Whakaoti i te 15m^2+m-6

$$15 m ^ { 2 } + m - 6$$

Tauwehe

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

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 15m^{2}+am+bm-6. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=1 ab=15\left(-6\right)=-90$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōrunga te a+b, he nui ake te uara pū o te tau tōrunga i tō te tōraro. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua -90.

$$-1,90 -2,45 -3,30 -5,18 -6,15 -9,10$$

Tātaihia te tapeke mō ia takirua.

$$-1+90=89 -2+45=43 -3+30=27 -5+18=13 -6+15=9 -9+10=1$$

Ko te otinga te takirua ka hoatu i te tapeke 1.

$$a=-9 b=10$$

Tuhia anō te 15m^{2}+m-6 hei \left(15m^{2}-9m\right)+\left(10m-6\right).

$$\left(15m^{2}-9m\right)+\left(10m-6\right)$$

Tauwehea te 3m i te tuatahi me te 2 i te rōpū tuarua.

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

Whakatauwehea atu te kīanga pātahi 5m-3 mā te whakamahi i te āhuatanga tātai tohatoha.

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

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.

$$15m^{2}+m-6=0$$

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.

$$m=\frac{-1±\sqrt{1^{2}-4\times 15\left(-6\right)}}{2\times 15}$$

Pūrua 1.

$$m=\frac{-1±\sqrt{1-4\times 15\left(-6\right)}}{2\times 15}$$

Whakareatia -4 ki te 15.

$$m=\frac{-1±\sqrt{1-60\left(-6\right)}}{2\times 15}$$

Whakareatia -60 ki te -6.

$$m=\frac{-1±\sqrt{1+360}}{2\times 15}$$

Tāpiri 1 ki te 360.

$$m=\frac{-1±\sqrt{361}}{2\times 15}$$

Tuhia te pūtakerua o te 361.

$$m=\frac{-1±19}{2\times 15}$$

Whakareatia 2 ki te 15.

$$m=\frac{-1±19}{30}$$

Nā, me whakaoti te whārite m=\frac{-1±19}{30} ina he tāpiri te ±. Tāpiri -1 ki te 19.

$$m=\frac{18}{30}$$

Whakahekea te hautanga \frac{18}{30} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 6.

$$m=\frac{3}{5}$$

Nā, me whakaoti te whārite m=\frac{-1±19}{30} ina he tango te ±. Tango 19 mai i -1.

$$m=-\frac{20}{30}$$

Whakahekea te hautanga \frac{-20}{30} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 10.

$$m=-\frac{2}{3}$$

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

$$15m^{2}+m-6=15\left(m-\frac{3}{5}\right)\left(m-\left(-\frac{2}{3}\right)\right)$$

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

$$15m^{2}+m-6=15\left(m-\frac{3}{5}\right)\left(m+\frac{2}{3}\right)$$

Tango \frac{3}{5} mai i m 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.

$$15m^{2}+m-6=15\times \frac{5m-3}{5}\left(m+\frac{2}{3}\right)$$

Tāpiri \frac{2}{3} ki te m 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.

$$15m^{2}+m-6=15\times \frac{5m-3}{5}\times \frac{3m+2}{3}$$

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

$$15m^{2}+m-6=15\times \frac{\left(5m-3\right)\left(3m+2\right)}{5\times 3}$$

Whakareatia 5 ki te 3.

$$15m^{2}+m-6=15\times \frac{\left(5m-3\right)\left(3m+2\right)}{15}$$

Whakakorea atu te tauwehe pūnoa nui rawa 15 i roto i te 15 me te 15.

$$15m^{2}+m-6=\left(5m-3\right)\left(3m+2\right)$$

Aromātai

$$15m^{2}+m-6$$