Whakaoti i te 1a^2+8a+15

$$1 a ^ { 2 } + 8 a + 15$$

Tauwehe

$$\left(a+3\right)\left(a+5\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei a^{2}+pa+qa+15. Hei kimi p me q, whakaritea tētahi pūnaha kia whakaoti.

$$p+q=8 pq=1\times 15=15$$

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. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 15.

$$1,15 3,5$$

Tātaihia te tapeke mō ia takirua.

$$1+15=16 3+5=8$$

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

$$p=3 q=5$$

Tuhia anō te a^{2}+8a+15 hei \left(a^{2}+3a\right)+\left(5a+15\right).

$$\left(a^{2}+3a\right)+\left(5a+15\right)$$

Tauwehea te a i te tuatahi me te 5 i te rōpū tuarua.

$$a\left(a+3\right)+5\left(a+3\right)$$

Whakatauwehea atu te kīanga pātahi a+3 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(a+3\right)\left(a+5\right)$$

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.

$$a^{2}+8a+15=0$$

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.

$$a=\frac{-8±\sqrt{8^{2}-4\times 15}}{2}$$

Pūrua 8.

$$a=\frac{-8±\sqrt{64-4\times 15}}{2}$$

Whakareatia -4 ki te 15.

$$a=\frac{-8±\sqrt{64-60}}{2}$$

Tāpiri 64 ki te -60.

$$a=\frac{-8±\sqrt{4}}{2}$$

Tuhia te pūtakerua o te 4.

$$a=\frac{-8±2}{2}$$

Nā, me whakaoti te whārite a=\frac{-8±2}{2} ina he tāpiri te ±. Tāpiri -8 ki te 2.

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

Whakawehe -6 ki te 2.

$$a=-3$$

Nā, me whakaoti te whārite a=\frac{-8±2}{2} ina he tango te ±. Tango 2 mai i -8.

$$a=-\frac{10}{2}$$

Whakawehe -10 ki te 2.

$$a=-5$$

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 -3 mō te x_{1} me te -5 mō te x_{2}.

$$a^{2}+8a+15=\left(a-\left(-3\right)\right)\left(a-\left(-5\right)\right)$$

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

$$a^{2}+8a+15=\left(a+3\right)\left(a+5\right)$$

Aromātai

$$\left(a+3\right)\left(a+5\right)$$