Whakaoti i te 3t^2+20t-32

$$3 t ^ { 2 } + 20 t - 32$$

Tauwehe

$$\left(3t-4\right)\left(t+8\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 3t^{2}+at+bt-32. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=20 ab=3\left(-32\right)=-96$$

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 -96.

$$-1,96 -2,48 -3,32 -4,24 -6,16 -8,12$$

Tātaihia te tapeke mō ia takirua.

$$-1+96=95 -2+48=46 -3+32=29 -4+24=20 -6+16=10 -8+12=4$$

Ko te otinga te takirua ka hoatu i te tapeke 20.

$$a=-4 b=24$$

Tuhia anō te 3t^{2}+20t-32 hei \left(3t^{2}-4t\right)+\left(24t-32\right).

$$\left(3t^{2}-4t\right)+\left(24t-32\right)$$

Tauwehea te t i te tuatahi me te 8 i te rōpū tuarua.

$$t\left(3t-4\right)+8\left(3t-4\right)$$

Whakatauwehea atu te kīanga pātahi 3t-4 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(3t-4\right)\left(t+8\right)$$

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.

$$3t^{2}+20t-32=0$$

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.

$$t=\frac{-20±\sqrt{20^{2}-4\times 3\left(-32\right)}}{2\times 3}$$

Pūrua 20.

$$t=\frac{-20±\sqrt{400-4\times 3\left(-32\right)}}{2\times 3}$$

Whakareatia -4 ki te 3.

$$t=\frac{-20±\sqrt{400-12\left(-32\right)}}{2\times 3}$$

Whakareatia -12 ki te -32.

$$t=\frac{-20±\sqrt{400+384}}{2\times 3}$$

Tāpiri 400 ki te 384.

$$t=\frac{-20±\sqrt{784}}{2\times 3}$$

Tuhia te pūtakerua o te 784.

$$t=\frac{-20±28}{2\times 3}$$

Whakareatia 2 ki te 3.

$$t=\frac{-20±28}{6}$$

Nā, me whakaoti te whārite t=\frac{-20±28}{6} ina he tāpiri te ±. Tāpiri -20 ki te 28.

$$t=\frac{8}{6}$$

Whakahekea te hautanga \frac{8}{6} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$t=\frac{4}{3}$$

Nā, me whakaoti te whārite t=\frac{-20±28}{6} ina he tango te ±. Tango 28 mai i -20.

$$t=-\frac{48}{6}$$

Whakawehe -48 ki te 6.

$$t=-8$$

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{4}{3} mō te x_{1} me te -8 mō te x_{2}.

$$3t^{2}+20t-32=3\left(t-\frac{4}{3}\right)\left(t-\left(-8\right)\right)$$

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

$$3t^{2}+20t-32=3\left(t-\frac{4}{3}\right)\left(t+8\right)$$

Tango \frac{4}{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.

$$3t^{2}+20t-32=3\times \frac{3t-4}{3}\left(t+8\right)$$

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

$$3t^{2}+20t-32=\left(3t-4\right)\left(t+8\right)$$

Aromātai

$$\left(3t-4\right)\left(t+8\right)$$