Whakaoti i te 10p^2+9p+2

$$10 p ^ { 2 } + 9 p + 2$$

Tauwehe

$$\left(2p+1\right)\left(5p+2\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 10p^{2}+ap+bp+2. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=9 ab=10\times 2=20$$

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 20.

$$1,20 2,10 4,5$$

Tātaihia te tapeke mō ia takirua.

$$1+20=21 2+10=12 4+5=9$$

Ko te otinga te takirua ka hoatu i te tapeke 9.

$$a=4 b=5$$

Tuhia anō te 10p^{2}+9p+2 hei \left(10p^{2}+4p\right)+\left(5p+2\right).

$$\left(10p^{2}+4p\right)+\left(5p+2\right)$$

Whakatauwehea atu 2p i te 10p^{2}+4p.

$$2p\left(5p+2\right)+5p+2$$

Whakatauwehea atu te kīanga pātahi 5p+2 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(5p+2\right)\left(2p+1\right)$$

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.

$$10p^{2}+9p+2=0$$

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.

$$p=\frac{-9±\sqrt{9^{2}-4\times 10\times 2}}{2\times 10}$$

Pūrua 9.

$$p=\frac{-9±\sqrt{81-4\times 10\times 2}}{2\times 10}$$

Whakareatia -4 ki te 10.

$$p=\frac{-9±\sqrt{81-40\times 2}}{2\times 10}$$

Whakareatia -40 ki te 2.

$$p=\frac{-9±\sqrt{81-80}}{2\times 10}$$

Tāpiri 81 ki te -80.

$$p=\frac{-9±\sqrt{1}}{2\times 10}$$

Tuhia te pūtakerua o te 1.

$$p=\frac{-9±1}{2\times 10}$$

Whakareatia 2 ki te 10.

$$p=\frac{-9±1}{20}$$

Nā, me whakaoti te whārite p=\frac{-9±1}{20} ina he tāpiri te ±. Tāpiri -9 ki te 1.

$$p=-\frac{8}{20}$$

Whakahekea te hautanga \frac{-8}{20} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

$$p=-\frac{2}{5}$$

Nā, me whakaoti te whārite p=\frac{-9±1}{20} ina he tango te ±. Tango 1 mai i -9.

$$p=-\frac{10}{20}$$

Whakahekea te hautanga \frac{-10}{20} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 10.

$$p=-\frac{1}{2}$$

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 -\frac{2}{5} mō te x_{1} me te -\frac{1}{2} mō te x_{2}.

$$10p^{2}+9p+2=10\left(p-\left(-\frac{2}{5}\right)\right)\left(p-\left(-\frac{1}{2}\right)\right)$$

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

$$10p^{2}+9p+2=10\left(p+\frac{2}{5}\right)\left(p+\frac{1}{2}\right)$$

Tāpiri \frac{2}{5} ki te p mā te kimi i te tauraro pātahi me te tāpiri i ngā taurunga. Ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$10p^{2}+9p+2=10\times \frac{5p+2}{5}\left(p+\frac{1}{2}\right)$$

Tāpiri \frac{1}{2} ki te p mā te kimi i te tauraro pātahi me te tāpiri i ngā taurunga. Ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$10p^{2}+9p+2=10\times \frac{5p+2}{5}\times \frac{2p+1}{2}$$

Whakareatia \frac{5p+2}{5} ki te \frac{2p+1}{2} mā te whakarea taurunga ki te taurunga me te tauraro ki te tauraro, ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$10p^{2}+9p+2=10\times \frac{\left(5p+2\right)\left(2p+1\right)}{5\times 2}$$

Whakareatia 5 ki te 2.

$$10p^{2}+9p+2=10\times \frac{\left(5p+2\right)\left(2p+1\right)}{10}$$

Whakakorea atu te tauwehe pūnoa nui rawa 10 i roto i te 10 me te 10.

$$10p^{2}+9p+2=\left(5p+2\right)\left(2p+1\right)$$

Aromātai

$$\left(2p+1\right)\left(5p+2\right)$$