Whakaoti i te 3x^2-10x+8

$$3x^2-10x+8$$

Tauwehe

$$\left(x-2\right)\left(3x-4\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 3x^{2}+ax+bx+8. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-10 ab=3\times 8=24$$

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 24.

$$-1,-24 -2,-12 -3,-8 -4,-6$$

Tātaihia te tapeke mō ia takirua.

$$-1-24=-25 -2-12=-14 -3-8=-11 -4-6=-10$$

Ko te otinga te takirua ka hoatu i te tapeke -10.

$$a=-6 b=-4$$

Tuhia anō te 3x^{2}-10x+8 hei \left(3x^{2}-6x\right)+\left(-4x+8\right).

$$\left(3x^{2}-6x\right)+\left(-4x+8\right)$$

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

$$3x\left(x-2\right)-4\left(x-2\right)$$

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

$$\left(x-2\right)\left(3x-4\right)$$

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.

$$3x^{2}-10x+8=0$$

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.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 3\times 8}}{2\times 3}$$

Pūrua -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 3\times 8}}{2\times 3}$$

Whakareatia -4 ki te 3.

$$x=\frac{-\left(-10\right)±\sqrt{100-12\times 8}}{2\times 3}$$

Whakareatia -12 ki te 8.

$$x=\frac{-\left(-10\right)±\sqrt{100-96}}{2\times 3}$$

Tāpiri 100 ki te -96.

$$x=\frac{-\left(-10\right)±\sqrt{4}}{2\times 3}$$

Tuhia te pūtakerua o te 4.

$$x=\frac{-\left(-10\right)±2}{2\times 3}$$

Ko te tauaro o -10 ko 10.

$$x=\frac{10±2}{2\times 3}$$

Whakareatia 2 ki te 3.

$$x=\frac{10±2}{6}$$

Nā, me whakaoti te whārite x=\frac{10±2}{6} ina he tāpiri te ±. Tāpiri 10 ki te 2.

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

Whakawehe 12 ki te 6.

$$x=2$$

Nā, me whakaoti te whārite x=\frac{10±2}{6} ina he tango te ±. Tango 2 mai i 10.

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

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

$$3x^{2}-10x+8=3\left(x-2\right)\left(x-\frac{4}{3}\right)$$

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.

$$3x^{2}-10x+8=3\left(x-2\right)\times \frac{3x-4}{3}$$

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

$$3x^{2}-10x+8=\left(x-2\right)\left(3x-4\right)$$

Aromātai

$$\left(x-2\right)\left(3x-4\right)$$