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

$$3 { x }^{ 2 } -10x-8$$

Tauwehe

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

Ngā Upane Otinga

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

$$a+b=-10 ab=3\left(-8\right)=-24$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōraro te a+b, he nui ake te uara pū o te tau tōraro i tō te tōrunga. 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.

$$1-24=-23 2-12=-10 3-8=-5 4-6=-2$$

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

$$a=-12 b=2$$

Tuhia anō te 3x^{2}-10x-8 hei \left(3x^{2}-12x\right)+\left(2x-8\right).

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

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

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

Whakatauwehea atu te kīanga pātahi x-4 mā te whakamahi i te āhuatanga tātai tohatoha.

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

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.

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

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.

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

Pūrua -10.

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

Whakareatia -4 ki te 3.

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

Whakareatia -12 ki te -8.

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

Tāpiri 100 ki te 96.

$$x=\frac{-\left(-10\right)±\sqrt{196}}{2\times 3}$$

Tuhia te pūtakerua o te 196.

$$x=\frac{-\left(-10\right)±14}{2\times 3}$$

Ko te tauaro o -10 ko 10.

$$x=\frac{10±14}{2\times 3}$$

Whakareatia 2 ki te 3.

$$x=\frac{10±14}{6}$$

Nā, me whakaoti te whārite x=\frac{10±14}{6} ina he tāpiri te ±. Tāpiri 10 ki te 14.

$$x=\frac{24}{6}$$

Whakawehe 24 ki te 6.

$$x=4$$

Nā, me whakaoti te whārite x=\frac{10±14}{6} ina he tango te ±. Tango 14 mai i 10.

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

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

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

Whakamāmātia ngā kīanga katoa o te āhua p-\left(-q\right) ki te p+q.

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

Tāpiri \frac{2}{3} ki te x mā te kimi i te tauraro pātahi me te tāpiri i ngā taurunga. Ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

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

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

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

Aromātai

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