Whakaoti i te 3n^2-16n+20

$$3 n ^ { 2 } - 16 n + 20$$

Tauwehe

$$\left(n-2\right)\left(3n-10\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 3n^{2}+an+bn+20. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-16 ab=3\times 20=60$$

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 60.

$$-1,-60 -2,-30 -3,-20 -4,-15 -5,-12 -6,-10$$

Tātaihia te tapeke mō ia takirua.

$$-1-60=-61 -2-30=-32 -3-20=-23 -4-15=-19 -5-12=-17 -6-10=-16$$

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

$$a=-10 b=-6$$

Tuhia anō te 3n^{2}-16n+20 hei \left(3n^{2}-10n\right)+\left(-6n+20\right).

$$\left(3n^{2}-10n\right)+\left(-6n+20\right)$$

Tauwehea te n i te tuatahi me te -2 i te rōpū tuarua.

$$n\left(3n-10\right)-2\left(3n-10\right)$$

Whakatauwehea atu te kīanga pātahi 3n-10 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(3n-10\right)\left(n-2\right)$$

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.

$$3n^{2}-16n+20=0$$

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.

$$n=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 3\times 20}}{2\times 3}$$

Pūrua -16.

$$n=\frac{-\left(-16\right)±\sqrt{256-4\times 3\times 20}}{2\times 3}$$

Whakareatia -4 ki te 3.

$$n=\frac{-\left(-16\right)±\sqrt{256-12\times 20}}{2\times 3}$$

Whakareatia -12 ki te 20.

$$n=\frac{-\left(-16\right)±\sqrt{256-240}}{2\times 3}$$

Tāpiri 256 ki te -240.

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

Tuhia te pūtakerua o te 16.

$$n=\frac{-\left(-16\right)±4}{2\times 3}$$

Ko te tauaro o -16 ko 16.

$$n=\frac{16±4}{2\times 3}$$

Whakareatia 2 ki te 3.

$$n=\frac{16±4}{6}$$

Nā, me whakaoti te whārite n=\frac{16±4}{6} ina he tāpiri te ±. Tāpiri 16 ki te 4.

$$n=\frac{20}{6}$$

Whakahekea te hautanga \frac{20}{6} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$n=\frac{10}{3}$$

Nā, me whakaoti te whārite n=\frac{16±4}{6} ina he tango te ±. Tango 4 mai i 16.

$$n=\frac{12}{6}$$

Whakawehe 12 ki te 6.

$$n=2$$

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{10}{3} mō te x_{1} me te 2 mō te x_{2}.

$$3n^{2}-16n+20=3\left(n-\frac{10}{3}\right)\left(n-2\right)$$

Tango \frac{10}{3} mai i n 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.

$$3n^{2}-16n+20=3\times \frac{3n-10}{3}\left(n-2\right)$$

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

$$3n^{2}-16n+20=\left(3n-10\right)\left(n-2\right)$$

Aromātai

$$\left(n-2\right)\left(3n-10\right)$$