Whakaoti i te 3z^2+16z+20

$$3 z ^ { 2 } + 16 z + 20$$

Tauwehe

$$\left(z+2\right)\left(3z+10\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 3z^{2}+az+bz+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 tōrunga te a+b, he tōrunga 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.

$$a=6 b=10$$

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

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

Tauwehea te 3z i te tuatahi me te 10 i te rōpū tuarua.

$$3z\left(z+2\right)+10\left(z+2\right)$$

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

$$\left(z+2\right)\left(3z+10\right)$$

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.

$$3z^{2}+16z+20=0$$

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.

$$z=\frac{-16±\sqrt{16^{2}-4\times 3\times 20}}{2\times 3}$$

Pūrua 16.

$$z=\frac{-16±\sqrt{256-4\times 3\times 20}}{2\times 3}$$

Whakareatia -4 ki te 3.

$$z=\frac{-16±\sqrt{256-12\times 20}}{2\times 3}$$

Whakareatia -12 ki te 20.

$$z=\frac{-16±\sqrt{256-240}}{2\times 3}$$

Tāpiri 256 ki te -240.

$$z=\frac{-16±\sqrt{16}}{2\times 3}$$

Tuhia te pūtakerua o te 16.

$$z=\frac{-16±4}{2\times 3}$$

Whakareatia 2 ki te 3.

$$z=\frac{-16±4}{6}$$

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

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

Whakawehe -12 ki te 6.

$$z=-2$$

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

$$z=-\frac{20}{6}$$

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

$$z=-\frac{10}{3}$$

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 -2 mō te x_{1} me te -\frac{10}{3} mō te x_{2}.

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

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

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

Tāpiri \frac{10}{3} ki te z 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.

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

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

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

Aromātai

$$\left(z+2\right)\left(3z+10\right)$$