Whakaoti i te 4u^2-5u-6

$$4 u ^ { 2 } - 5 u - 6$$

Tauwehe

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

Ngā Upane Otinga

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

$$a+b=-5 ab=4\left(-6\right)=-24$$

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 -5.

$$a=-8 b=3$$

Tuhia anō te 4u^{2}-5u-6 hei \left(4u^{2}-8u\right)+\left(3u-6\right).

$$\left(4u^{2}-8u\right)+\left(3u-6\right)$$

Tauwehea te 4u i te tuatahi me te 3 i te rōpū tuarua.

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

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

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

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.

$$4u^{2}-5u-6=0$$

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.

$$u=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 4\left(-6\right)}}{2\times 4}$$

Pūrua -5.

$$u=\frac{-\left(-5\right)±\sqrt{25-4\times 4\left(-6\right)}}{2\times 4}$$

Whakareatia -4 ki te 4.

$$u=\frac{-\left(-5\right)±\sqrt{25-16\left(-6\right)}}{2\times 4}$$

Whakareatia -16 ki te -6.

$$u=\frac{-\left(-5\right)±\sqrt{25+96}}{2\times 4}$$

Tāpiri 25 ki te 96.

$$u=\frac{-\left(-5\right)±\sqrt{121}}{2\times 4}$$

Tuhia te pūtakerua o te 121.

$$u=\frac{-\left(-5\right)±11}{2\times 4}$$

Ko te tauaro o -5 ko 5.

$$u=\frac{5±11}{2\times 4}$$

Whakareatia 2 ki te 4.

$$u=\frac{5±11}{8}$$

Nā, me whakaoti te whārite u=\frac{5±11}{8} ina he tāpiri te ±. Tāpiri 5 ki te 11.

$$u=\frac{16}{8}$$

Whakawehe 16 ki te 8.

$$u=2$$

Nā, me whakaoti te whārite u=\frac{5±11}{8} ina he tango te ±. Tango 11 mai i 5.

$$u=-\frac{6}{8}$$

Whakahekea te hautanga \frac{-6}{8} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$u=-\frac{3}{4}$$

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{3}{4} mō te x_{2}.

$$4u^{2}-5u-6=4\left(u-2\right)\left(u-\left(-\frac{3}{4}\right)\right)$$

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

$$4u^{2}-5u-6=4\left(u-2\right)\left(u+\frac{3}{4}\right)$$

Tāpiri \frac{3}{4} ki te u 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.

$$4u^{2}-5u-6=4\left(u-2\right)\times \frac{4u+3}{4}$$

Whakakorea atu te tauwehe pūnoa nui rawa 4 i roto i te 4 me te 4.

$$4u^{2}-5u-6=\left(u-2\right)\left(4u+3\right)$$

Aromātai

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