When 3x^3 + 2x^2 - 7x + 8 is divided by x + 2, what is the remainder?
2006-07-19 20:29:07 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
6
2006-07-19 20:44:00 · answer #1 · answered by jimbob 6 · 0⤊ 0⤋
By the remainder theorem:
f(-2) = 3(-2)^3 + 2(-2)^2 - 7(-2) +8
= -24 + 8 + 14 + 8
= 6
Therefore, the remainder is 6.
2006-07-19 20:42:39 · answer #2 · answered by emee_rocks 2 · 0⤊ 0⤋
(3x^3 + 2x² - 7x + 8) ÷ (x + 2) = 3x² - 4x + 1
(3x² - 4x + 1)(x + 2) = 3x^3 +2x² - 7x + 2
(3x^3 + 2x² - 7x + 8)-(3x^3 +2x² - 7x + 2) = 6
(that is, a remainder of 6).
2006-07-19 21:02:37 · answer #3 · answered by Brenmore 5 · 0⤊ 0⤋
using remainder theorem
if we assume x+2 be the factor of3x^3+2x^2-7x+8
then ut x=-2
answer is 6
2006-07-19 20:57:09 · answer #4 · answered by Anonymous · 0⤊ 0⤋