remainder theorem

小霞的fans · 發表於 3/3/2010 02:47 PM

when a polynomial P(x) is divided by x-k, the quotient is x+4 and the remainder is 3.when P(x) is divided by x-2k , the remainder is 9. find the possible values of k .

唔係太識點做 t^t "" 希望有人教教我 .. 代x=k 嗎??

sapphire · 發表於 3/3/2010 03:01 PM

1. When a polynomial P(x) is divided by x-k, the quotient is x+4 and the remainder is 3.

which implies
P(x) = (x - k)(x + 4) + 3
After calculation, P(x) = x^2 + (4 - k)x - 4k + 3

2. When P(x) is divided by x-2k , the remainder is 9.

which implies
P(2k) = (2k)^2 + (4 - k)(2k) - 4k + 3 = 9

For the rest, it is only a quadratic equation.

Ans : k = 1 or -3

小霞的fans · 發表於 3/3/2010 04:01 PM

1. When a polynomial P(x) is divided by x-k, the quotient is x+4 and the remainder is 3.

which implies
P(x) = (x - k)(x + 4) + 3
After calculation, P(x) = x^2 + (4 - k)x - 4k + 3

2. When P(x) ...
sapphire 發表於 2/3/2010 07:01 PM

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仲有1題variation我想問@@..

y^2 varies directly as x^2

咁應該寫成y^2=kx^2定係y^2=k^2x^2 ??

unlock-YM · 發表於 3/3/2010 04:45 PM

仲有1題variation我想問@@..

y^2 varies directly as x^2

咁應該寫成y^2=kx^2定係y^2=k^2x^2 ??
小霞的fans 發表於 3/3/2010 04:01 PM

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理論上兩個都冇所謂，反正k只係constant。
實際上你寫 y^2 = kx^2 已經得，無謂畫蛇添足。

Ivanhy92 · 發表於 3/3/2010 08:06 PM

理論上兩個都冇所謂，反正k只係constant。
實際上你寫 y^2 = kx^2 已經得，無謂畫蛇添足。
unlock-YM 發表於 3-3-2010 16:45

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y^2 = k x^2 (y^2 is DIRECTLY proportional to x^2)
y^2 = k^2 x^2 (y^2 is EXPONENTIALLY proportional to x^2)

unlock-YM · 發表於 3/3/2010 08:36 PM

y^2 = k x^2 (y^2 is DIRECTLY proportional to x^2)
y^2 = k^2 x^2 (y^2 is EXPONENTIALLY proportional to x^2)
arararchchch 發表於 3/3/2010 08:06 PM

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