Whakaoti i te p^2-11p+28

$$p ^ { 2 } - 11 p + 28$$

Tauwehe

$$\left(p-7\right)\left(p-4\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei p^{2}+ap+bp+28. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-11 ab=1\times 28=28$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōraro te a+b, he tōraro hoki a a me b. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 28.

$$-1,-28 -2,-14 -4,-7$$

Tātaihia te tapeke mō ia takirua.

$$-1-28=-29 -2-14=-16 -4-7=-11$$

Ko te otinga te takirua ka hoatu i te tapeke -11.

$$a=-7 b=-4$$

Tuhia anō te p^{2}-11p+28 hei \left(p^{2}-7p\right)+\left(-4p+28\right).

$$\left(p^{2}-7p\right)+\left(-4p+28\right)$$

Tauwehea te p i te tuatahi me te -4 i te rōpū tuarua.

$$p\left(p-7\right)-4\left(p-7\right)$$

Whakatauwehea atu te kīanga pātahi p-7 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(p-7\right)\left(p-4\right)$$

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.

$$p^{2}-11p+28=0$$

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.

$$p=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 28}}{2}$$

Pūrua -11.

$$p=\frac{-\left(-11\right)±\sqrt{121-4\times 28}}{2}$$

Whakareatia -4 ki te 28.

$$p=\frac{-\left(-11\right)±\sqrt{121-112}}{2}$$

Tāpiri 121 ki te -112.

$$p=\frac{-\left(-11\right)±\sqrt{9}}{2}$$

Tuhia te pūtakerua o te 9.

$$p=\frac{-\left(-11\right)±3}{2}$$

Ko te tauaro o -11 ko 11.

$$p=\frac{11±3}{2}$$

Nā, me whakaoti te whārite p=\frac{11±3}{2} ina he tāpiri te ±. Tāpiri 11 ki te 3.

$$p=\frac{14}{2}$$

Whakawehe 14 ki te 2.

$$p=7$$

Nā, me whakaoti te whārite p=\frac{11±3}{2} ina he tango te ±. Tango 3 mai i 11.

$$p=\frac{8}{2}$$

Whakawehe 8 ki te 2.

$$p=4$$

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 7 mō te x_{1} me te 4 mō te x_{2}.

$$p^{2}-11p+28=\left(p-7\right)\left(p-4\right)$$

Aromātai

$$\left(p-7\right)\left(p-4\right)$$