Whakaoti i te b^2+4b+3

$$b ^ { 2 } + 4 b + 3$$

Tauwehe

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

Ngā Upane Otinga

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

$$p+q=4 pq=1\times 3=3$$

I te mea kua tōrunga te pq, he ōrite te tohu o p me q. I te mea kua tōrunga te p+q, he tōrunga hoki a p me q. Ko te takirua anake pērā ko te otinga pūnaha.

$$p=1 q=3$$

Tuhia anō te b^{2}+4b+3 hei \left(b^{2}+b\right)+\left(3b+3\right).

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

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

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

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

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

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.

$$b^{2}+4b+3=0$$

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.

$$b=\frac{-4±\sqrt{4^{2}-4\times 3}}{2}$$

Pūrua 4.

$$b=\frac{-4±\sqrt{16-4\times 3}}{2}$$

Whakareatia -4 ki te 3.

$$b=\frac{-4±\sqrt{16-12}}{2}$$

Tāpiri 16 ki te -12.

$$b=\frac{-4±\sqrt{4}}{2}$$

Tuhia te pūtakerua o te 4.

$$b=\frac{-4±2}{2}$$

Nā, me whakaoti te whārite b=\frac{-4±2}{2} ina he tāpiri te ±. Tāpiri -4 ki te 2.

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

Whakawehe -2 ki te 2.

$$b=-1$$

Nā, me whakaoti te whārite b=\frac{-4±2}{2} ina he tango te ±. Tango 2 mai i -4.

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

Whakawehe -6 ki te 2.

$$b=-3$$

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 -1 mō te x_{1} me te -3 mō te x_{2}.

$$b^{2}+4b+3=\left(b-\left(-1\right)\right)\left(b-\left(-3\right)\right)$$

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

$$b^{2}+4b+3=\left(b+1\right)\left(b+3\right)$$

Aromātai

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