Whakaoti i te 10-11x-6x^2

$$10 - 11 x - 6 x ^ { 2 }$$

Tauwehe

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

Ngā Upane Otinga

Hurinahatia te pūrau ki te āhua tānga ngahuru. Whakaraupapahia ngā kīanga tau mai i te pū teitei rawa ki te mea iti rawa.

$$-6x^{2}-11x+10$$

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei -6x^{2}+ax+bx+10. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-11 ab=-6\times 10=-60$$

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 -60.

$$1,-60 2,-30 3,-20 4,-15 5,-12 6,-10$$

Tātaihia te tapeke mō ia takirua.

$$1-60=-59 2-30=-28 3-20=-17 4-15=-11 5-12=-7 6-10=-4$$

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

$$a=4 b=-15$$

Tuhia anō te -6x^{2}-11x+10 hei \left(-6x^{2}+4x\right)+\left(-15x+10\right).

$$\left(-6x^{2}+4x\right)+\left(-15x+10\right)$$

Tauwehea te 2x i te tuatahi me te 5 i te rōpū tuarua.

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

Whakatauwehea atu te kīanga pātahi -3x+2 mā te whakamahi i te āhuatanga tātai tohatoha.

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

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.

$$-6x^{2}-11x+10=0$$

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.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\left(-6\right)\times 10}}{2\left(-6\right)}$$

Pūrua -11.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\left(-6\right)\times 10}}{2\left(-6\right)}$$

Whakareatia -4 ki te -6.

$$x=\frac{-\left(-11\right)±\sqrt{121+24\times 10}}{2\left(-6\right)}$$

Whakareatia 24 ki te 10.

$$x=\frac{-\left(-11\right)±\sqrt{121+240}}{2\left(-6\right)}$$

Tāpiri 121 ki te 240.

$$x=\frac{-\left(-11\right)±\sqrt{361}}{2\left(-6\right)}$$

Tuhia te pūtakerua o te 361.

$$x=\frac{-\left(-11\right)±19}{2\left(-6\right)}$$

Ko te tauaro o -11 ko 11.

$$x=\frac{11±19}{2\left(-6\right)}$$

Whakareatia 2 ki te -6.

$$x=\frac{11±19}{-12}$$

Nā, me whakaoti te whārite x=\frac{11±19}{-12} ina he tāpiri te ±. Tāpiri 11 ki te 19.

$$x=\frac{30}{-12}$$

Whakahekea te hautanga \frac{30}{-12} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 6.

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

Nā, me whakaoti te whārite x=\frac{11±19}{-12} ina he tango te ±. Tango 19 mai i 11.

$$x=-\frac{8}{-12}$$

Whakahekea te hautanga \frac{-8}{-12} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

$$x=\frac{2}{3}$$

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

$$-6x^{2}-11x+10=-6\left(x-\left(-\frac{5}{2}\right)\right)\left(x-\frac{2}{3}\right)$$

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

$$-6x^{2}-11x+10=-6\left(x+\frac{5}{2}\right)\left(x-\frac{2}{3}\right)$$

Tāpiri \frac{5}{2} ki te x 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.

$$-6x^{2}-11x+10=-6\times \frac{-2x-5}{-2}\left(x-\frac{2}{3}\right)$$

Tango \frac{2}{3} mai i x 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.

$$-6x^{2}-11x+10=-6\times \frac{-2x-5}{-2}\times \frac{-3x+2}{-3}$$

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

$$-6x^{2}-11x+10=-6\times \frac{\left(-2x-5\right)\left(-3x+2\right)}{-2\left(-3\right)}$$

Whakareatia -2 ki te -3.

$$-6x^{2}-11x+10=-6\times \frac{\left(-2x-5\right)\left(-3x+2\right)}{6}$$

Whakakorea atu te tauwehe pūnoa nui rawa 6 i roto i te -6 me te 6.

$$-6x^{2}-11x+10=-\left(-2x-5\right)\left(-3x+2\right)$$

Aromātai

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