Whakaoti i te 3d^2+20d+12

$$3 d ^ { 2 } + 20 d + 12$$

Tauwehe

$$\left(d+6\right)\left(3d+2\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 3d^{2}+ad+bd+12. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=20 ab=3\times 12=36$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōrunga te a+b, he tōrunga hoki a a me b. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 36.

$$1,36 2,18 3,12 4,9 6,6$$

Tātaihia te tapeke mō ia takirua.

$$1+36=37 2+18=20 3+12=15 4+9=13 6+6=12$$

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

$$a=2 b=18$$

Tuhia anō te 3d^{2}+20d+12 hei \left(3d^{2}+2d\right)+\left(18d+12\right).

$$\left(3d^{2}+2d\right)+\left(18d+12\right)$$

Tauwehea te d i te tuatahi me te 6 i te rōpū tuarua.

$$d\left(3d+2\right)+6\left(3d+2\right)$$

Whakatauwehea atu te kīanga pātahi 3d+2 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(3d+2\right)\left(d+6\right)$$

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.

$$3d^{2}+20d+12=0$$

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.

$$d=\frac{-20±\sqrt{20^{2}-4\times 3\times 12}}{2\times 3}$$

Pūrua 20.

$$d=\frac{-20±\sqrt{400-4\times 3\times 12}}{2\times 3}$$

Whakareatia -4 ki te 3.

$$d=\frac{-20±\sqrt{400-12\times 12}}{2\times 3}$$

Whakareatia -12 ki te 12.

$$d=\frac{-20±\sqrt{400-144}}{2\times 3}$$

Tāpiri 400 ki te -144.

$$d=\frac{-20±\sqrt{256}}{2\times 3}$$

Tuhia te pūtakerua o te 256.

$$d=\frac{-20±16}{2\times 3}$$

Whakareatia 2 ki te 3.

$$d=\frac{-20±16}{6}$$

Nā, me whakaoti te whārite d=\frac{-20±16}{6} ina he tāpiri te ±. Tāpiri -20 ki te 16.

$$d=-\frac{4}{6}$$

Whakahekea te hautanga \frac{-4}{6} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$d=-\frac{2}{3}$$

Nā, me whakaoti te whārite d=\frac{-20±16}{6} ina he tango te ±. Tango 16 mai i -20.

$$d=-\frac{36}{6}$$

Whakawehe -36 ki te 6.

$$d=-6$$

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{2}{3} mō te x_{1} me te -6 mō te x_{2}.

$$3d^{2}+20d+12=3\left(d-\left(-\frac{2}{3}\right)\right)\left(d-\left(-6\right)\right)$$

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

$$3d^{2}+20d+12=3\left(d+\frac{2}{3}\right)\left(d+6\right)$$

Tāpiri \frac{2}{3} ki te d 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.

$$3d^{2}+20d+12=3\times \frac{3d+2}{3}\left(d+6\right)$$

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

$$3d^{2}+20d+12=\left(3d+2\right)\left(d+6\right)$$

Aromātai

$$\left(d+6\right)\left(3d+2\right)$$