Whakaoti i te =-x^2-x+2

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

Tauwehe

$$\left(1-x\right)\left(x+2\right)$$

Ngā Upane Otinga

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

$$a+b=-1 ab=-2=-2$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōraro te a+b, he nui ake te uara pū o te tau tōraro i tō te tōrunga. Ko te takirua anake pērā ko te otinga pūnaha.

$$a=1 b=-2$$

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

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

Tauwehea te x i te tuatahi me te 2 i te rōpū tuarua.

$$x\left(-x+1\right)+2\left(-x+1\right)$$

Whakatauwehea atu te kīanga pātahi -x+1 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(-x+1\right)\left(x+2\right)$$

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.

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

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.

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

Whakareatia -4 ki te -1.

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

Whakareatia 4 ki te 2.

$$x=\frac{-\left(-1\right)±\sqrt{1+8}}{2\left(-1\right)}$$

Tāpiri 1 ki te 8.

$$x=\frac{-\left(-1\right)±\sqrt{9}}{2\left(-1\right)}$$

Tuhia te pūtakerua o te 9.

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

Ko te tauaro o -1 ko 1.

$$x=\frac{1±3}{2\left(-1\right)}$$

Whakareatia 2 ki te -1.

$$x=\frac{1±3}{-2}$$

Nā, me whakaoti te whārite x=\frac{1±3}{-2} ina he tāpiri te ±. Tāpiri 1 ki te 3.

$$x=\frac{4}{-2}$$

Whakawehe 4 ki te -2.

$$x=-2$$

Nā, me whakaoti te whārite x=\frac{1±3}{-2} ina he tango te ±. Tango 3 mai i 1.

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

Whakawehe -2 ki te -2.

$$x=1$$

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 -2 mō te x_{1} me te 1 mō te x_{2}.

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

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

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

Aromātai

$$\left(1-x\right)\left(x+2\right)$$