Whakaoti i te 10c^2-19c-15

$$10 c ^ { 2 } - 19 c - 15$$

Tauwehe

$$\left(2c-5\right)\left(5c+3\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 10c^{2}+ac+bc-15. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-19 ab=10\left(-15\right)=-150$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōraro te a+b, he nui ake te uara pū o te tau tōraro i tō te tōrunga. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua -150.

$$1,-150 2,-75 3,-50 5,-30 6,-25 10,-15$$

Tātaihia te tapeke mō ia takirua.

$$1-150=-149 2-75=-73 3-50=-47 5-30=-25 6-25=-19 10-15=-5$$

Ko te otinga te takirua ka hoatu i te tapeke -19.

$$a=-25 b=6$$

Tuhia anō te 10c^{2}-19c-15 hei \left(10c^{2}-25c\right)+\left(6c-15\right).

$$\left(10c^{2}-25c\right)+\left(6c-15\right)$$

Tauwehea te 5c i te tuatahi me te 3 i te rōpū tuarua.

$$5c\left(2c-5\right)+3\left(2c-5\right)$$

Whakatauwehea atu te kīanga pātahi 2c-5 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(2c-5\right)\left(5c+3\right)$$

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.

$$10c^{2}-19c-15=0$$

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.

$$c=\frac{-\left(-19\right)±\sqrt{\left(-19\right)^{2}-4\times 10\left(-15\right)}}{2\times 10}$$

Pūrua -19.

$$c=\frac{-\left(-19\right)±\sqrt{361-4\times 10\left(-15\right)}}{2\times 10}$$

Whakareatia -4 ki te 10.

$$c=\frac{-\left(-19\right)±\sqrt{361-40\left(-15\right)}}{2\times 10}$$

Whakareatia -40 ki te -15.

$$c=\frac{-\left(-19\right)±\sqrt{361+600}}{2\times 10}$$

Tāpiri 361 ki te 600.

$$c=\frac{-\left(-19\right)±\sqrt{961}}{2\times 10}$$

Tuhia te pūtakerua o te 961.

$$c=\frac{-\left(-19\right)±31}{2\times 10}$$

Ko te tauaro o -19 ko 19.

$$c=\frac{19±31}{2\times 10}$$

Whakareatia 2 ki te 10.

$$c=\frac{19±31}{20}$$

Nā, me whakaoti te whārite c=\frac{19±31}{20} ina he tāpiri te ±. Tāpiri 19 ki te 31.

$$c=\frac{50}{20}$$

Whakahekea te hautanga \frac{50}{20} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 10.

$$c=\frac{5}{2}$$

Nā, me whakaoti te whārite c=\frac{19±31}{20} ina he tango te ±. Tango 31 mai i 19.

$$c=-\frac{12}{20}$$

Whakahekea te hautanga \frac{-12}{20} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

$$c=-\frac{3}{5}$$

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 \frac{5}{2} mō te x_{1} me te -\frac{3}{5} mō te x_{2}.

$$10c^{2}-19c-15=10\left(c-\frac{5}{2}\right)\left(c-\left(-\frac{3}{5}\right)\right)$$

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

$$10c^{2}-19c-15=10\left(c-\frac{5}{2}\right)\left(c+\frac{3}{5}\right)$$

Tango \frac{5}{2} mai i c mā te kimi i te tauraro pātahi me te tango i ngā taurunga, ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$10c^{2}-19c-15=10\times \frac{2c-5}{2}\left(c+\frac{3}{5}\right)$$

Tāpiri \frac{3}{5} ki te c 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.

$$10c^{2}-19c-15=10\times \frac{2c-5}{2}\times \frac{5c+3}{5}$$

Whakareatia \frac{2c-5}{2} ki te \frac{5c+3}{5} 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.

$$10c^{2}-19c-15=10\times \frac{\left(2c-5\right)\left(5c+3\right)}{2\times 5}$$

Whakareatia 2 ki te 5.

$$10c^{2}-19c-15=10\times \frac{\left(2c-5\right)\left(5c+3\right)}{10}$$

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

$$10c^{2}-19c-15=\left(2c-5\right)\left(5c+3\right)$$

Aromātai

$$\left(2c-5\right)\left(5c+3\right)$$