Whakaoti i te 3x^2-x-10

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

Tauwehe

$$\left(x-2\right)\left(3x+5\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 3x^{2}+ax+bx-10. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-1 ab=3\left(-10\right)=-30$$

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 -30.

$$1,-30 2,-15 3,-10 5,-6$$

Tātaihia te tapeke mō ia takirua.

$$1-30=-29 2-15=-13 3-10=-7 5-6=-1$$

Ko te otinga te takirua ka hoatu i te tapeke -1.

$$a=-6 b=5$$

Tuhia anō te 3x^{2}-x-10 hei \left(3x^{2}-6x\right)+\left(5x-10\right).

$$\left(3x^{2}-6x\right)+\left(5x-10\right)$$

Tauwehea te 3x i te tuatahi me te 5 i te rōpū tuarua.

$$3x\left(x-2\right)+5\left(x-2\right)$$

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

$$\left(x-2\right)\left(3x+5\right)$$

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.

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

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.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 3\left(-10\right)}}{2\times 3}$$

Whakareatia -4 ki te 3.

$$x=\frac{-\left(-1\right)±\sqrt{1-12\left(-10\right)}}{2\times 3}$$

Whakareatia -12 ki te -10.

$$x=\frac{-\left(-1\right)±\sqrt{1+120}}{2\times 3}$$

Tāpiri 1 ki te 120.

$$x=\frac{-\left(-1\right)±\sqrt{121}}{2\times 3}$$

Tuhia te pūtakerua o te 121.

$$x=\frac{-\left(-1\right)±11}{2\times 3}$$

Ko te tauaro o -1 ko 1.

$$x=\frac{1±11}{2\times 3}$$

Whakareatia 2 ki te 3.

$$x=\frac{1±11}{6}$$

Nā, me whakaoti te whārite x=\frac{1±11}{6} ina he tāpiri te ±. Tāpiri 1 ki te 11.

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

Whakawehe 12 ki te 6.

$$x=2$$

Nā, me whakaoti te whārite x=\frac{1±11}{6} ina he tango te ±. Tango 11 mai i 1.

$$x=-\frac{10}{6}$$

Whakahekea te hautanga \frac{-10}{6} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x=-\frac{5}{3}$$

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{5}{3} mō te x_{2}.

$$3x^{2}-x-10=3\left(x-2\right)\left(x-\left(-\frac{5}{3}\right)\right)$$

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

$$3x^{2}-x-10=3\left(x-2\right)\left(x+\frac{5}{3}\right)$$

Tāpiri \frac{5}{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.

$$3x^{2}-x-10=3\left(x-2\right)\times \frac{3x+5}{3}$$

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

$$3x^{2}-x-10=\left(x-2\right)\left(3x+5\right)$$

Aromātai

$$\left(x-2\right)\left(3x+5\right)$$