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

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

Tauwehe

$$\left(1-x\right)\left(2x+3\right)$$

Ngā Upane Otinga

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

$$a+b=-1 ab=-2\times 3=-6$$

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 -6.

$$1,-6 2,-3$$

Tātaihia te tapeke mō ia takirua.

$$1-6=-5 2-3=-1$$

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

$$a=2 b=-3$$

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

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

Tauwehea te 2x i te tuatahi me te 3 i te rōpū tuarua.

$$2x\left(-x+1\right)+3\left(-x+1\right)$$

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

$$\left(-x+1\right)\left(2x+3\right)$$

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.

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

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.

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

Whakareatia -4 ki te -2.

$$x=\frac{-\left(-1\right)±\sqrt{1+8\times 3}}{2\left(-2\right)}$$

Whakareatia 8 ki te 3.

$$x=\frac{-\left(-1\right)±\sqrt{1+24}}{2\left(-2\right)}$$

Tāpiri 1 ki te 24.

$$x=\frac{-\left(-1\right)±\sqrt{25}}{2\left(-2\right)}$$

Tuhia te pūtakerua o te 25.

$$x=\frac{-\left(-1\right)±5}{2\left(-2\right)}$$

Ko te tauaro o -1 ko 1.

$$x=\frac{1±5}{2\left(-2\right)}$$

Whakareatia 2 ki te -2.

$$x=\frac{1±5}{-4}$$

Nā, me whakaoti te whārite x=\frac{1±5}{-4} ina he tāpiri te ±. Tāpiri 1 ki te 5.

$$x=\frac{6}{-4}$$

Whakahekea te hautanga \frac{6}{-4} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x=-\frac{3}{2}$$

Nā, me whakaoti te whārite x=\frac{1±5}{-4} ina he tango te ±. Tango 5 mai i 1.

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

Whakawehe -4 ki te -4.

$$x=1$$

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 -\frac{3}{2} mō te x_{1} me te 1 mō te x_{2}.

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

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

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

Tāpiri \frac{3}{2} ki te x mā te kimi i te tauraro pātahi me te tāpiri i ngā taurunga. Ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

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

Whakakorea atu te tauwehe pūnoa nui rawa 2 i roto i te -2 me te 2.

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

Aromātai

$$\left(1-x\right)\left(2x+3\right)$$