - 3n)(m + 3n - 2)
*Both sets of parenthesis have "2" and one is negative, the negative sign is used for the final answer.
2007-01-08 05:13:39 · answer #1 · answered by ♪♥Annie♥♪ 6 · 0⤊ 1⤋
1. m^2 + 6n - 9n^2 - 2m
a. (m-3n)(m+3n-2)
is correct
2007-01-08 13:13:15 · answer #2 · answered by yupchagee 7 · 0⤊ 0⤋
should be (a)
Rearrange to
m^2-9n^2-2m+6n
Then
(m^2-(3n)^2)-2(m-3n)
think u should able to solve the rest
2007-01-08 13:24:58 · answer #3 · answered by solora2000 1 · 0⤊ 1⤋
a is the correct answer
multiply the two together and get the product you have.
2007-01-08 13:16:27 · answer #4 · answered by Ray 5 · 0⤊ 1⤋