perform the indicated operations:
(c - 2)(c + 4)(c - 6)
A. c3 - 4c2 - 20c + 48
B. c3 - 4c2 + 20c + 48
C. c3 - 4c2 - 20c - 48
D. c3 - 4c2 + 180c - 48
2007-07-05 04:13:22 · 7 answers · asked by how's it going 1 in Science & Mathematics ➔ Mathematics
The answer is A.
There is an easier way to do this, but let's stick to basic algebra.
(c - 2)(c + 4)(c - 6) =
(c^2 - 2c + 4c - 8)(c - 6) =
(c^2 + 2c - 8)(c - 6)=
c^3 + 2c^2 - 8c - 6c^2 - 12c + 48 =
c^3 - 4c^2 -20c + 48
2007-07-05 04:18:50 · answer #1 · answered by Anonymous · 0⤊ 0⤋
First you multiply (c-2)(c+4)
(c^2+4c-2c-8)(c-6)
=C^2+2c-8)(c-6)
=c^3=6c^2
=2c^2-12c-8c+48
=c^3-4c^2-20c+48
2007-07-05 04:28:22 · answer #2 · answered by Steven B 2 · 0⤊ 0⤋
(c-2)(c+4) = c^2 + 4c - 2c - 8 = c^2 + 2c - 8
(c^2 + 2c - 8)(c-6) = c^3 - 4c^2 - 20c + 48
2007-07-05 04:19:22 · answer #3 · answered by Kevin H 1 · 0⤊ 0⤋
you multiply factors by factors
(c-2) (c+4) = c^2+4c-2c-8) = c^2+2c-8
(c^2+2c-8)(c-6) = c^3+2c^2-8c-6c^2-12c+48)
result c^3-4c^2-20c+48
2007-07-05 04:20:57 · answer #4 · answered by maussy 7 · 0⤊ 0⤋
You have to simplify the first formula, right?
2007-07-05 04:21:16 · answer #5 · answered by Not Dave 2 · 0⤊ 0⤋
I came up with answer A.
2007-07-05 04:24:20 · answer #6 · answered by Anonymous · 0⤊ 0⤋
a
2007-07-05 04:23:44 · answer #7 · answered by snoopykins 2 · 0⤊ 0⤋