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

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

Tauwehe

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

Ngā Upane Otinga

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

$$a+b=3 ab=-2\times 2=-4$$

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 -4.

$$-1,4 -2,2$$

Tātaihia te tapeke mō ia takirua.

$$-1+4=3 -2+2=0$$

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

$$a=4 b=-1$$

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

$$\left(-2x^{2}+4x\right)+\left(-x+2\right)$$

Whakatauwehea atu 2x i te -2x^{2}+4x.

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

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

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

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.

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

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.

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

Pūrua 3.

$$x=\frac{-3±\sqrt{9-4\left(-2\right)\times 2}}{2\left(-2\right)}$$

Whakareatia -4 ki te -2.

$$x=\frac{-3±\sqrt{9+8\times 2}}{2\left(-2\right)}$$

Whakareatia 8 ki te 2.

$$x=\frac{-3±\sqrt{9+16}}{2\left(-2\right)}$$

Tāpiri 9 ki te 16.

$$x=\frac{-3±\sqrt{25}}{2\left(-2\right)}$$

Tuhia te pūtakerua o te 25.

$$x=\frac{-3±5}{2\left(-2\right)}$$

Whakareatia 2 ki te -2.

$$x=\frac{-3±5}{-4}$$

Nā, me whakaoti te whārite x=\frac{-3±5}{-4} ina he tāpiri te ±. Tāpiri -3 ki te 5.

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

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

$$x=-\frac{1}{2}$$

Nā, me whakaoti te whārite x=\frac{-3±5}{-4} ina he tango te ±. Tango 5 mai i -3.

$$x=-\frac{8}{-4}$$

Whakawehe -8 ki te -4.

$$x=2$$

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{1}{2} mō te x_{1} me te 2 mō te x_{2}.

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

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

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

Tāpiri \frac{1}{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.

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

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

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

Aromātai

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