maths~!~
2008-04-20 5:31 am
When a polynomial P(x) is divided by x-k,the quotient is x+4 and remainder is 3.When P(x) is divided by x-2k,the remainder is 9.Find the possible values of k.
回答 (2)
2008-04-20 5:55 am
✔ 最佳答案
P(x) = (x-k)(x+4) +3= x^2 +(4-k)x+3
By long division
P(x) =( x-2k)[x +(4+k)] + [2k^2 + 4k +3]
where 2k^2 +4k +3 is remainder = 9
2k^2 + 4k + 3 = 9
2k^2 + 4k - 6 = 0
k^2 + 2k - 3 = 0
(k-1) (k+3) = 0
k = 1 or k = -3
2008-04-20 5:55 am
P(x) = (x-k)(x+4) + 3
P(x) = x^2 + 4x - kx - 4k + 3
P(x) = x^2 + (4-k)x + (3-4k)
P(2k) = 9
P(2k) = (2k)^2 + (4-k)(2k) + (3-4k) = 9
4k^2 + 8k - 2k^2 + 3 - 4k = 9
2k^2 + 4k - 6 = 0
k^2 + 2k - 3 = 0
(k + 3)(k - 1) = 0
k = -3 or k = 1
P(x) = x^2 + 4x - kx - 4k + 3
P(x) = x^2 + (4-k)x + (3-4k)
P(2k) = 9
P(2k) = (2k)^2 + (4-k)(2k) + (3-4k) = 9
4k^2 + 8k - 2k^2 + 3 - 4k = 9
2k^2 + 4k - 6 = 0
k^2 + 2k - 3 = 0
(k + 3)(k - 1) = 0
k = -3 or k = 1
收錄日期: 2021-04-23 20:39:42
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https://hk.answers.yahoo.com/question/index?qid=20080419000051KK03367