多項式一問(中五)
2010-10-08 10:09 pm
回答 (2)
2010-10-08 10:23 pm
✔ 最佳答案
f(x) = (x^2 - x - 12)Q(x) - 2x + 5 where Q(x) is a polynomial in x.so f(x) = (x^2 - x - 12)Q(x) - 2x + 5 = (x + 3)P(x) + R where P(x) is another polynomial in x.
Put x = - 3
(9 + 3 - 12)Q(-3) - 2(-3) + 5 = (-3 + 3)P(-3) + R
6 + 5 = R
so remainder, R = 11.
2010-10-09 2:03 am
f(x) = (x^2 - x - 12)Q(x) - 2x + 5 where Q(x) is a polynomial in x.
so f(x) = (x^2 - x - 12)Q(x) - 2x + 5 = (x + 3)P(x) + R where P(x)
Put x = - 3
(9 + 3 - 12)Q(-3) - 2(-3) + 5 = (-3 + 3)P(-3) + R
6 + 5 = R
R = 11.
so f(x) = (x^2 - x - 12)Q(x) - 2x + 5 = (x + 3)P(x) + R where P(x)
Put x = - 3
(9 + 3 - 12)Q(-3) - 2(-3) + 5 = (-3 + 3)P(-3) + R
6 + 5 = R
R = 11.
收錄日期: 2021-04-25 22:40:28
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20101008000051KK00527