2007-04-27 17:55:25 · 7 answers · asked by Jamie 2 in Science & Mathematics ➔ Mathematics
3a^2(a – b) – 6a(a – b) + 21(a – b)?
2007-04-27 18:05:00 · update #1
3a^2(a – b) – 6a(a – b) + 21(a – b)
=(a-b)(3a^2-6a+21)
=3(a-b)(a^2-2a+7)
2007-04-27 18:03:53 · answer #1 · answered by iyiogrenci 6 · 0⤊ 0⤋
3a²(a – b) – 6a(a – b) + 21(a – b) =
(a-b) (3a² - 6a + 21) =
3(a-b) (a² - 2a + 7)
2007-04-27 18:15:54 · answer #2 · answered by M. Abuhelwa 5 · 0⤊ 0⤋
i believe what he/she means is 3a^2(a – b) – 6a(a – b) + 21(a – b)..
so, you can factor out (a-b)
(3a^2-6a+21)(a-b)
3(a^2 - 2a + 7)(a-b) <<<< final answer...
2007-04-27 18:10:21 · answer #3 · answered by Paolo Y 2 · 0⤊ 0⤋
3a^2(a – b) – 6a(a – b) + 21(a – b)
=(3a^2 - 6a + 21)(a-b)
=[a+1+(i)sqrt(6)]*[a+1-(i)sqrt(6)]*(a-b)
2007-04-27 18:16:23 · answer #4 · answered by someone 3 · 0⤊ 0⤋
given expression
=(a-b)(3a^2-6a+21)
={(a-b)}{3(a^2-2a+7)}
=3(a-b)(a^2-2a+7)
2007-04-27 18:06:04 · answer #5 · answered by alpha 7 · 0⤊ 0⤋
Hmmm..... The first two terms are the same, so the result is just the third term:
21(a-b)
You sure you copied that down right? It's not really a factoring problem!
2007-04-27 18:03:09 · answer #6 · answered by engqm743 2 · 0⤊ 1⤋
3a2(a-b) - 6a(a-b) + 21(a-b)
=(a-b)[6a-6a+21]
=21(a-b)
2007-04-27 18:03:50 · answer #7 · answered by iamdeyb 2 · 0⤊ 1⤋