Factor. 3a2(a – b) – 6a(a – b) + 21(a – b)
A) 3(a – b)(a2 – 2a + 7)
B) (3a – b)(a2 – 2a + 7)
C) (a – b)(3a2 – 6a + 21)
D) 3(a – b)(a2 – 6a + 21)
Choose correct answer
2007-07-12 17:54:59 · 6 answers · asked by Somebody 2 in Science & Mathematics ➔ Mathematics
I knew I shouldn't have taken the pillow to algebra class
2007-07-12 17:58:32 · answer #1 · answered by rob c 3 · 0⤊ 4⤋
Every term has an (a-b) factor, so we factor this out and write in the remaining parts of each term:
(a-b) (3a^2 - 6a + 21)
Now everything inside the second bracket is divisible by 3, so we can factor this out too to get
3(a-b) (a^2 - 2a + 7)
So the answer is A.
2007-07-13 01:00:23 · answer #2 · answered by Scarlet Manuka 7 · 0⤊ 0⤋
(a-b)(3a^2 - 6a + 21)
(a-b) 3( a^2 - 2a + 7)
3(a-b)(a^2 - 2a + 7)
is the answer
all you have to do is take 3(a-b) as common and the rest is easy
2007-07-13 03:06:26 · answer #3 · answered by Anonymous · 0⤊ 0⤋
3a^2(a - b) -6a(a - b) + 21(a - b)
=(a - b)(3a^2 - 6a + 21)
=(a - b)[3(a^2 - 2a + 7)]
=3(a - b)(a^2 - 2a + 7)
Answer is (A).
2007-07-13 01:06:25 · answer #4 · answered by Sarang 3 · 0⤊ 0⤋
C =(a-b)(3a2 - 6a +21)
2007-07-13 00:59:58 · answer #5 · answered by Snoopy 3 · 0⤊ 1⤋
there is a common (a-b) in all thus
C
2007-07-13 00:58:33 · answer #6 · answered by sin2acos2a1 2 · 0⤊ 1⤋