Whakaoti i te f(x)=3+2x-x^2

$$f ( x ) = 3 + 2 x - x ^ { 2 }$$

Tauwehe

$$-\left(x-3\right)\left(x+1\right)$$

Ngā Upane Otinga

Hurinahatia te pūrau ki te āhua tānga ngahuru. Whakaraupapahia ngā kīanga tau mai i te pū teitei rawa ki te mea iti rawa.

$$-x^{2}+2x+3$$

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

$$a+b=2 ab=-3=-3$$

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. Ko te takirua anake pērā ko te otinga pūnaha.

$$a=3 b=-1$$

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

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

Tauwehea te -x i te tuatahi me te -1 i te rōpū tuarua.

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

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

$$\left(x-3\right)\left(-x-1\right)$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Ka taea te huamaha pūrua te tauwehe mā te whakamahi i te huringa ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), ina ko x_{1} me x_{2} ngā otinga o te whārite pūrua ax^{2}+bx+c=0.

$$-x^{2}+2x+3=0$$

Ko ngā whārite katoa o te āhua ax^{2}+bx+c=0 ka taea te whakaoti mā te whakamahi i te tikanga tātai pūrua: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. E rua ngā otinga ka puta i te tikanga tātai pūrua, ko tētahi ina he tāpiri a ±, ā, ko tētahi ina he tango.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-1\right)\times 3}}{2\left(-1\right)}$$

Pūrua 2.

$$x=\frac{-2±\sqrt{4-4\left(-1\right)\times 3}}{2\left(-1\right)}$$

Whakareatia -4 ki te -1.

$$x=\frac{-2±\sqrt{4+4\times 3}}{2\left(-1\right)}$$

Whakareatia 4 ki te 3.

$$x=\frac{-2±\sqrt{4+12}}{2\left(-1\right)}$$

Tāpiri 4 ki te 12.

$$x=\frac{-2±\sqrt{16}}{2\left(-1\right)}$$

Tuhia te pūtakerua o te 16.

$$x=\frac{-2±4}{2\left(-1\right)}$$

Whakareatia 2 ki te -1.

$$x=\frac{-2±4}{-2}$$

Nā, me whakaoti te whārite x=\frac{-2±4}{-2} ina he tāpiri te ±. Tāpiri -2 ki te 4.

$$x=\frac{2}{-2}$$

Whakawehe 2 ki te -2.

$$x=-1$$

Nā, me whakaoti te whārite x=\frac{-2±4}{-2} ina he tango te ±. Tango 4 mai i -2.

$$x=-\frac{6}{-2}$$

Whakawehe -6 ki te -2.

$$x=3$$

Tauwehea te kīanga taketake mā te whakamahi i te ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Me whakakapi te -1 mō te x_{1} me te 3 mō te x_{2}.

$$-x^{2}+2x+3=-\left(x-\left(-1\right)\right)\left(x-3\right)$$

Whakamāmātia ngā kīanga katoa o te āhua p-\left(-q\right) ki te p+q.

$$-x^{2}+2x+3=-\left(x+1\right)\left(x-3\right)$$

Aromātai

$$-\left(x-3\right)\left(x+1\right)$$