Whakaoti i te 12c^2+11c-15

$$12 c ^ { 2 } + 11 c - 15$$

Tauwehe

$$\left(4c-3\right)\left(3c+5\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 12c^{2}+ac+bc-15. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=11 ab=12\left(-15\right)=-180$$

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 -180.

$$-1,180 -2,90 -3,60 -4,45 -5,36 -6,30 -9,20 -10,18 -12,15$$

Tātaihia te tapeke mō ia takirua.

$$-1+180=179 -2+90=88 -3+60=57 -4+45=41 -5+36=31 -6+30=24 -9+20=11 -10+18=8 -12+15=3$$

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

$$a=-9 b=20$$

Tuhia anō te 12c^{2}+11c-15 hei \left(12c^{2}-9c\right)+\left(20c-15\right).

$$\left(12c^{2}-9c\right)+\left(20c-15\right)$$

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

$$3c\left(4c-3\right)+5\left(4c-3\right)$$

Whakatauwehea atu te kīanga pātahi 4c-3 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(4c-3\right)\left(3c+5\right)$$

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.

$$12c^{2}+11c-15=0$$

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.

$$c=\frac{-11±\sqrt{11^{2}-4\times 12\left(-15\right)}}{2\times 12}$$

Pūrua 11.

$$c=\frac{-11±\sqrt{121-4\times 12\left(-15\right)}}{2\times 12}$$

Whakareatia -4 ki te 12.

$$c=\frac{-11±\sqrt{121-48\left(-15\right)}}{2\times 12}$$

Whakareatia -48 ki te -15.

$$c=\frac{-11±\sqrt{121+720}}{2\times 12}$$

Tāpiri 121 ki te 720.

$$c=\frac{-11±\sqrt{841}}{2\times 12}$$

Tuhia te pūtakerua o te 841.

$$c=\frac{-11±29}{2\times 12}$$

Whakareatia 2 ki te 12.

$$c=\frac{-11±29}{24}$$

Nā, me whakaoti te whārite c=\frac{-11±29}{24} ina he tāpiri te ±. Tāpiri -11 ki te 29.

$$c=\frac{18}{24}$$

Whakahekea te hautanga \frac{18}{24} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 6.

$$c=\frac{3}{4}$$

Nā, me whakaoti te whārite c=\frac{-11±29}{24} ina he tango te ±. Tango 29 mai i -11.

$$c=-\frac{40}{24}$$

Whakahekea te hautanga \frac{-40}{24} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 8.

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

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

$$12c^{2}+11c-15=12\left(c-\frac{3}{4}\right)\left(c-\left(-\frac{5}{3}\right)\right)$$

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

$$12c^{2}+11c-15=12\left(c-\frac{3}{4}\right)\left(c+\frac{5}{3}\right)$$

Tango \frac{3}{4} 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.

$$12c^{2}+11c-15=12\times \frac{4c-3}{4}\left(c+\frac{5}{3}\right)$$

Tāpiri \frac{5}{3} 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.

$$12c^{2}+11c-15=12\times \frac{4c-3}{4}\times \frac{3c+5}{3}$$

Whakareatia \frac{4c-3}{4} ki te \frac{3c+5}{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.

$$12c^{2}+11c-15=12\times \frac{\left(4c-3\right)\left(3c+5\right)}{4\times 3}$$

Whakareatia 4 ki te 3.

$$12c^{2}+11c-15=12\times \frac{\left(4c-3\right)\left(3c+5\right)}{12}$$

Whakakorea atu te tauwehe pūnoa nui rawa 12 i roto i te 12 me te 12.

$$12c^{2}+11c-15=\left(4c-3\right)\left(3c+5\right)$$

Aromātai

$$\left(4c-3\right)\left(3c+5\right)$$