急! F4 polynomicals 1題 #7

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f(x) = 2x³ - 9x + 8k, where k is a constant.

The remainder is k when f(x) is divided by x - k.

⇒ f(k) = k

⇒ 2k³ - 9k + 8k = k

⇒ 2k³ - 2k = 0

⇒ k³ - k = 0

⇒ k(k² - 1) = 0

⇒ k(k - 1)(k + 1) = 0

⇒ k = 0 or k = 1 or k = -1

The remainder when f(x) is divided by x - 2k is f(2k).

f(2k) = 2(2k)³ - 9(2k) + 8k
= 2(8k³) - 18k + 8k
= 16k³ - 10k

If k = 0, f(2k) = 0.
If k = 1, f(2k) = 16 - 10 = 6.
If k = -1, f(2k) = -16 + 10 = -6.

2014-02-28 19:52:12 補充：
螞蟻雄兵，是否有誤？？

題目應是　f(x) = 2x³ - 9x + 8k　。

2014-03-01 13:33:17 補充：
If k = 0, f(2k) = 0 = 6*0 = 6k
If k = 1, f(2k) = 16 - 10 = 6 = 6*1 = 6k
If k = -1, f(2k) = -16 + 10 = -6 = 6*(-1) = 6k

可見對於每一個 k 值的情況, 答案均是 6k。

2014-03-01 21:30:13 補充：
那些年大大的解答很好～

請容許我加入補充之中。

ღ(｡◕‿◠｡)ღ

2014-03-01 21:32:04 補充：
如果不想列出所有情況再反想答案為 6k，可以用那些年大大的方法。

以上從 k³ - k = 0　這一步可見 k³ = k。

The remainder when f(x) is divided by x - 2k is f(2k).

f(2k) = 2(2k)³ - 9(2k) + 8k
= 2(8k³) - 18k + 8k
= 16k³ - 10k
= 16k - 10k
= 6k

Using remainder theorem, f(k) = k, therefore,
2k^3 - 9k + 8k = k
==> k^3 = k

When f(x) is divided by (x - 2k), the remainder is :
f(2k)
= 2(2k)^3 - 9(2k) + 8k
= 16k^3 - 18k + 8k
= 16k - 18k + 8k ⋯⋯ (as k^3 = k)
= 6k