Whakaoti i te 6x^2-2x-4==

$$6 { x }^{ 2 } -2x-4==$$

Tauwehe

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

Ngā Upane Otinga

Tauwehea te 2.

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

Whakaarohia te 3x^{2}-x-2. 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.

$$a+b=-1 ab=3\left(-2\right)=-6$$

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.

$$a=-3 b=2$$

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

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

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

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

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

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

Me tuhi anō te kīanga whakatauwehe katoa.

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

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.

$$6x^{2}-2x-4=0$$

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.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 6\left(-4\right)}}{2\times 6}$$

Pūrua -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\times 6\left(-4\right)}}{2\times 6}$$

Whakareatia -4 ki te 6.

$$x=\frac{-\left(-2\right)±\sqrt{4-24\left(-4\right)}}{2\times 6}$$

Whakareatia -24 ki te -4.

$$x=\frac{-\left(-2\right)±\sqrt{4+96}}{2\times 6}$$

Tāpiri 4 ki te 96.

$$x=\frac{-\left(-2\right)±\sqrt{100}}{2\times 6}$$

Tuhia te pūtakerua o te 100.

$$x=\frac{-\left(-2\right)±10}{2\times 6}$$

Ko te tauaro o -2 ko 2.

$$x=\frac{2±10}{2\times 6}$$

Whakareatia 2 ki te 6.

$$x=\frac{2±10}{12}$$

Nā, me whakaoti te whārite x=\frac{2±10}{12} ina he tāpiri te ±. Tāpiri 2 ki te 10.

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

Whakawehe 12 ki te 12.

$$x=1$$

Nā, me whakaoti te whārite x=\frac{2±10}{12} ina he tango te ±. Tango 10 mai i 2.

$$x=-\frac{8}{12}$$

Whakahekea te hautanga \frac{-8}{12} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

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

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

$$6x^{2}-2x-4=6\left(x-1\right)\left(x-\left(-\frac{2}{3}\right)\right)$$

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

$$6x^{2}-2x-4=6\left(x-1\right)\left(x+\frac{2}{3}\right)$$

Tāpiri \frac{2}{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.

$$6x^{2}-2x-4=6\left(x-1\right)\times \frac{3x+2}{3}$$

Whakakorea atu te tauwehe pūnoa nui rawa 3 i roto i te 6 me te 3.

$$6x^{2}-2x-4=2\left(x-1\right)\left(3x+2\right)$$

Aromātai

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