Whakaoti i te 9x^2-15x+6

$$9 x ^ { 2 } - 15 x + 6$$

Tauwehe

$$3\left(x-1\right)\left(3x-2\right)$$

Ngā Upane Otinga

Tauwehea te 3.

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

Whakaarohia te 3x^{2}-5x+2. Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 3x^{2}+ax+bx+2. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-5 ab=3\times 2=6$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōraro te a+b, he tōraro hoki a a me b. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 6.

$$-1,-6 -2,-3$$

Tātaihia te tapeke mō ia takirua.

$$-1-6=-7 -2-3=-5$$

Ko te otinga te takirua ka hoatu i te tapeke -5.

$$a=-3 b=-2$$

Tuhia anō te 3x^{2}-5x+2 hei \left(3x^{2}-3x\right)+\left(-2x+2\right).

$$\left(3x^{2}-3x\right)+\left(-2x+2\right)$$

Tauwehea te 3x i te tuatahi me te -2 i te rōpū tuarua.

$$3x\left(x-1\right)-2\left(x-1\right)$$

Whakatauwehea atu te kīanga pātahi x-1 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-1\right)\left(3x-2\right)$$

Me tuhi anō te kīanga whakatauwehe katoa.

$$3\left(x-1\right)\left(3x-2\right)$$

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.

$$9x^{2}-15x+6=0$$

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.

$$x=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\times 9\times 6}}{2\times 9}$$

Pūrua -15.

$$x=\frac{-\left(-15\right)±\sqrt{225-4\times 9\times 6}}{2\times 9}$$

Whakareatia -4 ki te 9.

$$x=\frac{-\left(-15\right)±\sqrt{225-36\times 6}}{2\times 9}$$

Whakareatia -36 ki te 6.

$$x=\frac{-\left(-15\right)±\sqrt{225-216}}{2\times 9}$$

Tāpiri 225 ki te -216.

$$x=\frac{-\left(-15\right)±\sqrt{9}}{2\times 9}$$

Tuhia te pūtakerua o te 9.

$$x=\frac{-\left(-15\right)±3}{2\times 9}$$

Ko te tauaro o -15 ko 15.

$$x=\frac{15±3}{2\times 9}$$

Whakareatia 2 ki te 9.

$$x=\frac{15±3}{18}$$

Nā, me whakaoti te whārite x=\frac{15±3}{18} ina he tāpiri te ±. Tāpiri 15 ki te 3.

$$x=\frac{18}{18}$$

Whakawehe 18 ki te 18.

$$x=1$$

Nā, me whakaoti te whārite x=\frac{15±3}{18} ina he tango te ±. Tango 3 mai i 15.

$$x=\frac{12}{18}$$

Whakahekea te hautanga \frac{12}{18} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 6.

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

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

$$9x^{2}-15x+6=9\left(x-1\right)\left(x-\frac{2}{3}\right)$$

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.

$$9x^{2}-15x+6=9\left(x-1\right)\times \frac{3x-2}{3}$$

Whakakorea atu te tauwehe pūnoa nui rawa 3 i roto i te 9 me te 3.

$$9x^{2}-15x+6=3\left(x-1\right)\left(3x-2\right)$$

Aromātai

$$3\left(x-1\right)\left(3x-2\right)$$