Whakaoti i te 3c^2-16c+5

$$3 c ^ { 2 } - 16 c + 5$$

Tauwehe

$$\left(c-5\right)\left(3c-1\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 3c^{2}+ac+bc+5. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-16 ab=3\times 5=15$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōraro te a+b, he tōraro hoki a a me b. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 15.

$$-1,-15 -3,-5$$

Tātaihia te tapeke mō ia takirua.

$$-1-15=-16 -3-5=-8$$

Ko te otinga te takirua ka hoatu i te tapeke -16.

$$a=-15 b=-1$$

Tuhia anō te 3c^{2}-16c+5 hei \left(3c^{2}-15c\right)+\left(-c+5\right).

$$\left(3c^{2}-15c\right)+\left(-c+5\right)$$

Tauwehea te 3c i te tuatahi me te -1 i te rōpū tuarua.

$$3c\left(c-5\right)-\left(c-5\right)$$

Whakatauwehea atu te kīanga pātahi c-5 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(c-5\right)\left(3c-1\right)$$

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.

$$3c^{2}-16c+5=0$$

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.

$$c=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 3\times 5}}{2\times 3}$$

Pūrua -16.

$$c=\frac{-\left(-16\right)±\sqrt{256-4\times 3\times 5}}{2\times 3}$$

Whakareatia -4 ki te 3.

$$c=\frac{-\left(-16\right)±\sqrt{256-12\times 5}}{2\times 3}$$

Whakareatia -12 ki te 5.

$$c=\frac{-\left(-16\right)±\sqrt{256-60}}{2\times 3}$$

Tāpiri 256 ki te -60.

$$c=\frac{-\left(-16\right)±\sqrt{196}}{2\times 3}$$

Tuhia te pūtakerua o te 196.

$$c=\frac{-\left(-16\right)±14}{2\times 3}$$

Ko te tauaro o -16 ko 16.

$$c=\frac{16±14}{2\times 3}$$

Whakareatia 2 ki te 3.

$$c=\frac{16±14}{6}$$

Nā, me whakaoti te whārite c=\frac{16±14}{6} ina he tāpiri te ±. Tāpiri 16 ki te 14.

$$c=\frac{30}{6}$$

Whakawehe 30 ki te 6.

$$c=5$$

Nā, me whakaoti te whārite c=\frac{16±14}{6} ina he tango te ±. Tango 14 mai i 16.

$$c=\frac{2}{6}$$

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

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

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

$$3c^{2}-16c+5=3\left(c-5\right)\left(c-\frac{1}{3}\right)$$

Tango \frac{1}{3} mai i c 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.

$$3c^{2}-16c+5=3\left(c-5\right)\times \frac{3c-1}{3}$$

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

$$3c^{2}-16c+5=\left(c-5\right)\left(3c-1\right)$$

Aromātai

$$\left(c-5\right)\left(3c-1\right)$$