Whakaoti i te 6x^2-23x-4?

$$6 x ^ { 2 } - 23 x - 4 ?$$

Tauwehe

$$\left(x-4\right)\left(6x+1\right)$$

Ngā Upane Otinga

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

$$a+b=-23 ab=6\left(-4\right)=-24$$

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 -23.

$$a=-24 b=1$$

Tuhia anō te 6x^{2}-23x-4 hei \left(6x^{2}-24x\right)+\left(x-4\right).

$$\left(6x^{2}-24x\right)+\left(x-4\right)$$

Whakatauwehea atu 6x i te 6x^{2}-24x.

$$6x\left(x-4\right)+x-4$$

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

$$\left(x-4\right)\left(6x+1\right)$$

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.

$$6x^{2}-23x-4=0$$

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.

$$x=\frac{-\left(-23\right)±\sqrt{\left(-23\right)^{2}-4\times 6\left(-4\right)}}{2\times 6}$$

Pūrua -23.

$$x=\frac{-\left(-23\right)±\sqrt{529-4\times 6\left(-4\right)}}{2\times 6}$$

Whakareatia -4 ki te 6.

$$x=\frac{-\left(-23\right)±\sqrt{529-24\left(-4\right)}}{2\times 6}$$

Whakareatia -24 ki te -4.

$$x=\frac{-\left(-23\right)±\sqrt{529+96}}{2\times 6}$$

Tāpiri 529 ki te 96.

$$x=\frac{-\left(-23\right)±\sqrt{625}}{2\times 6}$$

Tuhia te pūtakerua o te 625.

$$x=\frac{-\left(-23\right)±25}{2\times 6}$$

Ko te tauaro o -23 ko 23.

$$x=\frac{23±25}{2\times 6}$$

Whakareatia 2 ki te 6.

$$x=\frac{23±25}{12}$$

Nā, me whakaoti te whārite x=\frac{23±25}{12} ina he tāpiri te ±. Tāpiri 23 ki te 25.

$$x=\frac{48}{12}$$

Whakawehe 48 ki te 12.

$$x=4$$

Nā, me whakaoti te whārite x=\frac{23±25}{12} ina he tango te ±. Tango 25 mai i 23.

$$x=-\frac{2}{12}$$

Whakahekea te hautanga \frac{-2}{12} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x=-\frac{1}{6}$$

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{1}{6} mō te x_{2}.

$$6x^{2}-23x-4=6\left(x-4\right)\left(x-\left(-\frac{1}{6}\right)\right)$$

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

$$6x^{2}-23x-4=6\left(x-4\right)\left(x+\frac{1}{6}\right)$$

Tāpiri \frac{1}{6} 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.

$$6x^{2}-23x-4=6\left(x-4\right)\times \frac{6x+1}{6}$$

Whakakorea atu te tauwehe pūnoa nui rawa 6 i roto i te 6 me te 6.

$$6x^{2}-23x-4=\left(x-4\right)\left(6x+1\right)$$

Aromātai

$$\left(x-4\right)\left(6x+1\right)$$