Whakaoti i te 3x^2-2x-5

$$3 { x }^{ 2 } -2x-5$$

Tauwehe

$$\left(3x-5\right)\left(x+1\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 3x^{2}+ax+bx-5. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-2 ab=3\left(-5\right)=-15$$

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 -15.

$$1,-15 3,-5$$

Tātaihia te tapeke mō ia takirua.

$$1-15=-14 3-5=-2$$

Ko te otinga te takirua ka hoatu i te tapeke -2.

$$a=-5 b=3$$

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

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

Whakatauwehea atu x i te 3x^{2}-5x.

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

Whakatauwehea atu te kīanga pātahi 3x-5 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(3x-5\right)\left(x+1\right)$$

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.

$$3x^{2}-2x-5=0$$

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.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 3\left(-5\right)}}{2\times 3}$$

Pūrua -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\times 3\left(-5\right)}}{2\times 3}$$

Whakareatia -4 ki te 3.

$$x=\frac{-\left(-2\right)±\sqrt{4-12\left(-5\right)}}{2\times 3}$$

Whakareatia -12 ki te -5.

$$x=\frac{-\left(-2\right)±\sqrt{4+60}}{2\times 3}$$

Tāpiri 4 ki te 60.

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

Tuhia te pūtakerua o te 64.

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

Ko te tauaro o -2 ko 2.

$$x=\frac{2±8}{2\times 3}$$

Whakareatia 2 ki te 3.

$$x=\frac{2±8}{6}$$

Nā, me whakaoti te whārite x=\frac{2±8}{6} ina he tāpiri te ±. Tāpiri 2 ki te 8.

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

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

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

Nā, me whakaoti te whārite x=\frac{2±8}{6} ina he tango te ±. Tango 8 mai i 2.

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

Whakawehe -6 ki te 6.

$$x=-1$$

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 \frac{5}{3} mō te x_{1} me te -1 mō te x_{2}.

$$3x^{2}-2x-5=3\left(x-\frac{5}{3}\right)\left(x-\left(-1\right)\right)$$

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

$$3x^{2}-2x-5=3\left(x-\frac{5}{3}\right)\left(x+1\right)$$

Tango \frac{5}{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.

$$3x^{2}-2x-5=3\times \frac{3x-5}{3}\left(x+1\right)$$

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

$$3x^{2}-2x-5=\left(3x-5\right)\left(x+1\right)$$

Aromātai

$$\left(3x-5\right)\left(x+1\right)$$