Whakaoti i te 3z^2+14z-5

$$3 z ^ { 2 } + 14 z - 5$$

Tauwehe

$$\left(3z-1\right)\left(z+5\right)$$

Ngā Upane Otinga

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

$$a+b=14 ab=3\left(-5\right)=-15$$

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 -15.

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

Tātaihia te tapeke mō ia takirua.

$$-1+15=14 -3+5=2$$

Ko te otinga te takirua ka hoatu i te tapeke 14.

$$a=-1 b=15$$

Tuhia anō te 3z^{2}+14z-5 hei \left(3z^{2}-z\right)+\left(15z-5\right).

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

Tauwehea te z i te tuatahi me te 5 i te rōpū tuarua.

$$z\left(3z-1\right)+5\left(3z-1\right)$$

Whakatauwehea atu te kīanga pātahi 3z-1 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(3z-1\right)\left(z+5\right)$$

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.

$$3z^{2}+14z-5=0$$

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.

$$z=\frac{-14±\sqrt{14^{2}-4\times 3\left(-5\right)}}{2\times 3}$$

Pūrua 14.

$$z=\frac{-14±\sqrt{196-4\times 3\left(-5\right)}}{2\times 3}$$

Whakareatia -4 ki te 3.

$$z=\frac{-14±\sqrt{196-12\left(-5\right)}}{2\times 3}$$

Whakareatia -12 ki te -5.

$$z=\frac{-14±\sqrt{196+60}}{2\times 3}$$

Tāpiri 196 ki te 60.

$$z=\frac{-14±\sqrt{256}}{2\times 3}$$

Tuhia te pūtakerua o te 256.

$$z=\frac{-14±16}{2\times 3}$$

Whakareatia 2 ki te 3.

$$z=\frac{-14±16}{6}$$

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

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

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

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

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

$$z=-\frac{30}{6}$$

Whakawehe -30 ki te 6.

$$z=-5$$

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{1}{3} mō te x_{1} me te -5 mō te x_{2}.

$$3z^{2}+14z-5=3\left(z-\frac{1}{3}\right)\left(z-\left(-5\right)\right)$$

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

$$3z^{2}+14z-5=3\left(z-\frac{1}{3}\right)\left(z+5\right)$$

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

$$3z^{2}+14z-5=3\times \frac{3z-1}{3}\left(z+5\right)$$

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

$$3z^{2}+14z-5=\left(3z-1\right)\left(z+5\right)$$

Aromātai

$$\left(3z-1\right)\left(z+5\right)$$