Whakaoti i te 12-y-y^2

$$12 - y - y ^ { 2 }$$

Tauwehe

$$\left(3-y\right)\left(y+4\right)$$

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.

$$-y^{2}-y+12$$

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei -y^{2}+ay+by+12. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-1 ab=-12=-12$$

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. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua -12.

$$1,-12 2,-6 3,-4$$

Tātaihia te tapeke mō ia takirua.

$$1-12=-11 2-6=-4 3-4=-1$$

Ko te otinga te takirua ka hoatu i te tapeke -1.

$$a=3 b=-4$$

Tuhia anō te -y^{2}-y+12 hei \left(-y^{2}+3y\right)+\left(-4y+12\right).

$$\left(-y^{2}+3y\right)+\left(-4y+12\right)$$

Tauwehea te y i te tuatahi me te 4 i te rōpū tuarua.

$$y\left(-y+3\right)+4\left(-y+3\right)$$

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

$$\left(-y+3\right)\left(y+4\right)$$

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.

$$-y^{2}-y+12=0$$

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.

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

Whakareatia -4 ki te -1.

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

Whakareatia 4 ki te 12.

$$y=\frac{-\left(-1\right)±\sqrt{1+48}}{2\left(-1\right)}$$

Tāpiri 1 ki te 48.

$$y=\frac{-\left(-1\right)±\sqrt{49}}{2\left(-1\right)}$$

Tuhia te pūtakerua o te 49.

$$y=\frac{-\left(-1\right)±7}{2\left(-1\right)}$$

Ko te tauaro o -1 ko 1.

$$y=\frac{1±7}{2\left(-1\right)}$$

Whakareatia 2 ki te -1.

$$y=\frac{1±7}{-2}$$

Nā, me whakaoti te whārite y=\frac{1±7}{-2} ina he tāpiri te ±. Tāpiri 1 ki te 7.

$$y=\frac{8}{-2}$$

Whakawehe 8 ki te -2.

$$y=-4$$

Nā, me whakaoti te whārite y=\frac{1±7}{-2} ina he tango te ±. Tango 7 mai i 1.

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

Whakawehe -6 ki te -2.

$$y=3$$

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 -4 mō te x_{1} me te 3 mō te x_{2}.

$$-y^{2}-y+12=-\left(y-\left(-4\right)\right)\left(y-3\right)$$

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

$$-y^{2}-y+12=-\left(y+4\right)\left(y-3\right)$$

Aromātai

$$\left(3-y\right)\left(y+4\right)$$