Whakaoti i te 3x^2-8x+4

$$3 { x }^{ 2 } -8x+4$$

Tauwehe

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

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 3x^{2}+ax+bx+4. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-8 ab=3\times 4=12$$

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 12.

$$-1,-12 -2,-6 -3,-4$$

Tātaihia te tapeke mō ia takirua.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

Ko te otinga te takirua ka hoatu i te tapeke -8.

$$a=-6 b=-2$$

Tuhia anō te 3x^{2}-8x+4 hei \left(3x^{2}-6x\right)+\left(-2x+4\right).

$$\left(3x^{2}-6x\right)+\left(-2x+4\right)$$

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

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

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

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

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.

$$3x^{2}-8x+4=0$$

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.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 3\times 4}}{2\times 3}$$

Pūrua -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 3\times 4}}{2\times 3}$$

Whakareatia -4 ki te 3.

$$x=\frac{-\left(-8\right)±\sqrt{64-12\times 4}}{2\times 3}$$

Whakareatia -12 ki te 4.

$$x=\frac{-\left(-8\right)±\sqrt{64-48}}{2\times 3}$$

Tāpiri 64 ki te -48.

$$x=\frac{-\left(-8\right)±\sqrt{16}}{2\times 3}$$

Tuhia te pūtakerua o te 16.

$$x=\frac{-\left(-8\right)±4}{2\times 3}$$

Ko te tauaro o -8 ko 8.

$$x=\frac{8±4}{2\times 3}$$

Whakareatia 2 ki te 3.

$$x=\frac{8±4}{6}$$

Nā, me whakaoti te whārite x=\frac{8±4}{6} ina he tāpiri te ±. Tāpiri 8 ki te 4.

$$x=\frac{12}{6}$$

Whakawehe 12 ki te 6.

$$x=2$$

Nā, me whakaoti te whārite x=\frac{8±4}{6} ina he tango te ±. Tango 4 mai i 8.

$$x=\frac{4}{6}$$

Whakahekea te hautanga \frac{4}{6} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

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

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

$$3x^{2}-8x+4=3\left(x-2\right)\left(x-\frac{2}{3}\right)$$

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.

$$3x^{2}-8x+4=3\left(x-2\right)\times \frac{3x-2}{3}$$

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

$$3x^{2}-8x+4=\left(x-2\right)\left(3x-2\right)$$

Aromātai

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