Whakaoti i te -3x^2+5x+2

$$- 3 x ^ { 2 } + 5 x + 2$$

Tauwehe

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

Ngā Upane Otinga

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

$$a+b=5 ab=-3\times 2=-6$$

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 -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 5.

$$a=6 b=-1$$

Tuhia anō te -3x^{2}+5x+2 hei \left(-3x^{2}+6x\right)+\left(-x+2\right).

$$\left(-3x^{2}+6x\right)+\left(-x+2\right)$$

Whakatauwehea atu 3x i te -3x^{2}+6x.

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

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

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

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.

$$-3x^{2}+5x+2=0$$

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.

$$x=\frac{-5±\sqrt{5^{2}-4\left(-3\right)\times 2}}{2\left(-3\right)}$$

Pūrua 5.

$$x=\frac{-5±\sqrt{25-4\left(-3\right)\times 2}}{2\left(-3\right)}$$

Whakareatia -4 ki te -3.

$$x=\frac{-5±\sqrt{25+12\times 2}}{2\left(-3\right)}$$

Whakareatia 12 ki te 2.

$$x=\frac{-5±\sqrt{25+24}}{2\left(-3\right)}$$

Tāpiri 25 ki te 24.

$$x=\frac{-5±\sqrt{49}}{2\left(-3\right)}$$

Tuhia te pūtakerua o te 49.

$$x=\frac{-5±7}{2\left(-3\right)}$$

Whakareatia 2 ki te -3.

$$x=\frac{-5±7}{-6}$$

Nā, me whakaoti te whārite x=\frac{-5±7}{-6} ina he tāpiri te ±. Tāpiri -5 ki te 7.

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

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

$$x=-\frac{1}{3}$$

Nā, me whakaoti te whārite x=\frac{-5±7}{-6} ina he tango te ±. Tango 7 mai i -5.

$$x=-\frac{12}{-6}$$

Whakawehe -12 ki te -6.

$$x=2$$

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}{3} mō te x_{1} me te 2 mō te x_{2}.

$$-3x^{2}+5x+2=-3\left(x-\left(-\frac{1}{3}\right)\right)\left(x-2\right)$$

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

$$-3x^{2}+5x+2=-3\left(x+\frac{1}{3}\right)\left(x-2\right)$$

Tāpiri \frac{1}{3} 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.

$$-3x^{2}+5x+2=-3\times \frac{-3x-1}{-3}\left(x-2\right)$$

Whakakorea atu te tauwehe pūnoa nui rawa 3 i roto i te -3 me te 3.

$$-3x^{2}+5x+2=\left(-3x-1\right)\left(x-2\right)$$

Aromātai

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