Whakaoti i te 8x^2-22x+12

$$8 x ^ { 2 } - 22 x + 12$$

Tauwehe

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

Ngā Upane Otinga

Tauwehea te 2.

$$2\left(4x^{2}-11x+6\right)$$

Whakaarohia te 4x^{2}-11x+6. Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 4x^{2}+ax+bx+6. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-11 ab=4\times 6=24$$

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 -11.

$$a=-8 b=-3$$

Tuhia anō te 4x^{2}-11x+6 hei \left(4x^{2}-8x\right)+\left(-3x+6\right).

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

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

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

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

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

Me tuhi anō te kīanga whakatauwehe katoa.

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

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.

$$8x^{2}-22x+12=0$$

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.

$$x=\frac{-\left(-22\right)±\sqrt{\left(-22\right)^{2}-4\times 8\times 12}}{2\times 8}$$

Pūrua -22.

$$x=\frac{-\left(-22\right)±\sqrt{484-4\times 8\times 12}}{2\times 8}$$

Whakareatia -4 ki te 8.

$$x=\frac{-\left(-22\right)±\sqrt{484-32\times 12}}{2\times 8}$$

Whakareatia -32 ki te 12.

$$x=\frac{-\left(-22\right)±\sqrt{484-384}}{2\times 8}$$

Tāpiri 484 ki te -384.

$$x=\frac{-\left(-22\right)±\sqrt{100}}{2\times 8}$$

Tuhia te pūtakerua o te 100.

$$x=\frac{-\left(-22\right)±10}{2\times 8}$$

Ko te tauaro o -22 ko 22.

$$x=\frac{22±10}{2\times 8}$$

Whakareatia 2 ki te 8.

$$x=\frac{22±10}{16}$$

Nā, me whakaoti te whārite x=\frac{22±10}{16} ina he tāpiri te ±. Tāpiri 22 ki te 10.

$$x=\frac{32}{16}$$

Whakawehe 32 ki te 16.

$$x=2$$

Nā, me whakaoti te whārite x=\frac{22±10}{16} ina he tango te ±. Tango 10 mai i 22.

$$x=\frac{12}{16}$$

Whakahekea te hautanga \frac{12}{16} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

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

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{3}{4} mō te x_{2}.

$$8x^{2}-22x+12=8\left(x-2\right)\left(x-\frac{3}{4}\right)$$

Tango \frac{3}{4} 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.

$$8x^{2}-22x+12=8\left(x-2\right)\times \frac{4x-3}{4}$$

Whakakorea atu te tauwehe pūnoa nui rawa 4 i roto i te 8 me te 4.

$$8x^{2}-22x+12=2\left(x-2\right)\left(4x-3\right)$$

Aromātai

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