Whakaoti i te 1-t-6t^2

$$1 - t - 6 t ^ { 2 }$$

Tauwehe

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

Ngā Upane Otinga

Hurinahatia te pūrau ki te āhua tānga ngahuru. Whakaraupapahia ngā kīanga tau mai i te pū teitei rawa ki te mea iti rawa.

$$-6t^{2}-t+1$$

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei -6t^{2}+at+bt+1. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-1 ab=-6=-6$$

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 -6t^{2}-t+1 hei \left(-6t^{2}+2t\right)+\left(-3t+1\right).

$$\left(-6t^{2}+2t\right)+\left(-3t+1\right)$$

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

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

Whakatauwehea atu te kīanga pātahi -3t+1 mā te whakamahi i te āhuatanga tātai tohatoha.

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

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.

$$-6t^{2}-t+1=0$$

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.

$$t=\frac{-\left(-1\right)±\sqrt{1-4\left(-6\right)}}{2\left(-6\right)}$$

Whakareatia -4 ki te -6.

$$t=\frac{-\left(-1\right)±\sqrt{1+24}}{2\left(-6\right)}$$

Tāpiri 1 ki te 24.

$$t=\frac{-\left(-1\right)±\sqrt{25}}{2\left(-6\right)}$$

Tuhia te pūtakerua o te 25.

$$t=\frac{-\left(-1\right)±5}{2\left(-6\right)}$$

Ko te tauaro o -1 ko 1.

$$t=\frac{1±5}{2\left(-6\right)}$$

Whakareatia 2 ki te -6.

$$t=\frac{1±5}{-12}$$

Nā, me whakaoti te whārite t=\frac{1±5}{-12} ina he tāpiri te ±. Tāpiri 1 ki te 5.

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

Whakahekea te hautanga \frac{6}{-12} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 6.

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

Nā, me whakaoti te whārite t=\frac{1±5}{-12} ina he tango te ±. Tango 5 mai i 1.

$$t=-\frac{4}{-12}$$

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

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

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

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

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

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

Tāpiri \frac{1}{2} 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.

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

Tango \frac{1}{3} mai i t mā te kimi i te tauraro pātahi me te tango i ngā taurunga, ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$-6t^{2}-t+1=-6\times \frac{-2t-1}{-2}\times \frac{-3t+1}{-3}$$

Whakareatia \frac{-2t-1}{-2} ki te \frac{-3t+1}{-3} mā te whakarea taurunga ki te taurunga me te tauraro ki te tauraro, ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

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

Whakareatia -2 ki te -3.

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

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

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

Aromātai

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