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)
2007-01-05 16:17:48 · 5 answers · asked by wahoo!!! 1 in Science & Mathematics ➔ Mathematics
Factor completely.
2007-01-05 16:19:17 · update #1
m^2 + 6n - 9n^2 - 2m
= m^2 - 9n^2 - 2m + 6n, rearrange terms
= (m+3n)(m-3n) - 2(m-3n), factor the first two and the last two terms.
= (m-3n)(m+3n-2), take the common factor (m+3n) out
Therefore, the right answer is A.
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Since one of the four answers must be true, you should have a negative sign before 9n^2.
2007-01-05 16:28:29 · answer #1 · answered by sahsjing 7 · 0⤊ 1⤋
1) Rearrange the terms so you have a difference of squares grouped together.
m^2 - 9n^2 - 2m + 6n
2) Factor the first group using the "difference of squares" and factor a -2 from the second group.
(m - 3n)(m + 2n) - 2(m - 3n)
3) Factor out the (m - 3n) term.
(m - 3n)(m + 2n - 2)
The answer is A.
2007-01-06 00:44:36 · answer #2 · answered by purpicita_LM_es_fg_MDK 2 · 0⤊ 0⤋
Check the question and rewrite. What sense does 6n 9n^2 make?
2007-01-06 00:27:57 · answer #3 · answered by ? 6 · 0⤊ 1⤋
Are you sure about that middle term?
2007-01-06 00:26:57 · answer #4 · answered by keely_66 3 · 0⤊ 1⤋
Something is not right ? Are you sure about this equation ?
2007-01-06 00:29:07 · answer #5 · answered by Anonymous · 0⤊ 1⤋