Whakaoti i te 2x^2+10x+12

$$2 x ^ { 2 } + 10 x + 12$$

Tauwehe

$$2\left(x+2\right)\left(x+3\right)$$

Ngā Upane Otinga

Tauwehea te 2.

$$2\left(x^{2}+5x+6\right)$$

Whakaarohia te x^{2}+5x+6. Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei x^{2}+ax+bx+6. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=5 ab=1\times 6=6$$

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 6.

$$1,6 2,3$$

Tātaihia te tapeke mō ia takirua.

$$1+6=7 2+3=5$$

Ko te otinga te takirua ka hoatu i te tapeke 5.

$$a=2 b=3$$

Tuhia anō te x^{2}+5x+6 hei \left(x^{2}+2x\right)+\left(3x+6\right).

$$\left(x^{2}+2x\right)+\left(3x+6\right)$$

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

$$x\left(x+2\right)+3\left(x+2\right)$$

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

$$\left(x+2\right)\left(x+3\right)$$

Me tuhi anō te kīanga whakatauwehe katoa.

$$2\left(x+2\right)\left(x+3\right)$$

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.

$$2x^{2}+10x+12=0$$

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.

$$x=\frac{-10±\sqrt{10^{2}-4\times 2\times 12}}{2\times 2}$$

Pūrua 10.

$$x=\frac{-10±\sqrt{100-4\times 2\times 12}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$x=\frac{-10±\sqrt{100-8\times 12}}{2\times 2}$$

Whakareatia -8 ki te 12.

$$x=\frac{-10±\sqrt{100-96}}{2\times 2}$$

Tāpiri 100 ki te -96.

$$x=\frac{-10±\sqrt{4}}{2\times 2}$$

Tuhia te pūtakerua o te 4.

$$x=\frac{-10±2}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{-10±2}{4}$$

Nā, me whakaoti te whārite x=\frac{-10±2}{4} ina he tāpiri te ±. Tāpiri -10 ki te 2.

$$x=-\frac{8}{4}$$

Whakawehe -8 ki te 4.

$$x=-2$$

Nā, me whakaoti te whārite x=\frac{-10±2}{4} ina he tango te ±. Tango 2 mai i -10.

$$x=-\frac{12}{4}$$

Whakawehe -12 ki te 4.

$$x=-3$$

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 -3 mō te x_{2}.

$$2x^{2}+10x+12=2\left(x-\left(-2\right)\right)\left(x-\left(-3\right)\right)$$

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

$$2x^{2}+10x+12=2\left(x+2\right)\left(x+3\right)$$

Aromātai

$$2\left(x+2\right)\left(x+3\right)$$