Whakaoti i te 3x^2-13x+12

$$3 x ^ { 2 } - 13 x + 12$$

Tauwehe

$$\left(x-3\right)\left(3x-4\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 3x^{2}+ax+bx+12. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-13 ab=3\times 12=36$$

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 36.

$$-1,-36 -2,-18 -3,-12 -4,-9 -6,-6$$

Tātaihia te tapeke mō ia takirua.

$$-1-36=-37 -2-18=-20 -3-12=-15 -4-9=-13 -6-6=-12$$

Ko te otinga te takirua ka hoatu i te tapeke -13.

$$a=-9 b=-4$$

Tuhia anō te 3x^{2}-13x+12 hei \left(3x^{2}-9x\right)+\left(-4x+12\right).

$$\left(3x^{2}-9x\right)+\left(-4x+12\right)$$

Tauwehea te 3x i te tuatahi me te -4 i te rōpū tuarua.

$$3x\left(x-3\right)-4\left(x-3\right)$$

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

$$\left(x-3\right)\left(3x-4\right)$$

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.

$$3x^{2}-13x+12=0$$

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.

$$x=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 3\times 12}}{2\times 3}$$

Pūrua -13.

$$x=\frac{-\left(-13\right)±\sqrt{169-4\times 3\times 12}}{2\times 3}$$

Whakareatia -4 ki te 3.

$$x=\frac{-\left(-13\right)±\sqrt{169-12\times 12}}{2\times 3}$$

Whakareatia -12 ki te 12.

$$x=\frac{-\left(-13\right)±\sqrt{169-144}}{2\times 3}$$

Tāpiri 169 ki te -144.

$$x=\frac{-\left(-13\right)±\sqrt{25}}{2\times 3}$$

Tuhia te pūtakerua o te 25.

$$x=\frac{-\left(-13\right)±5}{2\times 3}$$

Ko te tauaro o -13 ko 13.

$$x=\frac{13±5}{2\times 3}$$

Whakareatia 2 ki te 3.

$$x=\frac{13±5}{6}$$

Nā, me whakaoti te whārite x=\frac{13±5}{6} ina he tāpiri te ±. Tāpiri 13 ki te 5.

$$x=\frac{18}{6}$$

Whakawehe 18 ki te 6.

$$x=3$$

Nā, me whakaoti te whārite x=\frac{13±5}{6} ina he tango te ±. Tango 5 mai i 13.

$$x=\frac{8}{6}$$

Whakahekea te hautanga \frac{8}{6} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

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

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

$$3x^{2}-13x+12=3\left(x-3\right)\left(x-\frac{4}{3}\right)$$

Tango \frac{4}{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.

$$3x^{2}-13x+12=3\left(x-3\right)\times \frac{3x-4}{3}$$

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

$$3x^{2}-13x+12=\left(x-3\right)\left(3x-4\right)$$

Aromātai

$$\left(x-3\right)\left(3x-4\right)$$