Whakaoti i te B^2+6B+8

$$B ^ { 2 } + 6 B + 8$$

Tauwehe

$$\left(B+2\right)\left(B+4\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei B^{2}+aB+bB+8. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=6 ab=1\times 8=8$$

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 8.

$$1,8 2,4$$

Tātaihia te tapeke mō ia takirua.

$$1+8=9 2+4=6$$

Ko te otinga te takirua ka hoatu i te tapeke 6.

$$a=2 b=4$$

Tuhia anō te B^{2}+6B+8 hei \left(B^{2}+2B\right)+\left(4B+8\right).

$$\left(B^{2}+2B\right)+\left(4B+8\right)$$

Tauwehea te B i te tuatahi me te 4 i te rōpū tuarua.

$$B\left(B+2\right)+4\left(B+2\right)$$

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

$$\left(B+2\right)\left(B+4\right)$$

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.

$$B^{2}+6B+8=0$$

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.

$$B=\frac{-6±\sqrt{6^{2}-4\times 8}}{2}$$

Pūrua 6.

$$B=\frac{-6±\sqrt{36-4\times 8}}{2}$$

Whakareatia -4 ki te 8.

$$B=\frac{-6±\sqrt{36-32}}{2}$$

Tāpiri 36 ki te -32.

$$B=\frac{-6±\sqrt{4}}{2}$$

Tuhia te pūtakerua o te 4.

$$B=\frac{-6±2}{2}$$

Nā, me whakaoti te whārite B=\frac{-6±2}{2} ina he tāpiri te ±. Tāpiri -6 ki te 2.

$$B=-\frac{4}{2}$$

Whakawehe -4 ki te 2.

$$B=-2$$

Nā, me whakaoti te whārite B=\frac{-6±2}{2} ina he tango te ±. Tango 2 mai i -6.

$$B=-\frac{8}{2}$$

Whakawehe -8 ki te 2.

$$B=-4$$

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 -4 mō te x_{2}.

$$B^{2}+6B+8=\left(B-\left(-2\right)\right)\left(B-\left(-4\right)\right)$$

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

$$B^{2}+6B+8=\left(B+2\right)\left(B+4\right)$$

Aromātai

$$\left(B+2\right)\left(B+4\right)$$