Whakaoti i te 3x^2-14x-5

$$3 x ^ { 2 } - 14 x - 5$$

Tauwehe

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

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.

$$a+b=-14 ab=3\left(-5\right)=-15$$

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 -14.

$$a=-15 b=1$$

Tuhia anō te 3x^{2}-14x-5 hei \left(3x^{2}-15x\right)+\left(x-5\right).

$$\left(3x^{2}-15x\right)+\left(x-5\right)$$

Whakatauwehea atu 3x i te 3x^{2}-15x.

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

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

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

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.

$$3x^{2}-14x-5=0$$

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.

$$x=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\times 3\left(-5\right)}}{2\times 3}$$

Pūrua -14.

$$x=\frac{-\left(-14\right)±\sqrt{196-4\times 3\left(-5\right)}}{2\times 3}$$

Whakareatia -4 ki te 3.

$$x=\frac{-\left(-14\right)±\sqrt{196-12\left(-5\right)}}{2\times 3}$$

Whakareatia -12 ki te -5.

$$x=\frac{-\left(-14\right)±\sqrt{196+60}}{2\times 3}$$

Tāpiri 196 ki te 60.

$$x=\frac{-\left(-14\right)±\sqrt{256}}{2\times 3}$$

Tuhia te pūtakerua o te 256.

$$x=\frac{-\left(-14\right)±16}{2\times 3}$$

Ko te tauaro o -14 ko 14.

$$x=\frac{14±16}{2\times 3}$$

Whakareatia 2 ki te 3.

$$x=\frac{14±16}{6}$$

Nā, me whakaoti te whārite x=\frac{14±16}{6} ina he tāpiri te ±. Tāpiri 14 ki te 16.

$$x=\frac{30}{6}$$

Whakawehe 30 ki te 6.

$$x=5$$

Nā, me whakaoti te whārite x=\frac{14±16}{6} ina he tango te ±. Tango 16 mai i 14.

$$x=-\frac{2}{6}$$

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

$$x=-\frac{1}{3}$$

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

$$3x^{2}-14x-5=3\left(x-5\right)\left(x-\left(-\frac{1}{3}\right)\right)$$

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

$$3x^{2}-14x-5=3\left(x-5\right)\left(x+\frac{1}{3}\right)$$

Tāpiri \frac{1}{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.

$$3x^{2}-14x-5=3\left(x-5\right)\times \frac{3x+1}{3}$$

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

$$3x^{2}-14x-5=\left(x-5\right)\left(3x+1\right)$$

Aromātai

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