Whakaoti i te 3b^2+8b-3

$$3 b ^ { 2 } + 8 b - 3$$

Tauwehe

$$\left(3b-1\right)\left(b+3\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 3b^{2}+pb+qb-3. Hei kimi p me q, whakaritea tētahi pūnaha kia whakaoti.

$$p+q=8 pq=3\left(-3\right)=-9$$

I te mea kua tōraro te pq, he tauaro ngā tohu o p me q. I te mea kua tōrunga te p+q, 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 -9.

$$-1,9 -3,3$$

Tātaihia te tapeke mō ia takirua.

$$-1+9=8 -3+3=0$$

Ko te otinga te takirua ka hoatu i te tapeke 8.

$$p=-1 q=9$$

Tuhia anō te 3b^{2}+8b-3 hei \left(3b^{2}-b\right)+\left(9b-3\right).

$$\left(3b^{2}-b\right)+\left(9b-3\right)$$

Tauwehea te b i te tuatahi me te 3 i te rōpū tuarua.

$$b\left(3b-1\right)+3\left(3b-1\right)$$

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

$$\left(3b-1\right)\left(b+3\right)$$

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.

$$3b^{2}+8b-3=0$$

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.

$$b=\frac{-8±\sqrt{8^{2}-4\times 3\left(-3\right)}}{2\times 3}$$

Pūrua 8.

$$b=\frac{-8±\sqrt{64-4\times 3\left(-3\right)}}{2\times 3}$$

Whakareatia -4 ki te 3.

$$b=\frac{-8±\sqrt{64-12\left(-3\right)}}{2\times 3}$$

Whakareatia -12 ki te -3.

$$b=\frac{-8±\sqrt{64+36}}{2\times 3}$$

Tāpiri 64 ki te 36.

$$b=\frac{-8±\sqrt{100}}{2\times 3}$$

Tuhia te pūtakerua o te 100.

$$b=\frac{-8±10}{2\times 3}$$

Whakareatia 2 ki te 3.

$$b=\frac{-8±10}{6}$$

Nā, me whakaoti te whārite b=\frac{-8±10}{6} ina he tāpiri te ±. Tāpiri -8 ki te 10.

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

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

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

Nā, me whakaoti te whārite b=\frac{-8±10}{6} ina he tango te ±. Tango 10 mai i -8.

$$b=-\frac{18}{6}$$

Whakawehe -18 ki te 6.

$$b=-3$$

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

$$3b^{2}+8b-3=3\left(b-\frac{1}{3}\right)\left(b-\left(-3\right)\right)$$

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

$$3b^{2}+8b-3=3\left(b-\frac{1}{3}\right)\left(b+3\right)$$

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

$$3b^{2}+8b-3=3\times \frac{3b-1}{3}\left(b+3\right)$$

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

$$3b^{2}+8b-3=\left(3b-1\right)\left(b+3\right)$$

Aromātai

$$\left(3b-1\right)\left(b+3\right)$$