Without doing the long division, find the remainder when 3x^4 - 5x^2 + 4 is divided by x^2 + 2.
2006-12-26 01:05:53 · 4 answers · asked by iqnabeel 1 in Science & Mathematics ➔ Mathematics
Actually answer #2 is incorrect. You would need to factor x^2+2 = (x-sqrt(-2)) (x+sqrt(-2)) and (somehow) use p(sqrt(-2)) and p(-sqrt(-2)).
Easier way to solve this is using modular arithmetic. You might have heard of this with regular numbers: "mod 9" is sometimes called "throwing out nines" so 12 is congruent to 3 mod 9. Well, modular arithmetic works also for polynomials. So you set x^2+2 congruent to 0, which means x^2 congruent to -2 mod (x^2+2), so then
3x^4-5x^2+4 = 3(x^2)^2-5x^2+4 = 3(-2)^2-5*-2+4 = 26. The remainder is 26.
> rem(3*x^4-5*x^2+4,x^2+2,x);
26
2006-12-26 02:00:59 · answer #1 · answered by a_math_guy 5 · 1⤊ 1⤋
Let p(x) = 3x^4 - 5x^2 + 4.
In order to find the remainder, all you have to do is plug in the negative of +2 (since (x^2 + 2) is a factor). That is, p(-2).
p(-2) = 3(-2)^4 - 5(-2)^2 + 4 = 3(16) - 5(4) + 4 = 48 - 20 + 4 = 32
2006-12-26 09:31:47 · answer #2 · answered by Puggy 7 · 0⤊ 1⤋
let 3 x^4 -5x^2 +4 = p(x)
use factor theorum after that
2006-12-26 09:10:50 · answer #3 · answered by Anonymous · 0⤊ 0⤋
if x^2+2 is the divisor the remainder will be
f(rt2i)=3(rt2i)^4-5(rt2i)^2+4
=3(4)+10+4=26
f(-rt2i)=2(-rt2i)^4-5(-rt2i)^2...
=26
so the raminderis 26
2006-12-26 10:12:45 · answer #4 · answered by raj 7 · 0⤊ 1⤋