Whakaoti i te x^2+5x+6

$${ x }^{ 2 } +5x+6$$

Tauwehe

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

Ngā Upane Otinga

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.

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

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.

$$x^{2}+5x+6=0$$

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.

$$x=\frac{-5±\sqrt{5^{2}-4\times 6}}{2}$$

Pūrua 5.

$$x=\frac{-5±\sqrt{25-4\times 6}}{2}$$

Whakareatia -4 ki te 6.

$$x=\frac{-5±\sqrt{25-24}}{2}$$

Tāpiri 25 ki te -24.

$$x=\frac{-5±\sqrt{1}}{2}$$

Tuhia te pūtakerua o te 1.

$$x=\frac{-5±1}{2}$$

Nā, me whakaoti te whārite x=\frac{-5±1}{2} ina he tāpiri te ±. Tāpiri -5 ki te 1.

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

Whakawehe -4 ki te 2.

$$x=-2$$

Nā, me whakaoti te whārite x=\frac{-5±1}{2} ina he tango te ±. Tango 1 mai i -5.

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

Whakawehe -6 ki te 2.

$$x=-3$$

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}.

$$x^{2}+5x+6=\left(x-\left(-2\right)\right)\left(x-\left(-3\right)\right)$$

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

$$x^{2}+5x+6=\left(x+2\right)\left(x+3\right)$$

Aromātai

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