Whakaoti i te 4x^3-27x+27

$$4 x ^ { 3 } - 27 x + 27$$

Tauwehe

$$\left(x+3\right)\left(2x-3\right)^{2}$$

Ngā Upane Otinga

Tā te Rational Root Theorem, ko ngā pūtake whakahau katoa o tētahi pūrau kei te āhua o \frac{p}{q}, ina wehea e p te kīanga pūmau 27, ā, ka wehea e q te whakarea arahanga 4. Ko tetahi pūtake pērā ko \frac{3}{2}. Tauwehea te pūrau mā te whakawehe mā te 2x-3.

$$\left(2x-3\right)\left(2x^{2}+3x-9\right)$$

Whakaarohia te 2x^{2}+3x-9. Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 2x^{2}+ax+bx-9. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=3 ab=2\left(-9\right)=-18$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōrunga te a+b, he nui ake te uara pū o te tau tōrunga i tō te tōraro. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua -18.

$$-1,18 -2,9 -3,6$$

Tātaihia te tapeke mō ia takirua.

$$-1+18=17 -2+9=7 -3+6=3$$

Ko te otinga te takirua ka hoatu i te tapeke 3.

$$a=-3 b=6$$

Tuhia anō te 2x^{2}+3x-9 hei \left(2x^{2}-3x\right)+\left(6x-9\right).

$$\left(2x^{2}-3x\right)+\left(6x-9\right)$$

Tauwehea te x i te tuatahi me te 3 i te rōpū tuarua.

$$x\left(2x-3\right)+3\left(2x-3\right)$$

Whakatauwehea atu te kīanga pātahi 2x-3 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(2x-3\right)\left(x+3\right)$$

Me tuhi anō te kīanga whakatauwehe katoa.

$$\left(x+3\right)\left(2x-3\right)^{2}$$

Aromātai

$$\left(x+3\right)\left(2x-3\right)^{2}$$