Whakaoti i te 3t^2-2t-1

$$3 t ^ { 2 } - 2 t - 1$$

Tauwehe

$$\left(t-1\right)\left(3t+1\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 3t^{2}+at+bt-1. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

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

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. Ko te takirua anake pērā ko te otinga pūnaha.

$$a=-3 b=1$$

Tuhia anō te 3t^{2}-2t-1 hei \left(3t^{2}-3t\right)+\left(t-1\right).

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

Whakatauwehea atu 3t i te 3t^{2}-3t.

$$3t\left(t-1\right)+t-1$$

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

$$\left(t-1\right)\left(3t+1\right)$$

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.

$$3t^{2}-2t-1=0$$

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.

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

Pūrua -2.

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

Whakareatia -4 ki te 3.

$$t=\frac{-\left(-2\right)±\sqrt{4-12\left(-1\right)}}{2\times 3}$$

Whakareatia -12 ki te -1.

$$t=\frac{-\left(-2\right)±\sqrt{4+12}}{2\times 3}$$

Tāpiri 4 ki te 12.

$$t=\frac{-\left(-2\right)±\sqrt{16}}{2\times 3}$$

Tuhia te pūtakerua o te 16.

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

Ko te tauaro o -2 ko 2.

$$t=\frac{2±4}{2\times 3}$$

Whakareatia 2 ki te 3.

$$t=\frac{2±4}{6}$$

Nā, me whakaoti te whārite t=\frac{2±4}{6} ina he tāpiri te ±. Tāpiri 2 ki te 4.

$$t=\frac{6}{6}$$

Whakawehe 6 ki te 6.

$$t=1$$

Nā, me whakaoti te whārite t=\frac{2±4}{6} ina he tango te ±. Tango 4 mai i 2.

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

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

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

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{1}{3} mō te x_{2}.

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

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

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

Tāpiri \frac{1}{3} ki te t 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.

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

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

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

Aromātai

$$\left(t-1\right)\left(3t+1\right)$$